Determine the Integral $\int_{-\infty}^{\infty} e^{-x^2} \cos(2bx)dx$

Question

Please do not mark this question as a duplicate. I have to solve this with a different method that I don't believe has been discussed about this particular question (at least to my knowledge).

I am confronted with computing the integral below:

$$\int_{-\infty}^{\infty} e^{-x^2} \cos(2bx)dx, b \in \mathbb R, b \gt0$$

I understand that this is a question in which Complex Analysis has applications to Real Analysis. Specifically, Cauchy's Theorem will be used. That being said, the hint I was given (and the method I am attempting to use) is the following:

Integrate $e^{-z^2}$ over this curve:

The curve is a rectangle such that its length is $2R$ and its width is $b$. Also, the length on the lower side of the rectangle lies on the x-axis. Lastly, the direction of curvature is counter-clockwise.

My question is this: why integrate $e^{-z^2}$, and not $e^{-z^2}\cos(2bz)$? That is, the actual integral we're computing? I know that the former would be easier to integrate, but where did the $\cos(2bz)$ term go, and where will it come into play again?

Also, it's important to say that $\int_{-\infty}^{\infty} e^{-t^2}dt = \sqrt\pi$ will be useful here as well.

Lastly, I know that from the hint suggested, there will be four curves to evaluate: the two lengths and the two widths of the rectangle.

Answer 1

If we set: $$ f(b) = \int_{\mathbb{R}} e^{-x^2}\cos(2bx)\,dx \tag{1}$$ we have: $$ f'(b) = -\int_{\mathbb{R}} 2x\,e^{-x^2} \sin(2bx)\,dx \stackrel{\text{IBP}}{=}-2b\int_{\mathbb{R}}e^{-x^2}\cos(2bx)\,dx=-2b\,f(b).\tag{2}$$ So we have that $f$ is a solution of a separable differential equation and $$ f(b) = f(0)\, e^{-b^2}.\tag{3}$$ Since $f(0)=\sqrt{\pi}$,

$$ \int_{\mathbb{R}} e^{-x^2}\cos(2bx)\,dx = \color{red}{\sqrt{\pi}\, e^{-b^2}}\tag{4}$$

follows.

Answer 2

$$ \begin{aligned} \int_{-\infty}^{\infty} e^{-x^2} \cos (2bx) d x & = \Re \int_{-\infty}^{\infty} e^{-x^2} \cdot e^{2b x i} d x \\ & =\Re\int_{-\infty}^{\infty} e^{-\left(x-bi\right)^2-b^2} d x \\ & =e^{-b^2} \Re \int_{-\infty}^{\infty} e^{-\left(x-b i\right)^2} d x \\ & =\sqrt{\pi} e^{-b^2} \end{aligned} $$

Answer 3

$$\eqalign{ & b > 0:I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}\cos \left( {2bx} \right)dx} = 2\int\limits_0^\infty {{e^{ - {x^2}}}\cos \left( {2bx} \right)dx} \cr & = 2\int\limits_0^\infty {{e^{ - {x^2}}}\left( {\underbrace {\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}{{\left( {2bx} \right)}^{2n}}} }_{{\text{Series of }}\cos \left( {2bx} \right)}} \right)dx} \cr & = 2\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}{{\left( {2b} \right)}^{2n}}} \underbrace {\int\limits_0^\infty {{e^{ - {x^2}}}{x^{2n}}dx} }_{{\text{Gamma function}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}{{\left( {2b} \right)}^{2n}}\Gamma \left( {n + \frac{1}{2}} \right)} \cr & \because \Gamma \left( {n + \frac{1}{2}} \right) = \frac{{\left( {2n} \right)!}}{{{4^n}n!}}\sqrt \pi \Rightarrow I = \sqrt \pi \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}{{\left( {2b} \right)}^{2n}}\frac{{\left( {2n} \right)!}}{{{4^n}n!}}} = \sqrt \pi \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - {b^2}} \right)}^n}}}{{n!}}} \cr & \because \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}} = {e^x}} \Rightarrow I = \sqrt \pi {e^{ - {b^2}}} \cr} $$