Computing $\sum_{n=-\infty}^\infty \int_{-\pi}^\pi e^{-x^2} \cos(nx) dx $

Question

I'm trying to find $$ \sum_{n=-\infty}^\infty \int_{-\pi}^\pi e^{-x^2} \cos(nx) dx. $$

About the only thing I can think of is the well-known identity

$$ \sum_{n=-k}^k \cos(nx) = \sum_{n=-k}^k e^{inx} = \frac{\sin\left((k+\frac{1}{2})x\right)}{\sin\left(\frac{x}{2}\right)}. $$

But this doesn't seem to simplify things much.

The integrand is even, so you can reduce it to a sum over the positive integers and an integral over $[0,\pi]$. That's all I could work out here. — Math1000, Jul 29 '15 at 01:23

score 5 · Accepted Answer · answered Jul 29 '15 at 01:57

Because $e^{-x^{2}}$ is an even function, the Fourier series for $e^{-x^{2}}$ is $$ e^{-x^{2}}\sim \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-t^2}dt+\sum_{n=1}^{\infty}\frac{1}{\pi}\int_{-\pi}^{\pi}e^{-t^{2}}\cos(nt)dt\cos(nx). $$ Evaluating at $x=0$ gives you what you want: $$ \begin{align} 1 & = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-t^{2}}dt+\frac{1}{\pi}\sum_{n=1}^{\infty}\int_{-\pi}^{\pi}e^{-t^{2}}\cos(nt)dt \\ & = \frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi}e^{-x^{2}}\cos(nt)dt \end{align} $$ Looks like the answer is $2\pi$.

score 2 · Answer 2 · answered Jul 29 '15 at 04:05

2

Well, there were two solid answers posted. So, let's try this using the complex Fourier series.

We have

$$e^{-x^2}=\sum_{n=-\infty}^{\infty}C_ne^{inx}\tag 1$$

where

$$C_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-t^2}e^{-int}dt \tag 2$$

Using $(2)$ in $(1)$ with $x=1$ yields

$$1=\sum_{n=-\infty}^{\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-t^2}e^{-int}dt \tag 3$$

whereupon multiplying both sides of $(3)$ by $2\pi$ and exploiting the fact that the sine function is odd reveals

$$\bbox[5px,border:2px solid #C0A000]{2\pi=\sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi}e^{-t^2}\cos (nt)\,dt }$$

answered Jul 29 '15 at 04:05

Mark Viola

179,405

This is how I ended up working the problem. – Philip Hoskins Jul 29 '15 at 04:07
@PhilipHoskins I think this way is just a bit faster, don't you? We don't have to worry about separating the constant term and unfolding a summation. Clean and efficient! – Mark Viola Jul 29 '15 at 04:09
Agreed. Also, it's the only way I really know how to compute a Fourier series – Philip Hoskins Jul 29 '15 at 04:11
@PhilipHoskins Well it isn't the only way, but it is quite compact compared to the cosine-sine series. No floating constant term with $\frac{a_0}{2}$, and a single $\frac{1}{2\pi}$ normalization. – Mark Viola Jul 29 '15 at 04:14
Certainly, but it's the only way I bothered to commit to memory – Philip Hoskins Jul 29 '15 at 04:28

Jack D'Aurizio · Answer 3 · 2015-07-29T04:12:35.760

1

$f(x)=e^{-x^2}$ is a function that belongs to the Schwarz space, and its Fourier series is uniformly converging to $f(x)$. So, if we set: $$ c_k = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(kx)\,dx $$ we have: $$ e^{-x^2} = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx + \sum_{n\geq 1} c_n \cos(nx) $$ and: $$ \sum_{n\geq 1}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx = \pi\sum_{n\geq 1}c_n=\pi\, f(0)-\frac{1}{2}\int_{-\pi}^{\pi}f(x)\,dx $$ as well as: $$ \sum_{n\in\mathbb{Z}}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx=2\pi\, f(0)=\color{red}{2\pi}.$$

That also follows from the convergence in distribution of the Dirichlet kernel to the Dirac delta function.

edited Jul 29 '15 at 04:12

answered Jul 29 '15 at 03:31

Jack D'Aurizio

353,855

I think its just a small bit faster using the complex FS. – Mark Viola Jul 29 '15 at 04:06
Yes, in fact $\sum_{n=-\infty}^{\infty}\cos (nx)\cos (nt)\sim 2\pi \delta(x-t)$ as a generalized function! And this is true for any orthogonal basis in $\mathscr{L}^2$!! Neat stuff. – Mark Viola Jul 29 '15 at 04:17

Computing $\sum_{n=-\infty}^\infty \int_{-\pi}^\pi e^{-x^2} \cos(nx) dx $

3 Answers3