Zurück zu Bestimmte Integrale
∫ 0 1 ∫ 0 1 ( x y ) x y d x d y {\displaystyle \int _{0}^{1}\int _{0}^{1}(xy)^{xy}\,dx\,dy} ist nach Substitution x ↦ x y {\displaystyle x\mapsto {\frac {x}{y}}} gleich ∫ 0 1 ∫ 0 y x x d x y d y {\displaystyle \int _{0}^{1}\int _{0}^{y}x^{x}\,{\frac {dx}{y}}\,dy} . Nach dem Vertauschen der Integrationsreihenfolge ist dies ∫ 0 1 ∫ x 1 x x d y y d x = ∫ 0 1 x x ( − log x ) d x {\displaystyle \int _{0}^{1}\int _{x}^{1}x^{x}\,{\frac {dy}{y}}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx} . Wegen ∫ 0 1 x x ( 1 + log x ) d x = [ x x ] 0 1 = 0 {\displaystyle \int _{0}^{1}x^{x}\,(1+\log x)\,dx=\left[x^{x}\right]_{0}^{1}=0} ist ∫ 0 1 x x d x = ∫ 0 1 x x ( − log x ) d x {\displaystyle \int _{0}^{1}x^{x}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx} .
Teile das Einheitsquadrat Q = [ 0 , 1 ] 2 {\displaystyle Q=[0,1]^{2}\,} in zwei kongruente Dreiecke, jeweils mit dem Ursprung als Spitze. Q 1 = { ( x , y ) ∈ Q : x ≥ y } , Q 2 = { ( x , y ) ∈ Q : y ≥ x } {\displaystyle Q_{1}=\{(x,y)\in Q\,:\,x\geq y\}\,,\,Q_{2}=\{(x,y)\in Q\,:\,y\geq x\}} I := ∬ Q x 2 + y 2 d x d y = 2 ∬ Q 1 x 2 + y 2 d x d y {\displaystyle I:=\iint _{Q}{\sqrt {x^{2}+y^{2}}}\,dx\,dy=2\iint _{Q_{1}}{\sqrt {x^{2}+y^{2}}}\,dx\,dy} Verwende nun Polarkoordinaten: [ x = r cos φ y = r sin φ ] d x d y = r d r d φ r = x 2 + y 2 {\displaystyle {\begin{bmatrix}x=r\cos \varphi \\y=r\sin \varphi \\\end{bmatrix}}\quad \qquad dx\,dy=r\,dr\,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}}}} Wegen x ≥ y ⟺ r cos φ ≥ r sin φ ⟺ cos φ ≥ sin φ ⟺ 0 ≤ φ ≤ π 4 {\displaystyle x\geq y\iff r\cos \varphi \geq r\sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}} x ≤ 1 ⟺ 0 ≤ r ≤ 1 cos φ {\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\cos \varphi }}} ist I = 2 ∫ 0 π 4 ∫ 0 1 cos φ r 2 d r d φ = 2 3 ∫ 0 π 4 d φ cos 3 φ {\displaystyle I=2\int _{0}^{\frac {\pi }{4}}\int _{0}^{\frac {1}{\cos \varphi }}r^{2}\,dr\,d\varphi ={\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{\cos ^{3}\varphi }}} = 1 3 [ sin φ cos 2 φ + log ( sec φ + tan φ ) ] 0 π 4 = 1 3 ( 2 + log ( 2 + 1 ) ) {\displaystyle ={\frac {1}{3}}\left[{\frac {\sin \varphi }{\cos ^{2}\varphi }}+\log(\sec \varphi +\tan \varphi )\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)} .
Teile den Einheitswürfel V = [ 0 , 1 ] 3 {\displaystyle V=[0,1]^{3}\,} in drei kongruente Pyramiden mit dem Ursprung als Spitze. V 1 = { ( x , y , z ) ∈ V : x ≥ y , z } , V 2 = { ( x , y , z ) ∈ V : y ≥ x , z } , V 3 = { ( x , y , z ) ∈ V : z ≥ x , y } {\displaystyle V_{1}=\{(x,y,z)\in V\,:\,x\geq y,z\}\,,\,V_{2}=\{(x,y,z)\in V\,:\,y\geq x,z\}\,,\,V_{3}=\{(x,y,z)\in V\,:\,z\geq x,y\}} I := ∭ V x 2 + y 2 + z 2 d x d y d z = 3 ∭ V 1 x 2 + y 2 + z 2 d x d y d z {\displaystyle I:=\iiint _{V}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=3\iiint _{V_{1}}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz} Verwende nun Kugelkoordinaten: [ x = r sin θ cos φ y = r sin θ sin φ z = r cos θ ] d x d y d z = r 2 sin θ d r d θ d φ r = x 2 + y 2 + z 2 {\displaystyle {\begin{bmatrix}x=r\sin \theta \cos \varphi \\y=r\sin \theta \sin \varphi \\z=r\cos \theta \end{bmatrix}}\quad \qquad dx\,dy\,dz=r^{2}\sin \theta \,dr\,d\theta \,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}+z^{2}}}} Wegen x ≥ y ⟺ r sin θ cos φ ≥ r sin θ sin φ ⟺ cos φ ≥ sin φ ⟺ 0 ≤ φ ≤ π 4 {\displaystyle x\geq y\iff r\sin \theta \cos \varphi \geq r\sin \theta \sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}} x ≥ z ⟺ r sin θ cos φ ≥ r cos θ ⟺ tan θ ≥ sec φ ⟺ arctan ( sec φ ) ≤ θ ≤ π 2 {\displaystyle x\geq z\iff r\sin \theta \cos \varphi \geq r\cos \theta \iff \tan \theta \geq \sec \varphi \iff \arctan(\sec \varphi )\leq \theta \leq {\frac {\pi }{2}}} x ≤ 1 ⟺ 0 ≤ r ≤ 1 sin θ cos φ {\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\sin \theta \,\cos \varphi }}} ist I = 3 ∫ 0 π 4 ∫ arctan ( sec φ ) π 2 ∫ 0 1 sin θ cos φ r ⋅ r 2 sin θ d r d θ d φ {\displaystyle I=3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\int _{0}^{\frac {1}{\sin \theta \,\cos \varphi }}r\cdot r^{2}\,\sin \theta \,dr\,d\theta \,d\varphi } = 3 ∫ 0 π 4 ∫ arctan ( sec φ ) π 2 [ r 4 4 ] 0 1 sin θ cos φ sin θ d θ d φ = 3 4 ∫ 0 π 4 ∫ arctan ( sec φ ) π 2 d θ sin 3 θ d φ cos 4 φ {\displaystyle =3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\left[{\frac {r^{4}}{4}}\right]_{0}^{\frac {1}{\sin \theta \,\cos \varphi }}\,\sin \theta \,d\theta \,d\varphi ={\frac {3}{4}}\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}\,{\frac {d\varphi }{\cos ^{4}\varphi }}} . Substituiere θ = arccot s {\displaystyle \theta ={\text{arccot}}\,s} , dann ist ∫ arctan ( sec φ ) π 2 d θ sin 3 θ = ∫ 0 cos φ 1 + s 2 d s = [ s 2 1 + s 2 + 1 2 arsinh s ] 0 cos φ {\displaystyle \int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}=\int _{0}^{\cos \varphi }{\sqrt {1+s^{2}}}\,ds=\left[{\frac {s}{2}}{\sqrt {1+s^{2}}}+{\frac {1}{2}}\,{\text{arsinh}}\,s\right]_{0}^{\cos \varphi }} . Also ist I = 3 8 ∫ 0 π 4 1 + cos 2 φ cos 3 φ d φ + 3 8 ∫ 0 π 4 arsinh ( cos φ ) cos 4 φ d φ {\displaystyle I={\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi +{\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi } , wobei ∫ 0 π 4 1 + cos 2 φ cos 3 φ d φ = ∫ 0 1 1 + 1 1 + x 2 1 1 + x 2 3 d x 1 + x 2 = ∫ 0 1 2 + x 2 d x {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\sqrt {1+{\frac {1}{1+x^{2}}}}}{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\sqrt {2+x^{2}}}\,dx} = [ x 2 2 + x 2 + arsinh x 2 ] 0 1 = 3 2 + arsinh 1 2 {\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}+{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}} ist. Und ∫ 0 π 4 arsinh ( cos φ ) cos 4 φ d φ {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi } = [ arsinh ( cos φ ) ( 1 3 sin φ cos 3 φ + 2 3 sin φ cos φ ) ] 0 π 4 + ∫ 0 π 4 sin φ 1 + cos 2 φ ( 1 3 sin φ cos 3 φ + 2 3 sin φ cos φ ) d φ {\displaystyle =\left[{\text{arsinh}}(\cos \varphi )\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\sin \varphi }{\sqrt {1+\cos ^{2}\varphi }}}\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)d\varphi } = 4 3 arsinh 1 2 + 1 3 ∫ 0 π 4 sin 2 φ 1 + cos 2 φ cos 3 φ d φ + 2 3 ∫ 0 π 4 sin 2 φ 1 + cos 2 φ cos φ d φ {\displaystyle ={\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi +{\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi } , wobei ∫ 0 π 4 sin 2 φ 1 + cos 2 φ cos 3 φ d φ = ∫ 0 1 x 2 1 + x 2 1 + 1 1 + x 2 1 1 + x 2 3 d x 1 + x 2 = ∫ 0 1 x 2 2 + x 2 d x {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\frac {x^{2}}{\sqrt {2+x^{2}}}}\,dx} = [ x 2 2 + x 2 − arsinh x 2 ] 0 1 = 3 2 − arsinh 1 2 {\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}-{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}} ist, und ∫ 0 π 4 sin 2 φ 1 + cos 2 φ cos φ d φ = ∫ 0 1 x 2 1 + x 2 1 + 1 1 + x 2 1 1 + x 2 d x 1 + x 2 {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{\sqrt {1+x^{2}}}}}}\,{\frac {dx}{1+x^{2}}}} = ∫ 0 1 x 2 2 + x 2 ( 1 + x 2 ) d x = [ arsinh x 2 − arctan x 2 + x 2 ] 0 1 = arsinh 1 2 − π 6 {\displaystyle =\int _{0}^{1}{\frac {x^{2}}{{\sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=\left[{\text{arsinh}}{\frac {x}{\sqrt {2}}}-\arctan {\frac {x}{\sqrt {2+x^{2}}}}\right]_{0}^{1}={\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}} . Also ist I = 3 8 ( 3 2 + arsinh 1 2 ) + 3 8 ( 4 3 arsinh 1 2 + 1 3 ( 3 2 − arsinh 1 2 ) + 2 3 ( arsinh 1 2 − π 6 ) ) {\displaystyle I={\frac {3}{8}}\left({\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {3}{8}}\left({\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\left({\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {2}{3}}\left({\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}\right)\right)} = arsinh 1 2 + 3 4 − π 24 {\displaystyle ={\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {\sqrt {3}}{4}}-{\frac {\pi }{24}}} , wobei arsinh 1 2 = log ( 3 + 1 ) − log 2 2 {\displaystyle {\text{arsinh}}{\frac {1}{\sqrt {2}}}=\log({\sqrt {3}}+1)-{\frac {\log 2}{2}}} ist.
Es sei V = [ 0 , 1 ] n {\displaystyle V=[0,1]^{n}\,} der n {\displaystyle n\,} -dimensionale Einheitswürfel und V j = { ( x 1 , . . . , x n ) ∈ V | x i α i < x j α j ∀ i ≠ j } {\displaystyle V_{j}=\left\{(x_{1},...,x_{n})\in V\,{\Big |}\,x_{i}^{\alpha _{i}}<x_{j}^{\alpha _{j}}\quad \forall i\neq j\right\}} . Für alle Tupel ( x 1 , . . . , x n ) ∈ V j {\displaystyle (x_{1},...,x_{n})\in V_{j}\,} ist dann x j α j = max { x 1 α 1 , . . . , x n α n } {\displaystyle x_{j}^{\alpha _{j}}=\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}} und x i < x j α j / α i ∀ i ≠ j {\displaystyle x_{i}<x_{j}^{\alpha _{j}/\alpha _{i}}\,\,\forall i\neq j} . V {\displaystyle V\,} ist die abgeschlossene Hülle der disjunkten Vereinigung aller V j {\displaystyle V_{j}} 's. Daher ist ∫ V = ∫ V 1 + . . . + ∫ V n {\displaystyle \int _{V}=\int _{V_{1}}+...+\int _{V_{n}}} . ∫ V j max { x 1 α 1 , . . . , x n α n } d x 1 ⋯ d x n = ∫ V j x j α j d x 1 ⋯ d x n {\displaystyle \int _{V_{j}}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}dx_{1}\cdots dx_{n}=\int _{V_{j}}x_{j}^{\alpha _{j}}dx_{1}\cdots dx_{n}} = ∫ 0 1 ∫ 0 x j α j / α n ⋯ ∫ 0 x j α j / α j ^ ⋯ ∫ 0 x j α j / α 1 x j α j d x 1 ⋯ d x j ^ ⋯ d x n ⋅ d x j {\displaystyle =\int _{0}^{1}\,\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{n}}}\cdots {\widehat {\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{j}}}}}\cdots \int _{0}^{x_{j}^{\alpha _{j}/\alpha _{1}}}x_{j}^{\alpha _{j}}\,dx_{1}\cdots {\widehat {dx_{j}}}\cdots dx_{n}\cdot dx_{j}} = ∫ 0 1 x j α j x j α j / α 1 ⋯ x j α j / α j ^ ⋯ x j α j / α n d x j = ∫ 0 1 x j α j ( 1 + 1 α 1 + . . . + 1 α j ^ + . . . + 1 α n ) d x j {\displaystyle =\int _{0}^{1}x_{j}^{\alpha _{j}}\,\,x_{j}^{\alpha _{j}/\alpha _{1}}\cdots {\widehat {x_{j}^{\alpha _{j}/\alpha _{j}}}}\cdots x_{j}^{\alpha _{j}/\alpha _{n}}\,dx_{j}=\int _{0}^{1}x_{j}^{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)}\,dx_{j}} = 1 α j ( 1 + 1 α 1 + . . . + 1 α j ^ + . . . + 1 α n ) + 1 = 1 α j 1 + 1 α 1 + . . . + 1 α j ^ + . . . + 1 α n + 1 α j = 1 α j 1 + 1 α 1 + . . . + 1 α n {\displaystyle ={\frac {1}{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)+1}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}+{\frac {1}{\alpha _{j}}}}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}} . Also ist ∫ 0 1 ⋯ ∫ 0 1 max { x 1 α 1 , . . . , x n α n } d x 1 ⋯ d x n = ∫ V 1 + . . . + ∫ V n = 1 α 1 + . . . + 1 α n 1 + 1 α 1 + . . . + 1 α n {\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}\,dx_{1}\cdots dx_{n}=\int _{V_{1}}+...+\int _{V_{n}}={\frac {{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}} .
