Integrating an exponential function

Question

How do I integrating this function- $$I=\int \exp{\mathrm{\frac{(-1)(z^2-k)^2}{2}}}dz$$ limit is from $-\infty$ to $+\infty.$ I tried a lot by nothing worker!Pls help!

Answer 1

the first thing is to simplify the integrand: $$ \int_{-\infty}^{\infty} e^{-\frac{z^2-k}2 dz} = 2 \sqrt{2}e^{\frac{k}2}\int_0^{\infty}e^{-x^2}dx $$ there are two ways of approaching the integral on the right-hand side, which are both instructive exercises.

firstly you may make the substitution $w=x^2$. this gives $$ I=\int_0^{\infty}e^{-x^2}dx =\frac12\int_0^{\infty}w^{-\frac12}e^{-w}dw = \frac{\sqrt{\pi}}{2} $$ using an elementary result from the theory of the gamma function

the second approach uses a "trick" of evaluating the square of the original integral, using a symmetry which can be exploited by a change to polar coordinates, noting that $|\frac{\partial(x,y)}{\partial(r,\theta)}|=r$: $$ I^2= \int_0^{\infty}e^{-x^2}dx\int_0^{\infty}e^{-y^2}dy =\int_0^{\infty}\int_0^{\infty}e^{-(x^2+y^2)}dxdy \\ =\int_0^{\frac{\pi}2}\int_0^{\infty}e^{-r^2}rdrd\theta \\ = \frac{\pi}2 \left[-\frac12e^{-r^2} \right]_0^{\infty} \\ =\frac{\pi}4 $$