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I'm studying a paper where an integral of similar form to this one appears:

$$\int_{-\infty}^{\infty}\text{sech}^2(x)\cos(x)\,dx$$

The authors only show the result, which involves a hyperbolic cosecant function with a $\pi$ in its argument. So, I assume that they considered a closed path and used the residuals theorem. I know that the poles of the integrand function are $z=i\left(\frac{\pi}{2}+k\pi\right),\,k\in\mathbb{Z}$. I tried to solve the integral by myself, but I don't know which path to choose. Do you have any hints to solve it?

Élio Pereira
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1 Answers1

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Hint:

Use $\operatorname{Re} e^{ix} =\cos x$. Then choose a contour that encircles the upper half-plane.

By using the residue theorem, we get the result $$I= \int_{-\infty}^{\infty} \frac{\cos x}{\cosh^2 x}\,dx = \operatorname{Re} \sum_{k=0}^\infty 2\pi i\,\operatorname{Res}_{x=i\pi (k+1/2)} \left(\frac{e^{ix}}{\cosh^2 x}\right).$$ We can calculate the residues as (the poles are of second order and $\cosh^{-2}(x) \approx -(x-x_0)^{-2}$ close to the pole at $x_0$.) $$\operatorname{Res}_{x=i\pi (k+1/2)} \left(\frac{e^{ix}}{\cosh^2 x}\right) = - \frac{d}{dx} e^{ix} \Big|_{x=i\pi (k+1/2)} = -i e^{-\pi(k+1/2)}\,.$$ The integral then assumes the form $$ I= 2\pi \sum_{k=0}^\infty e^{-\pi(k+1/2)} = \frac{2\pi e^{-\pi/2} }{1-e^{-\pi}} = \frac{\pi}{\sinh(\pi/2)}\,.$$

Fabian
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  • Aren't the poles of second order? The hyperbolic cosine function is squared. – Élio Pereira Jan 09 '19 at 13:15
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    @ÉlioPereira: thanks, I will correct it. – Fabian Jan 09 '19 at 13:24
  • Thank you very much for your help. – Élio Pereira Jan 09 '19 at 13:32
  • Can you help me with just one more thing? How could I show that the integral on the arc is null? – Élio Pereira Jan 09 '19 at 16:32
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    @ÉlioPereira: there are different ways to show that. The most straightforward way might be that $|e^{i z}/\cosh(z)^2| = 2e^{-y}/(\cos 2 y + \cosh 2x) \leq 2 e^{-y}/(\cosh2 x -1)$ for $z=x+i y$. On the arc, we have that $y\to+\infty$ fast enough that the integral vanishes. To show this technically, you might have to distinguish points $z=r e^{i\theta}$ with $|\theta -\pi/2| < \epsilon$ from the rest. – Fabian Jan 09 '19 at 17:56
  • How could I prove that the integral defined at $|\theta-\pi/2|<\epsilon$ goes to zero when $\epsilon \to 0$? There's an indetermination. – Élio Pereira Jan 09 '19 at 19:15
  • @ÉlioPereira: that integral is fine (it vanishes for $R\to\infty$). Just make sure that you choose for the arc specific sequence of radii $R$. For example $R_n= \pi n$ with $n\in\mathbb{N}$ and let $n\to\infty$. – Fabian Jan 09 '19 at 21:16
  • How do you solve this integral?: $2R\int_{\frac{\pi}{2}-\varepsilon}^{\frac{\pi}{2}+\varepsilon}\frac{e^{-R\sin\theta}}{\cosh\left(2R\cos\theta\right)-1},d\theta$. When $\theta=\frac{\pi}{2}$ we get 0 in the denominator of the integrand function. Can you elaborate a little bit more this integral please? It is my final doubt. – Élio Pereira Jan 09 '19 at 22:39
  • You should not use this approximation for this part. Stick to the exact expression $2e^{y}/(\cos2y+\cosh 2x)$ and then do approximations on it. We have $\cos(2y) \approx 1$ and $\cosh 2x \approx 1$. (for the choice of $R$ given above) – Fabian Jan 10 '19 at 10:05
  • I now understand. Thank you! – Élio Pereira Jan 10 '19 at 11:46