I := ∫ 0 1 ∫ 0 1 ( − log x y ) s − 2 1 − x y ( 1 − x ) d y d x {\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {(-\log xy)^{s-2}}{1-xy}}\,(1-x)\,dy\,dx} ist nach Substitution u = x y {\displaystyle u=xy\,} gleich ∫ 0 1 ∫ 0 x ( − log u ) s − 2 1 − u ( 1 − x ) d u x d x {\displaystyle \int _{0}^{1}\int _{0}^{x}{\frac {(-\log u)^{s-2}}{1-u}}\,(1-x)\,{\frac {du}{x}}\,dx} . Vertauscht man die Integrationsreihenfolge, so ist I = ∫ 0 1 ∫ u 1 ( 1 x − 1 ) d x ( − log u ) s − 2 1 − u d u {\displaystyle I=\int _{0}^{1}\int _{u}^{1}\left({\frac {1}{x}}-1\right)dx\,{\frac {(-\log u)^{s-2}}{1-u}}\,du} = ∫ 0 1 ( − log u − ( 1 − u ) ) ( − log u ) s − 2 1 − u d u = ∫ 0 1 ( ( − log u ) s − 1 1 − u − ( − log u ) s − 2 ) d u {\displaystyle =\int _{0}^{1}\left(-\log u-(1-u)\right){\frac {(-\log u)^{s-2}}{1-u}}\,du=\int _{0}^{1}\left({\frac {(-\log u)^{s-1}}{1-u}}-(-\log u)^{s-2}\right)du} . Substituiert man u = e − t {\displaystyle u=e^{-t}\,} , so ist I = ∫ 0 ∞ ( t s − 1 e t − 1 − t s − 2 e − t ) d t {\displaystyle I=\int _{0}^{\infty }\left({\frac {t^{s-1}}{e^{t}-1}}-t^{s-2}\,e^{-t}\right)dt} . Für Re ( s ) > 1 {\displaystyle {\text{Re}}(s)>1\,} ist das Γ ( s ) ζ ( s ) − Γ ( s − 1 ) = Γ ( s ) ( ζ ( s ) − 1 s − 1 ) {\displaystyle \Gamma (s)\zeta (s)-\Gamma (s-1)=\Gamma (s)\left(\zeta (s)-{\frac {1}{s-1}}\right)} . Wegen der analytischen Fortsetzbarkeit gilt dies auch für Re ( s ) > 0 {\displaystyle {\text{Re}}(s)>0\,} , wobei der Fall s = 1 {\displaystyle s=1\,} als Grenzwert γ {\displaystyle \gamma \,} zu interpretieren ist.
I := ∫ 0 π ∫ 0 π ∫ 0 π d x d y d z 1 − cos x cos y cos z {\displaystyle I:=\int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}} ist nach den Substitutionen [ x ↦ 2 arctan x y ↦ 2 arctan y z ↦ 2 arctan z ] {\displaystyle \left[{\begin{matrix}x\mapsto 2\arctan x\\y\mapsto 2\arctan y\\z\mapsto 2\arctan z\end{matrix}}\right]} gleich ∫ 0 ∞ ∫ 0 ∞ ∫ 0 ∞ 2 d x 1 + x 2 ⋅ 2 d y 1 + y 2 ⋅ 2 d z 1 + z 2 1 − 1 − x 2 1 + x 2 ⋅ 1 − y 2 1 + y 2 ⋅ 1 − z 2 1 + z 2 = 8 ∫ 0 ∞ ∫ 0 ∞ ∫ 0 ∞ d x d y d z ( 1 + x 2 ) ( 1 + y 2 ) ( 1 + z 2 ) − ( 1 − x 2 ) ( 1 − y 2 ) ( 1 − z 2 ) {\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {{\frac {2dx}{1+x^{2}}}\cdot {\frac {2dy}{1+y^{2}}}\cdot {\frac {2dz}{1+z^{2}}}}{1-{\frac {1-x^{2}}{1+x^{2}}}\cdot {\frac {1-y^{2}}{1+y^{2}}}\cdot {\frac {1-z^{2}}{1+z^{2}}}}}=8\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{(1+x^{2})(1+y^{2})(1+z^{2})-(1-x^{2})(1-y^{2})(1-z^{2})}}} = 4 ∫ 0 ∞ ∫ 0 ∞ ∫ 0 ∞ d x d y d z x 2 + y 2 + z 2 + x 2 y 2 z 2 {\displaystyle =4\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}}}} . Wechselt man von kartesischen Koordinaten zu Kugelkoordinaten [ x = r sin θ cos φ y = r sin θ sin φ z = r cos θ ] {\displaystyle \left[{\begin{matrix}x=r\,\sin \theta \,\cos \varphi \\y=r\,\sin \theta \,\sin \varphi \\z=r\,\cos \theta \qquad \;\end{matrix}}\right]} , so ist I = 4 ∫ 0 π 2 ∫ 0 π 2 ∫ 0 ∞ r 2 sin θ d r d θ d φ r 2 + r 2 sin 2 θ cos 2 φ ⋅ r 2 sin 2 θ sin 2 φ ⋅ r 2 cos 2 θ {\displaystyle I=4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {r^{2}\,\sin \theta \;dr\,d\theta \,d\varphi }{r^{2}+r^{2}\,\sin ^{2}\theta \,\cos ^{2}\varphi \cdot r^{2}\,\sin ^{2}\theta \,\sin ^{2}\varphi \cdot r^{2}\,\cos ^{2}\theta }}} = 4 ∫ 0 π 2 ∫ 0 π 2 ∫ 0 ∞ d r 1 + ( r sin θ cos θ cos φ sin φ ) 4 sin θ d θ d φ {\displaystyle =4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dr}{1+{\Big (}r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\cos \varphi \,\sin \varphi }}{\Big )}^{4}}}\,\sin \theta \;d\theta \,d\varphi } Das ist nach Substitution t = r sin θ cos θ sin φ cos φ {\displaystyle t=r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}} gleich 4 ∫ 0 π 2 ∫ 0 π 2 ∫ 0 ∞ d t d θ d φ ( 1 + t 4 ) cos θ sin φ cos φ {\displaystyle 4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dt\,d\theta \,d\varphi }{(1+t^{4})\,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}}}} = 4 ∫ 0 ∞ d t 1 + t 4 ⋅ ∫ 0 π 2 d θ cos θ ⋅ ∫ 0 π 2 d φ sin φ cos φ = 4 ⋅ 2 π 4 ⋅ Γ 2 ( 1 4 ) 2 2 π ⋅ Γ 2 ( 1 4 ) 2 π = 1 4 [ Γ ( 1 4 ) ] 4 {\displaystyle =4\int _{0}^{\infty }{\frac {dt}{1+t^{4}}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {\cos \theta }}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {\sin \varphi \,\cos \varphi }}}=4\cdot {\frac {{\sqrt {2}}\,\pi }{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2\,{\sqrt {2\pi }}}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2{\sqrt {\pi }}}}={\frac {1}{4}}\,\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}} .
