Definite integral of exponentials and trig functions

Question

Wikipedia has

$$ \int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}} $$

and

$$ \int_{-\infty}^{\infty} e^{-ax^2} e^{-2bx} dx= \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{a}} $$

https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Integrals_involving_exponential_and_trigonometric_functions

I encountered

$$ \int_{-\infty}^{\infty} e^{-x^2/a^2} cos(2kx) dx $$

where $k=2\pi n/L$. Wolfram alpha integrates this as

$$ \sqrt{\pi}a \cdot e^{-\pi^2 n^2 a^2/L^2} $$

https://www.wolframalpha.com/input/?i=integrate+e%5E%28-x%5E2%2Fa%5E2%29+cos%282+pi+n+x%2FL%29+from+-infinity+to+infinity

This seems to line up ALMOST perfectly with Wikipedia's

$$ \int_{-\infty}^{\infty} e^{-ax^2+bx} dx= \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{4a}} $$

Given this hint from Wikipedia and Wolfram's answer does the following hold?

$$ \int_{-\infty}^{\infty} e^{-ax^2}cos(bx) dx= \sqrt{\frac{\pi}{a}} e^{-\frac{b^2}{4a}} $$

Answer 1

Note that

$$ \int_{-\infty}^{\infty} e^{-ax^2+c x} dx= \sqrt{\frac{\pi}{a}} e^{\frac{c^2}{4a}} $$

holds for any complex coefficient $c$. Thus, let $c= ib$ to arrive at

$$ \int_{-\infty}^{\infty} e^{-ax^2+i b x} dx=\int_{-\infty}^{\infty} e^{-ax^2}cos(bx) dx= \sqrt{\frac{\pi}{a}} e^{-\frac{b^2}{4a}} $$

Answer 2

You have to evaluate the following integral: $$\int_{-\infty}^{\infty}e^{-ax^{2}}e^{ibx}dx = \int_{-\infty}^{\infty}e^{-ax^{2}+ibx}dx.$$ Note that, because $e^{ibx} = \cos(x)+i\sin(x)$ and $\sin(x)$ is an odd function, the product $\sin(x)e^{-ax^{2}}$ is also odd, so the imaginary part of the integral is zero and your integral $\int_{-\infty}^{\infty}e^{-ax^{2}}\cos(bx)dx$ is precisely the above integral. Now, we have: $$ ax^{2}-ibx = ax^{2}-ibx -\frac{b^{2}}{4a}+\frac{b^{2}}{4a} = \bigg{(}\sqrt{a}x-i\frac{b}{2\sqrt{a}}\bigg{)}^{2}+\frac{b^{2}}{4a}$$ Thus: $$\int_{-\infty}^{\infty}e^{-ax^{2}+ibx}dx = \int_{-\infty}^{\infty}e^{-\bigg{(}\sqrt{a}x-i\frac{b}{2\sqrt{a}}\bigg{)}^{2}-\frac{b^{2}}{4a}}dx = e^{-\frac{b^{2}}{4a}}\int_{-\infty}^{\infty}e^{-\bigg{(}\sqrt{a}x-i\frac{b}{2\sqrt{a}}\bigg{)}}dx$$ Setting $u = \sqrt{a}x-i\frac{b}{2\sqrt{a}}$, $du = \sqrt{a}dx$ and the former integral becomes: $$\frac{1}{\sqrt{a}}e^{-\frac{b^{2}}{4a}}\overbrace{\int_{-\infty}^{\infty}e^{-u^{2}}du}^{=\sqrt{\pi}} = \sqrt{\frac{\pi}{a}}e^{-\frac{b^{2}}{4a}}$$