Prove that $$\int_0^1 F(x)\cos(2\pi x)\, dx = \pi e^{-2\pi}.$$

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Here is this week's POTW:

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Let $F : \Bbb R \to \Bbb R$ be given by

$$F(x) = \sum_{n\, =\, -\infty}^\infty \frac{1}{1 + (x + n)^2}.$$

Prove that

$$\int_0^1 F(x)\cos(2\pi x)\, dx = \pi e^{-2\pi}.$$

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $M > 0$. For all real numbers $x$ and integers $n$, $|x| \le M$ and $|n| > 2M$ implies that

$$1 + (x + n)^2 \ge 1 + (|n| - |x|)^2 > 1 + \left(\frac{|n|}{2}\right)^2 = 1 + \frac{n^2}{4}.$$ Hence, on $[-M,M]$, $$0 < \frac{1}{1 + (x + n)^2} \le \frac{1}{1 +\frac{n^2}{4}},\quad |n| > 2M.$$

Since $\displaystyle\sum_{n\in \Bbb Z} \frac{1}{1 + n^2/4}$ converges, by the Weierstrass test, the series

$$\sum_{|n| > 2M} \frac{1}{1 + (x + n)^2}$$

converges uniformly on $[-M,M]$. So the full series

$$\sum_{n\, =\, -\infty}^\infty \frac{1}{1 + (x + n)^2}$$

converges uniformly on $[-M,M]$. Then $F$ is continuous on $[-M,M]$. As $M$ was arbitrary, $F$ is continuous on $\Bbb R$ and

$$\int_0^1 F(x)\cos(2\pi x) \, dx = \sum_{n\, =\, -\infty}^\infty \int_0^1 \frac{\cos(2\pi x)}{1 + (x + n)^2}\, dx = \sum_{n\, =\, -\infty}^\infty \int_n^{n+1} \frac{\cos[2\pi(x - n)]}{1 + x^2}\, dx = \int_{-\infty}^\infty \frac{\cos(2\pi x)}{1 + x^2}\, dx.$$

The last integral may be evaluated by method of residues. Indeed, since the function $h(z) = \frac{1}{1 + z^2}$ is holomorphic in $\Bbb H\setminus\{i\}$ (where $\Bbb H$ denotes the upper half plane) such that $h(z) = O(|z|^{-2})$ as $|z| \to \infty$, then

$$\int_{-\infty}^\infty \frac{\cos(2\pi x)}{1 + x^2}\, dx = \Re[2\pi i \cdot \operatorname{Res}_{z = i} e^{2\pi iz}h(z)] = \Re\left[2\pi i \cdot \frac{e^{2\pi i^2}}{2i}\right] = \pi e^{-2\pi}.$$