1. Beweis
Verwende die Reihenentwicklung
cos
(
2
a
x
)
=
∑
k
=
0
∞
(
−
1
)
k
(
2
k
)
!
(
2
a
x
)
2
k
{\displaystyle \cos(2ax)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}\,(2ax)^{2k}}
I
:=
∫
−
∞
∞
e
−
x
2
cos
(
2
a
x
)
d
x
=
∑
k
=
0
∞
(
−
1
)
k
(
2
k
)
!
(
2
a
)
2
k
∫
−
∞
∞
x
2
k
e
−
x
2
d
x
{\displaystyle I:=\int _{-\infty }^{\infty }e^{-x^{2}}\,\cos(2ax)\,dx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}\,(2a)^{2k}\,\int _{-\infty }^{\infty }x^{2k}\,e^{-x^{2}}\,dx}
∫
−
∞
∞
x
2
k
e
−
x
2
d
x
=
2
∫
0
∞
x
2
k
e
−
x
2
d
x
=
2
∫
0
∞
y
k
e
−
y
d
y
2
y
=
Γ
(
k
+
1
2
)
=
π
2
2
k
−
1
Γ
(
2
k
)
Γ
(
k
)
=
π
2
2
k
(
2
k
)
!
k
!
{\displaystyle \int _{-\infty }^{\infty }x^{2k}\,e^{-x^{2}}\,dx=2\int _{0}^{\infty }x^{2k}\,e^{-x^{2}}\,dx=2\int _{0}^{\infty }y^{k}\,e^{-y}\,{\frac {dy}{2{\sqrt {y}}}}=\Gamma \left(k+{\frac {1}{2}}\right)={\frac {\sqrt {\pi }}{2^{2k-1}}}\,{\frac {\Gamma (2k)}{\Gamma (k)}}={\frac {\sqrt {\pi }}{2^{2k}}}\,{\frac {(2k)!}{k!}}}
I
=
∑
k
=
0
∞
(
−
1
)
k
(
2
k
)
!
(
2
a
)
2
k
π
2
2
k
(
2
k
)
!
k
!
=
π
⋅
∑
k
=
0
∞
(
−
1
)
k
k
!
a
2
k
=
π
⋅
e
−
a
2
{\displaystyle I=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}\,(2a)^{2k}\,{\frac {\sqrt {\pi }}{2^{2k}}}\,{\frac {(2k)!}{k!}}={\sqrt {\pi }}\cdot \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}\,a^{2k}={\sqrt {\pi }}\cdot e^{-a^{2}}}
![{\displaystyle \Rightarrow \,\int _{-\infty }^{\infty }f(x+ia)\,dx=e^{a^{2}}\cdot {\Bigg [}\int _{-\infty }^{\infty }e^{-x^{2}}\cos(2ax)\,dx-i\underbrace {\int _{-\infty }^{\infty }e^{-x^{2}}\sin(2ax)\,dx} _{=0}{\Bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/526b7e011d7f5936c0f4dd2b0f62c5f5f42a506b)
