# Gaussian Integral

$I\overset { let }{ = } \int _{ -\infty }^{ \infty }{ { e }^{ -{ x }^{ 2 } }dx } =\int _{ -\infty }^{ \infty }{ { e }^{ -{ y }^{ 2 } }dy }$

\begin{aligned} { I }^{ 2 } & = (I)(I) \\ \quad & = (\int _{ -\infty }^{ \infty }{ { e }^{ -{ x }^{ 2 } }dx } )(\int _{ -\infty }^{ \infty }{ { e }^{ -{ y }^{ 2 } }dy } ) \\ \quad & = \int _{ -\infty }^{ \infty }{ \int _{ -\infty }^{ \infty }{ { e }^{ -{ (x }^{ 2 }+{ y }^{ 2 }) }dx } dy } \\ \quad & = \int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ \infty }{ r{ e }^{ -{ r }^{ 2 } }dr } d\theta } \\ \quad & = \int _{ 0 }^{ 2\pi }{ { \left[ -\frac { { e }^{ -{ r }^{ 2 } } }{ 2 } \right] }_{ 0 }^{ \infty }d\theta } \\ \quad & = \int _{ 0 }^{ 2\pi }{ \frac { 1 }{ 2 } d\theta } \\ \quad & = { \left[ \frac { \theta }{ 2 } \right] }_{ 0 }^{ 2\pi } \\ { I }^{ 2 } & = \pi \\ I & =\sqrt { \pi } \quad (I>0) \end{aligned}

$\boxed { \displaystyle \int _{ -\infty }^{ \infty }{ { e }^{ -{ x }^{ 2 } }dx } =\sqrt { \pi } }$

Note by Gandoff Tan
2 years, 1 month ago

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