Complex Analysis

Question

Gaussian Integral: compute

\int_{- \infty}^{\infty} e^{- a x^{2}} d x

Solution. Let

I = \int_{- \infty}^{\infty} e^{- a x^{2}} d x

Then

I^{2} = (\int_{- \infty}^{\infty} e^{- a x^{2}} d x) (\int_{- \infty}^{\infty} e^{- a x^{2}})

switting to polar coordinates we have

I^{2} = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- a (x^{2} + y^{2})} d x d y = \int_{0}^{2 π} \int_{0}^{\infty} e^{- a r^{2}} \cdot r d r d θ

since

det D a = det [\begin{matrix} - r \sin θ & \cos θ \\ r \cos θ & \sin θ \end{matrix}] = r

Therefore we have by change of variables theorem

\int_{0}^{2 π} \int_{0}^{\infty} e^{- a r^{2}} \cdot r d r d θ = \int_{0}^{2 π} (\frac{1}{2 a}) d θ = \frac{π}{a}

Therefore taking square roots we have

\sqrt{\frac{π}{a}}