Setze R ( λ ) = ∫ 0 ∞ ∫ 0 ∞ e − ( x + y + λ 3 x y ) ⋅ x 1 3 − 1 ⋅ y 2 3 − 1 d x d y {\displaystyle R(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy} . Differenziere nach λ {\displaystyle \lambda \,} : R ′ ( λ ) = ∫ 0 ∞ ∫ 0 ∞ e − ( x + y + λ 3 x y ) ⋅ 3 λ 2 x y ⋅ x 1 3 − 1 ⋅ y 2 3 − 1 d x d y {\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot {\frac {3\lambda ^{2}}{xy}}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy} Substituiere z = λ 3 x y ⇒ x = λ 3 y z ⇒ d x = − λ 3 y z 2 d z {\displaystyle z={\frac {\lambda ^{3}}{xy}}\,\Rightarrow \,x={\frac {\lambda ^{3}}{yz}}\,\Rightarrow \,dx=-{\frac {\lambda ^{3}}{yz^{2}}}\,dz} : R ′ ( λ ) = ∫ 0 ∞ ∫ ∞ 0 e − ( λ 3 y z + y + z ) ⋅ 3 z λ ⋅ λ − 2 y 1 3 − 1 ⋅ z 1 3 − 1 ⋅ y 2 3 − 1 ⋅ − λ 3 y z 2 d z d y {\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{\infty }^{0}e^{-\left({\frac {\lambda ^{3}}{yz}}+y+z\right)}\cdot {\frac {3z}{\lambda }}\cdot {\frac {\lambda ^{-2}}{y^{{\frac {1}{3}}-1}\cdot z^{{\frac {1}{3}}-1}}}\cdot y^{{\frac {2}{3}}-1}\cdot {\frac {-\lambda ^{3}}{yz^{2}}}\,dz\,dy} = 3 ∫ 0 ∞ ∫ 0 ∞ e − ( y + z + λ 3 y z ) ⋅ y 1 3 − 1 ⋅ z 2 3 − 1 d z d y = 3 ⋅ R ( λ ) {\displaystyle =3\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(y+z+{\frac {\lambda ^{3}}{yz}}\right)}\cdot y^{{\frac {1}{3}}-1}\cdot z^{{\frac {2}{3}}-1}\,dz\,dy=3\cdot R(\lambda )} ⇒ R ( λ ) = R ( 0 ) ⋅ e − 3 λ {\displaystyle \Rightarrow \,R(\lambda )=R(0)\cdot e^{-3\lambda }} mit R ( 0 ) = ∫ 0 ∞ ∫ 0 ∞ e − ( x + y ) ⋅ x 1 3 − 1 ⋅ y 2 3 − 1 d x d y {\displaystyle R(0)=\int _{0}^{\infty }\int _{0}^{\infty }e^{-(x+y)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy} = ∫ 0 ∞ x 1 3 − 1 e − x d x ⋅ ∫ 0 ∞ y 2 3 − 1 e − y d y = Γ ( 1 3 ) ⋅ Γ ( 2 3 ) = π sin π 3 = 2 π 3 {\displaystyle =\int _{0}^{\infty }x^{{\frac {1}{3}}-1}\,e^{-x}\,dx\cdot \int _{0}^{\infty }y^{{\frac {2}{3}}-1}\,e^{-y}\,dy=\Gamma \left({\frac {1}{3}}\right)\cdot \Gamma \left({\frac {2}{3}}\right)={\frac {\pi }{\sin {\frac {\pi }{3}}}}={\frac {2\pi }{\sqrt {3}}}}
Setze F ( z ) = ∫ 0 ∞ ⋯ ∫ 0 ∞ e − ( x 1 + x 2 + . . . + x n − 1 + z n x 1 x 2 ⋯ x n − 1 ) x 1 1 n − 1 x 2 2 n − 1 ⋯ x n − 1 n − 1 n − 1 d x 1 d x 2 ⋯ d x n − 1 {\displaystyle F(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}} . F ′ ( z ) = ∫ 0 ∞ ⋯ ∫ 0 ∞ e − ( x 1 + x 2 + . . . + x n − 1 + z n x 1 x 2 ⋯ x n − 1 ) − n z n − 1 x 1 x 2 ⋯ x n − 1 x 1 1 n − 1 x 2 2 n − 1 ⋯ x n − 1 n − 1 n − 1 d x 1 d x 2 ⋯ d x n − 1 {\displaystyle F'(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,{\frac {-n\,z^{n-1}}{x_{1}\,x_{2}\cdots x_{n-1}}}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}} = n ∫ 0 ∞ ⋯ ∫ 0 ∞ e − ( z n x 1 x 2 ⋯ x n − 1 + x 2 + . . . + x n − 1 + x 1 ) x 2 2 n − 1 ⋯ x n − 1 n − 1 n − 1 ⋅ x 1 1 n z ⋅ − z n x 1 2 x 2 ⋯ x n − 1 d x 1 d x 2 ⋯ d x n − 1 {\displaystyle =n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left({\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}+x_{2}+...+x_{n-1}+x_{1}\right)}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\cdot {\frac {x_{1}^{\frac {1}{n}}}{z}}\cdot {\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}\,dx_{2}\cdots dx_{n-1}} Nach Substitution u 1 = z n x 1 x 2 ⋯ x n − 1 {\displaystyle u_{1}={\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}} ist d u 1 = − z n x 1 2 x 2 ⋯ x n − 1 d x 1 {\displaystyle du_{1}={\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}} und wegen x 1 = z n u 1 ⋅ x 2 ⋯ x n − 1 {\displaystyle x_{1}={\frac {z^{n}}{u_{1}\cdot x_{2}\cdots x_{n-1}}}} ist x 1 1 n = z u 1 1 n x 2 1 n ⋯ x n − 1 1 n {\displaystyle x_{1}^{\frac {1}{n}}={\frac {z}{u_{1}^{\frac {1}{n}}\,x_{2}^{\frac {1}{n}}\cdots x_{n-1}^{\frac {1}{n}}}}} . F ′ ( z ) = − n ∫ 0 ∞ ⋯ ∫ 0 ∞ e − ( u 1 + x 2 + . . . + x n − 1 + z n u 1 x 2 ⋯ x n − 1 ) x 2 1 n − 1 ⋯ x n − 1 n − 2 n − 1 u 1 n − 1 n − 1 d u 1 d x 2 ⋯ d x n − 1 {\displaystyle F'(z)=-n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(u_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{u_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{2}^{{\frac {1}{n}}-1}\cdots x_{n-1}^{{\frac {n-2}{n}}-1}\,u_{1}^{{\frac {n-1}{n}}-1}\,du_{1}\,dx_{2}\cdots dx_{n-1}} Also ist F ′ ( z ) = − n F ( z ) {\displaystyle F'(z)=-n\,F(z)} , woraus F ( z ) = C ⋅ e − n z {\displaystyle F(z)=C\cdot e^{-nz}} folgt. Und C = F ( 0 ) = ∫ 0 ∞ ⋯ ∫ 0 ∞ e − ( x 1 + x 2 + . . . + x n − 1 ) x 1 1 n − 1 x 2 2 n − 1 ⋯ x n − 1 n − 1 n − 1 d x 1 d x 2 ⋯ d x n − 1 {\displaystyle C=F(0)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-(x_{1}+x_{2}+...+x_{n-1})}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}} = ∫ 0 ∞ x 1 1 n − 1 e − x 1 d x 1 ⋅ ∫ 0 ∞ x 2 2 n − 1 e − x 2 d x 2 ⋯ ∫ 0 ∞ x n − 1 n − 1 n − 1 e − x n − 1 d x n − 1 {\displaystyle =\int _{0}^{\infty }x_{1}^{{\frac {1}{n}}-1}\,e^{-x_{1}}\,dx_{1}\cdot \int _{0}^{\infty }x_{2}^{{\frac {2}{n}}-1}\,e^{-x_{2}}\,dx_{2}\cdots \int _{0}^{\infty }x_{n-1}^{{\frac {n-1}{n}}-1}\,e^{-x_{n-1}}\,dx_{n-1}} = Γ ( 1 n ) ⋅ Γ ( 2 n ) ⋯ Γ ( n − 1 n ) = 1 n 2 π n − 1 {\displaystyle =\Gamma \left({\frac {1}{n}}\right)\cdot \Gamma \left({\frac {2}{n}}\right)\cdots \Gamma \left({\frac {n-1}{n}}\right)={\frac {1}{\sqrt {n}}}\,{\sqrt {2\pi }}^{\,n-1}} .
Substituiere u = x y {\displaystyle u=xy} : I = ∫ 0 1 ∫ 0 y 1 ( 1 − u ) u y ( 1 − y ) d u y d y = ∫ 0 1 ∫ 0 y d u ( 1 − u ) u d y y 1 − y {\displaystyle I=\int _{0}^{1}\int _{0}^{y}{\frac {1}{(1-u)\,{\sqrt {{\frac {u}{y}}\,(1-y)}}}}\,{\frac {du}{y}}\,dy=\int _{0}^{1}\int _{0}^{y}{\frac {du}{(1-u)\,{\sqrt {u}}}}\,{\frac {dy}{{\sqrt {y}}\,{\sqrt {1-y}}}}} Vertausche die Integrationsreihenfolge: I = ∫ 0 1 ∫ u 1 d y y 1 − y d u ( 1 − u ) u = ∫ 0 1 2 arccos ( u ) d u ( 1 − u ) u {\displaystyle I=\int _{0}^{1}\int _{u}^{1}{\frac {dy}{{\sqrt {y}}{\sqrt {1-y}}}}\,{\frac {du}{(1-u)\,{\sqrt {u}}}}=\int _{0}^{1}2\arccos({\sqrt {u}}\,)\,{\frac {du}{(1-u)\,{\sqrt {u}}}}} Substituiere u = w 2 {\displaystyle u=w^{2}} : I = ∫ 0 1 2 arccos w 2 w d w ( 1 − w 2 ) w = 4 ∫ 0 1 arccos w 1 − w 2 d w {\displaystyle I=\int _{0}^{1}2\arccos w\,{\frac {2w\,dw}{(1-w^{2})\,w}}=4\int _{0}^{1}{\frac {\arccos w}{1-w^{2}}}\,dw} Substituiere w = cos t {\displaystyle w=\cos t} : I = 4 ∫ π 2 0 t 1 − cos 2 t ( − sin t ) d t = 4 ∫ 0 π 2 t sin t d t {\displaystyle I=4\int _{\frac {\pi }{2}}^{0}{\frac {t}{1-\cos ^{2}t}}\,(-\sin t)\,dt=4\int _{0}^{\frac {\pi }{2}}{\frac {t}{\sin t}}\,dt} Substituiere t = 2 arctan s {\displaystyle t=2\arctan s} : I = 8 ∫ 0 1 arctan s 2 s 1 + s 2 2 d s 1 + s 2 = 8 ∫ 0 1 arctan s s d s = 8 G {\displaystyle I=8\int _{0}^{1}{\frac {\arctan s}{\frac {2s}{1+s^{2}}}}\,{\frac {2ds}{1+s^{2}}}=8\int _{0}^{1}{\frac {\arctan s}{s}}\,ds=8G}
∫ 0 1 ∫ 0 1 1 4 ( x + y ) ( 1 − x ) ( 1 − y ) d x d y {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{4\,(x+y)\,{\sqrt {(1-x)(1-y)}}}}\,dx\,dy} ist nach der Substitution x = 1 − u 2 , y = 1 − v 2 {\displaystyle x=1-u^{2}\,,\,y=1-v^{2}} gleich ∫ 0 1 ∫ 0 1 2 u ⋅ 2 v 4 ( 2 − u 2 − v 2 ) u 2 v 2 d u d v = ∫ 0 1 ∫ 0 1 1 2 − u 2 − v 2 d u d v = G {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {2u\cdot 2v}{4\,(2-u^{2}-v^{2})\,{\sqrt {u^{2}v^{2}}}}}\,du\,dv=\int _{0}^{1}\int _{0}^{1}{\frac {1}{2-u^{2}-v^{2}}}\,du\,dv=G} .
