First look at the double integral
∫
−
∞
∞
∫
−
∞
∞
e
−
x
2
−
y
2
d
x
d
y
.
{\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy.}
Separating it,
∫
−
∞
∞
e
−
x
2
d
x
∫
−
∞
∞
e
−
y
2
d
y
=
(
∫
−
∞
∞
e
−
x
2
d
x
)
2
{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}dx\int _{-\infty }^{\infty }e^{-y^{2}}dy=(\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}}
So, the double integral is merely the square of the Gaussian integral.
Now, do the double integral in polar co-ordinates.
−
x
2
−
y
2
=
−
(
x
2
+
y
2
)
=
−
r
2
{\displaystyle -x^{2}-y^{2}=-(x^{2}+y^{2})=-r^{2}}
d
x
d
y
=
r
d
r
d
θ
{\displaystyle dxdy=rdrd\theta }
∫
−
∞
∞
∫
−
∞
∞
e
−
x
2
−
y
2
d
x
d
y
=
∫
0
2
π
∫
0
∞
e
−
r
2
r
d
r
d
θ
{\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy=\int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}rdrd\theta }
=
(
∫
0
2
π
d
θ
)
(
∫
0
∞
e
−
r
2
r
d
r
)
{\displaystyle =(\int _{0}^{2\pi }d\theta )(\int _{0}^{\infty }e^{-r^{2}}rdr)}
=
2
π
∫
0
∞
e
−
r
2
r
d
r
.
{\displaystyle =2\pi \int _{0}^{\infty }e^{-r^{2}}rdr.}
Substituting
z
=
r
2
{\displaystyle z=r^{2}}
integral :
=
2
π
∫
0
∞
e
−
z
(
z
1
/
2
)
(
d
z
2
z
1
/
2
)
{\displaystyle =2\pi \int _{0}^{\infty }e^{-z}(z^{1/2})({\frac {dz}{2z^{1/2}}})}
=
π
∫
0
∞
e
−
z
d
z
=
π
[
−
e
−
z
]
0
∞
=
π
.
{\displaystyle =\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .}
Therefore,
(
∫
−
∞
∞
e
−
x
2
d
x
)
2
=
π
{\displaystyle (\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}=\pi }
∫
−
∞
∞
e
−
x
2
d
x
=
π
.
{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}dx={\sqrt {\pi }}.}
![{\displaystyle =\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5336cdbf12de9441ca5a5914951eb17aa7a128b)