I := ∫ 0 1 ∫ 0 1 1 2 − x 2 − y 2 d x d y = 2 ⋅ ∫ 0 1 ∫ 0 y 1 2 − x 2 − y 2 d x d y {\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=2\cdot \int _{0}^{1}\int _{0}^{y}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy} Verwende Polarkoordinaten x = r cos φ , y = r sin φ {\displaystyle x=r\cos \varphi \,,\,y=r\sin \varphi } : I = ∫ 0 π 4 ∫ 0 sec φ 2 r 2 − r 2 d r d φ = ∫ 0 π 4 [ − log ( 2 − r 2 ) ] 0 sec φ d φ = ∫ 0 π 4 ( log 2 − log ( 2 − sec 2 φ ) ) d φ {\displaystyle I=\int _{0}^{\frac {\pi }{4}}\int _{0}^{\sec \varphi }{\frac {2r}{2-r^{2}}}\,dr\,d\varphi =\int _{0}^{\frac {\pi }{4}}\left[-\log(2-r^{2})\right]_{0}^{\sec \varphi }d\varphi =\int _{0}^{\frac {\pi }{4}}\left(\log 2-\log \left(2-\sec ^{2}\varphi \right)\right)d\varphi } = π 4 log 2 − ∫ 0 π 4 [ log ( 2 cos 2 φ − 1 ) − log ( cos 2 φ ) ] d φ {\displaystyle ={\frac {\pi }{4}}\log 2-\int _{0}^{\frac {\pi }{4}}\left[\log \left(2\cos ^{2}\varphi -1\right)-\log \left(\cos ^{2}\varphi \right)\right]d\varphi } , wobei 2 cos 2 φ − 1 = cos 2 φ {\displaystyle 2\cos ^{2}\varphi -1=\cos 2\varphi } ist. Also ist I = π 4 log 2 − ∫ 0 π 2 log ( cos φ ) d φ 2 ⏟ = − π 4 log 2 + ∫ 0 π 2 log ( cos 2 φ 2 ) d φ 2 = π 2 log 2 + ∫ 0 π 2 log ( cos φ 2 ) d φ = G {\displaystyle I={\frac {\pi }{4}}\log 2-\underbrace {\int _{0}^{\frac {\pi }{2}}\log(\cos \varphi )\,{\frac {d\varphi }{2}}} _{=-{\frac {\pi }{4}}\log 2}+\int _{0}^{\frac {\pi }{2}}\log \left(\cos ^{2}{\frac {\varphi }{2}}\right){\frac {d\varphi }{2}}={\frac {\pi }{2}}\log 2+\int _{0}^{\frac {\pi }{2}}\log \left(\cos {\frac {\varphi }{2}}\right)d\varphi =G} .
∫ 0 1 ∫ 0 1 x α − 1 y β − 1 1 + x y d x d y = ∑ k = 0 ∞ ( − 1 ) k ∫ 0 1 ∫ 0 1 x k + α − 1 y k + β − 1 d x d y = ∑ k = 0 ∞ ( − 1 ) k 1 k + α ⋅ 1 k + β {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\,\int _{0}^{1}\int _{0}^{1}x^{k+\alpha -1}\,y^{k+\beta -1}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {1}{k+\alpha }}\cdot {\frac {1}{k+\beta }}} = 1 α − β ( ∑ k = 0 ∞ ( − 1 ) k k + β − ∑ k = 0 ∞ ( − 1 ) k k + α ) {\displaystyle ={\frac {1}{\alpha -\beta }}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+\beta }}-\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+\alpha }}\right)} = 1 2 1 α − β ( [ ψ ( 1 2 + β 2 ) − ψ ( β 2 ) ] − [ ψ ( 1 2 + α 2 ) − ψ ( α 2 ) ] ) {\displaystyle ={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}
Aus der Formel ∫ 0 1 ∫ 0 1 x α − 1 y β − 1 1 + x y d x d y = 1 2 1 α − β ( [ ψ ( 1 2 + β 2 ) − ψ ( β 2 ) ] − [ ψ ( 1 2 + α 2 ) − ψ ( α 2 ) ] ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)} folgt ∫ 0 1 ∫ 0 1 x α − 1 y β − 1 ( x y ) γ 1 + x y d x d y = 1 2 1 α − β ( [ ψ ( 1 2 + β 2 + γ 2 ) − ψ ( β 2 + γ 2 ) ] − [ ψ ( 1 2 + α 2 + γ 2 ) − ψ ( α 2 + γ 2 ) ] ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)} Integriere nach γ {\displaystyle \gamma \,} : ∫ 0 1 ∫ 0 1 x α − 1 y β − 1 ( x y ) γ ( 1 + x y ) log ( x y ) d x d y = 1 α − β ( [ log Γ ( 1 2 + β 2 + γ 2 ) − log Γ ( β 2 + γ 2 ) ] − [ log Γ ( 1 2 + α 2 + γ 2 ) − log Γ ( α 2 + γ 2 ) ] ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)} Setze γ = 0 {\displaystyle \gamma =0\,} : ∫ 0 1 ∫ 0 1 x α − 1 y β − 1 ( 1 + x y ) log ( x y ) d x d y = 1 α − β ( [ log Γ ( 1 2 + β 2 ) − log Γ ( β 2 ) ] − [ log Γ ( 1 2 + α 2 ) − log Γ ( α 2 ) ] ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}\right)\right]\right)}
∫ 0 1 ∫ 0 1 ( 1 − x ) ⋅ ( x y ) α − 1 d x d y = ∫ 0 1 ∫ 0 1 x α − 1 y α − 1 d x d y − ∫ 0 1 ∫ 0 1 x α y α − 1 d x d y {\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\cdot (xy)^{\alpha -1}\,dx\,dy=\int _{0}^{1}\int _{0}^{1}x^{\alpha -1}\,y^{\alpha -1}\,dx\,dy-\int _{0}^{1}\int _{0}^{1}x^{\alpha }\,y^{\alpha -1}\,dx\,dy} = 1 α 2 − 1 α + 1 ⋅ 1 α = 1 α 2 − 1 α + 1 α + 1 {\displaystyle ={\frac {1}{\alpha ^{2}}}-{\frac {1}{\alpha +1}}\cdot {\frac {1}{\alpha }}={\frac {1}{\alpha ^{2}}}-{\frac {1}{\alpha }}+{\frac {1}{\alpha +1}}} Integriere zweimal hintereinander nach α : {\displaystyle \alpha :} ∫ 0 1 ∫ 0 1 ( 1 − x ) ( x y ) α − 1 log ( x y ) d x d y = − 1 α − log ( α ) + log ( α + 1 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\,{\frac {(xy)^{\alpha -1}}{\log(xy)}}\,dx\,dy=-{\frac {1}{\alpha }}-\log(\alpha )+\log(\alpha +1)} ∫ 0 1 ∫ 0 1 ( 1 − x ) ( x y ) α − 1 log 2 ( x y ) d x d y = ( α + 1 ) ⋅ log ( 1 + 1 α ) − 1 {\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\,{\frac {(xy)^{\alpha -1}}{\log ^{2}(xy)}}\,dx\,dy=(\alpha +1)\cdot \log \left(1+{\frac {1}{\alpha }}\right)-1} (Die Integrationskonstanten sind null, da beide Terme auf der rechten Seite gegen null gehen für α → ∞ {\displaystyle \alpha \to \infty } .) Substituiere x ↦ x 2 , y ↦ y 2 : {\displaystyle x\mapsto x^{2}\,,\,y\mapsto y^{2}\,:} ∫ 0 1 ∫ 0 1 ( 1 − x 2 ) ( x y ) 2 α − 1 log 2 ( x y ) d x d y = ( α + 1 ) ⋅ log ( 1 + 1 α ) − 1 {\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x^{2})\,{\frac {(xy)^{2\alpha -1}}{\log ^{2}(xy)}}\,dx\,dy=(\alpha +1)\cdot \log \left(1+{\frac {1}{\alpha }}\right)-1} Substituiere α = 2 k + 1 2 : {\displaystyle \alpha ={\frac {2k+1}{2}}\,:} ∫ 0 1 ∫ 0 1 ( 1 − x 2 ) ( x y ) 2 k log 2 ( x y ) d x d y = 2 k + 3 2 ⋅ log ( 1 + 2 2 k + 1 ) − 1 {\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x^{2})\,{\frac {(xy)^{2k}}{\log ^{2}(xy)}}\,dx\,dy={\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1} Generiere eine geometrische Reihe: ∫ 0 1 ∫ 0 1 1 − x 2 ( 1 + x 2 y 2 ) log 2 ( x y ) d x d y = ∑ k = 0 ∞ ( − 1 ) k [ 2 k + 3 2 ⋅ log ( 1 + 2 2 k + 1 ) − 1 ] {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1\right]} = ∑ k = 1 ∞ ( − 1 ) k − 1 [ 2 k + 1 2 ⋅ log ( 2 k + 1 2 k − 1 ) − 1 ] {\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\left[{\frac {2k+1}{2}}\cdot \log \left({\frac {2k+1}{2k-1}}\right)-1\right]} = ∑ k = 1 ∞ ( − 1 ) k − 1 [ 2 k ⋅ 1 2 log ( 2 k + 1 2 k − 1 ) − 1 ] ⏟ − 1 2 + 2 G π + 1 2 ∑ k = 1 ∞ ( − 1 ) k − 1 log ( 2 k + 1 2 k − 1 ) ⏟ − 2 log ( 2 ⋅ Γ ( 3 4 ) Γ ( 1 4 ) ) {\displaystyle =\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {2k+1}{2k-1}}\right)-1\right]} _{-{\frac {1}{2}}+{\frac {2G}{\pi }}}+{\frac {1}{2}}\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)} _{-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}} Nun müssen noch die unterschweiften Ausdrücke nachgerechnet werden. F ( x ) := ∑ k = 1 ∞ ( − 1 ) k − 1 [ 2 k ⋅ 1 2 log ( 1 + x 2 k 1 − x 2 k ) − x ] {\displaystyle F(x):=\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {1+{\frac {x}{2k}}}{1-{\frac {x}{2k}}}}\right)-x\right]} = ∑ k = 1 ∞ ( − 1 ) k − 1 ∑ n = 1 ∞ 1 2 n + 1 x 2 n + 1 ( 2 k ) 2 n = ∑ n = 1 ∞ 1 2 n + 1 ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 k ) 2 n x 2 n + 1 = ∑ n = 1 ∞ 1 2 n + 1 η ( 2 n ) 2 2 n x 2 n + 1 {\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\,{\frac {x^{2n+1}}{(2k)^{2n}}}=\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{(2k)^{2n}}}\,x^{2n+1}=\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\,{\frac {\eta (2n)}{2^{2n}}}\,x^{2n+1}} F ′ ( x ) = ∑ n = 1 ∞ η ( 2 n ) ( x 2 ) 2 n = − 1 2 + 1 4 ⋅ π x sin π x 2 {\displaystyle F'(x)=\sum _{n=1}^{\infty }\eta (2n)\,\left({\frac {x}{2}}\right)^{2n}=-{\frac {1}{2}}+{\frac {1}{4}}\cdot {\frac {\pi x}{\sin {\frac {\pi x}{2}}}}} Wegen F ( 0 ) = 0 {\displaystyle F(0)=0} ist F ( 1 ) = ∫ 0 1 F ′ ( x ) d x = − 1 2 + 1 4 ∫ 0 1 π x sin π x 2 d x = − 1 2 + 2 G π {\displaystyle F(1)=\int _{0}^{1}F'(x)\,dx=-{\frac {1}{2}}+{\frac {1}{4}}\int _{0}^{1}{\frac {\pi x}{\sin {\frac {\pi x}{2}}}}\,dx=-{\frac {1}{2}}+{\frac {2G}{\pi }}} . ∑ k = 1 ∞ ( − 1 ) k − 1 log ( 2 k + 1 2 k − 1 ) = ∑ k = 0 ∞ ( − 1 ) k log ( 2 k + 3 2 k + 1 ) {\displaystyle \sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\,\log \left({\frac {2k+3}{2k+1}}\right)} = ∑ k = 0 ∞ ( log 4 k + 3 4 k + 1 − log 4 k + 5 4 k + 3 ) = ∑ k = 0 ∞ log ( 4 k + 3 ) ( 4 k + 3 ) ( 4 k + 1 ) ( 4 k + 5 ) {\displaystyle =\sum _{k=0}^{\infty }\left(\log {\frac {4k+3}{4k+1}}-\log {\frac {4k+5}{4k+3}}\right)=\sum _{k=0}^{\infty }\log {\frac {(4k+3)(4k+3)}{(4k+1)(4k+5)}}} = log ∏ k = 0 ∞ ( k + 3 4 ) ( k + 3 4 ) ( k + 1 4 ) ( k + 5 4 ) = log Γ ( 1 4 ) Γ ( 5 4 ) Γ ( 3 4 ) Γ ( 3 4 ) = log ( 1 4 ⋅ Γ 2 ( 1 4 ) Γ 2 ( 3 4 ) ) {\displaystyle =\log \prod _{k=0}^{\infty }{\frac {\left(k+{\frac {3}{4}}\right)\left(k+{\frac {3}{4}}\right)}{\left(k+{\frac {1}{4}}\right)\left(k+{\frac {5}{4}}\right)}}=\log {\frac {\Gamma \left({\frac {1}{4}}\right)\,\Gamma \left({\frac {5}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)\,\Gamma \left({\frac {3}{4}}\right)}}=\log \left({\frac {1}{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}\right)} = 2 log ( 1 2 ⋅ Γ ( 1 4 ) Γ ( 3 4 ) ) = − 2 log ( 2 ⋅ Γ ( 3 4 ) Γ ( 1 4 ) ) {\displaystyle =2\log \left({\frac {1}{2}}\cdot {\frac {\Gamma \left({\frac {1}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)}}\right)=-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
![{\displaystyle V=[0,1]^{3}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8a636d21e68115492a9532363801c450f6ee6ef)
in drei kongruente Pyramiden mit dem Ursprung als Spitze.
![{\displaystyle =3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\left[{\frac {r^{4}}{4}}\right]_{0}^{\frac {1}{\sin \theta \,\cos \varphi }}\,\sin \theta \,d\theta \,d\varphi ={\frac {3}{4}}\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}\,{\frac {d\varphi }{\cos ^{4}\varphi }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc4d209427b5be32b0d1f946f3360dad15ecbb23)
.
, dann ist ![{\displaystyle \int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}=\int _{0}^{\cos \varphi }{\sqrt {1+s^{2}}}\,ds=\left[{\frac {s}{2}}{\sqrt {1+s^{2}}}+{\frac {1}{2}}\,{\text{arsinh}}\,s\right]_{0}^{\cos \varphi }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7eec724f5c45f282ff4afda34f611ce44ec70641)
.
,
![{\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}+{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/390f9ac48ee1f2e35fbc16a550c2d723fd1ccd3b)
ist. Und 
![{\displaystyle =\left[{\text{arsinh}}(\cos \varphi )\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\sin \varphi }{\sqrt {1+\cos ^{2}\varphi }}}\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)d\varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fad2ce2508b3b8813ce5de899befba95e02109e)
,
![{\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}-{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88d5de84e6db2c0947b84904d949a19d62e42beb)
ist,
![{\displaystyle =\int _{0}^{1}{\frac {x^{2}}{{\sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=\left[{\text{arsinh}}{\frac {x}{\sqrt {2}}}-\arctan {\frac {x}{\sqrt {2+x^{2}}}}\right]_{0}^{1}={\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd9de040eefd7730bd9c7793b88b1113cd3f3318)
.
, wobei 
ist.
![{\displaystyle \int _{0}^{1}x^{x}\,(1+\log x)\,dx=\left[x^{x}\right]_{0}^{1}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57718d6651b1fc0a213a4631c56f83ed493a1792)
![{\displaystyle Q=[0,1]^{2}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec922533ef8b06f150a23b594f91f65489cc1147)
![{\displaystyle ={\frac {1}{3}}\left[{\frac {\sin \varphi }{\cos ^{2}\varphi }}+\log(\sec \varphi +\tan \varphi )\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79774e783c20611846537b094e3d67d1c658cee2)
![{\displaystyle V=[0,1]^{n}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbaf81bef3689a2ba6690a5702a13d61c231036d)
![{\displaystyle \int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}={\frac {1}{4}}\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b856c33409fb8f8d34b2c2db59ea60c59e65f638)
![{\displaystyle \left[{\begin{matrix}x\mapsto 2\arctan x\\y\mapsto 2\arctan y\\z\mapsto 2\arctan z\end{matrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddbdb3a86339c505c807ae6b38f400882540ef68)
![{\displaystyle \left[{\begin{matrix}x=r\,\sin \theta \,\cos \varphi \\y=r\,\sin \theta \,\sin \varphi \\z=r\,\cos \theta \qquad \;\end{matrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd18de9868b8e813c001442d3ca93fedeb4cfec7)
![{\displaystyle =4\int _{0}^{\infty }{\frac {dt}{1+t^{4}}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {\cos \theta }}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {\sin \varphi \,\cos \varphi }}}=4\cdot {\frac {{\sqrt {2}}\,\pi }{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2\,{\sqrt {2\pi }}}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2{\sqrt {\pi }}}}={\frac {1}{4}}\,\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3047b5dd7f67ab152268d21efc7d2343cd16cc19)
![{\displaystyle I=\int _{0}^{\frac {\pi }{4}}\int _{0}^{\sec \varphi }{\frac {2r}{2-r^{2}}}\,dr\,d\varphi =\int _{0}^{\frac {\pi }{4}}\left[-\log(2-r^{2})\right]_{0}^{\sec \varphi }d\varphi =\int _{0}^{\frac {\pi }{4}}\left(\log 2-\log \left(2-\sec ^{2}\varphi \right)\right)d\varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cae6a31f43547f466bcf1d33ce2b40ac3cb8e70)
![{\displaystyle ={\frac {\pi }{4}}\log 2-\int _{0}^{\frac {\pi }{4}}\left[\log \left(2\cos ^{2}\varphi -1\right)-\log \left(\cos ^{2}\varphi \right)\right]d\varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbd57e6fdffb07b59663a8e3a03454ec5044a0c7)
![{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f328f1ba94750f5fe68d7cca2f671b8dce995cb9)
![{\displaystyle ={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/972365b22c65bb2b43649f0b6aac907c681a963a)
![{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3cdcc9abb9d336a7bf463d26418d085603b29e9b)
![{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08f63677a5d8475e20e1c393d69aa867e4779a2f)
![{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}\right)\right]\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d0a3aaacd9c7e26fedd4833f0ab0b9dc3d2cb3f)
![{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b93193789b4b08c0b5f8f84aac5dada8b3fac64)
![{\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\left[{\frac {2k+1}{2}}\cdot \log \left({\frac {2k+1}{2k-1}}\right)-1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c93b37a275656d2996a5416d9819d1f808fcdfdf)
![{\displaystyle =\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {2k+1}{2k-1}}\right)-1\right]} _{-{\frac {1}{2}}+{\frac {2G}{\pi }}}+{\frac {1}{2}}\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)} _{-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/615874753c9c8ddedcc38948a89c206c330414aa)
![{\displaystyle F(x):=\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {1+{\frac {x}{2k}}}{1-{\frac {x}{2k}}}}\right)-x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1573772c89221efd8d7c5a0bd63a70df0c31c36f)
