# Residue of $p.v.\int_{-\infty}^{\infty}\frac{e^{2x}}{\cosh(\pi x)}dx=\text{sec}1$

Show that $$p.v.\int_{-\infty}^{\infty}\frac{e^{2x}}{\cosh(\pi x)}dx=\text{sec}1$$ by integrating $\frac{e^{2z}}{\cosh(\pi z)}$ around rectangles with vertices at $z=\pm p,p+i,-p+i.$

I asked this similar question, and was recommended this link stackexchange, but the answer provided is confusing.

I am not sure how they got those vertices, but with those vertices I got the parameterization:

$\begin{cases} \Gamma_1 = t, & -p\leq t\leq p \\\\ \Gamma_2= p+it, &0\leq t \leq 1 \\\\ \Gamma_3 = i-t, & -p\leq t\leq p \\\\ \Gamma_4=i(1-t)-p, & 0\leq t\leq 1. \end{cases}$

Thus I have $\Gamma_p =\Gamma_1 +\Gamma_2+\Gamma_3+\Gamma_4 .$

$f(z)=\frac{e^{2z}}{\cosh(\pi z)} = \frac{2e^{z(2+i\pi)}}{e^{2i\pi z}+1}$, where the denominator has a $\text{pole}=(2n+1)\frac{1}{2}$.

And this is as far as I can get.

• There's a lot more in that answer than just the vertices. I think you should be more specific about what it is in the answer you accepted that you don't understand. – joriki Apr 13 '13 at 3:32

Consider the integral

$$\oint_C dz \frac{e^{2 z}}{\cosh{\pi z}}$$

where $C$ is the above-described rectangle. On the one hand, this integral is equal to the integral about the individual legs of the contour, viz.:

$$\int_{-p}^p dx \frac{e^{2 x}}{\cosh{\pi x}} + i \int_0^1 dy \frac{e^{2 (p+i y)}}{\cos{\pi (p+i y)}} + \int_{p}^{-p} dx \frac{e^{2 (x+i)}}{\cosh{\pi (x+i)}} + i \int_1^0 dy \frac{e^{2 (-p+i y)}}{\cos{\pi (-p+i y)}}$$

Take the limit as $p \rightarrow \infty$. It should be clear that the 2nd and 4th integrals - those over the vertical legs of $C$ - will vanish in this limit. That leaves the 1st and 3rd integrals over the horizontal sections; these may be combined to produce

$$\left(1+e^{i 2}\right) \int_{-\infty}^{\infty} dx \frac{e^{2 x}}{\cosh{\pi x}}$$

This equals, by the residue theorem, the residue at the only pole within $C$, namely, $z=i/2$:

$$i 2 \pi \lim_{z->i/2} \frac{e^{2 z}}{\cosh{\pi z}} = i 2 \pi\frac{e^{i}}{\pi \sinh{(i \pi/2)}} = 2 e^{i}$$

Therefore

$$\int_{\infty}^{\infty} dx \frac{e^{2 x}}{\cosh{\pi x}} = \frac{2 e^{i}}{1+e^{i 2}}= \sec{1}$$

You should note that no $PV$ was needed here.

• Thank you again, this is much more clear and concise. I have a couple of questions, why is the pole $i/2$ and not $1/2$? And how does $i 2 \pi \lim_{z->i/2} \frac{e^{2 z}}{\cosh{\pi z}} = i 2 \pi\frac{e^{i}}{\pi \sinh{(i \pi/2)}}$ is that from a identity? – Q.matin Apr 13 '13 at 5:30
• The pole is at $i/2$ because $\cosh{i y}=\cos{y}$. Also, $\text{Res}_{z=z_0} f(z)/(g(z) = f(z_0)/g'(z_0)$. – Ron Gordon Apr 13 '13 at 5:44

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\large\mbox{A}\ \Re eal\mbox{-Approach}}$$: \begin{align} &\bbox[5px,#ffd] {\int_{-\infty}^{\infty}{\expo{2x} \over \cosh\pars{\pi x}} \,\dd x} = 2\int_{0}^{\infty}{\cosh\pars{2x} \over \cosh\pars{\pi x}} \,\dd x \\[5mm] = &\ 4\int_{0}^{\infty}{\sinh\pars{\pi x}\cosh\pars{2x} \over \sinh\pars{2\pi x}}\,\dd x = {\tt J}\pars{\pi + 2} + {\tt J}\pars{\pi - 2} \label{1}\tag{1} \\[2mm] &\ \mbox{where}\ {\tt J}\pars{\alpha} \equiv 2\int_{0}^{\infty}{\sinh\pars{\alpha x} \over \sinh\pars{2\pi x}} \,\dd x \end{align}

\begin{align} {\tt J}\pars{\alpha} & = 2\int_{0}^{\infty} {\expo{-\pars{2\pi - \alpha}x} - \expo{-\pars{2\pi + \alpha}x} \over 1 - \expo{-4\pi x}}\,\dd x \\[5mm] \stackrel{4\pi x\ \mapsto\ x}{=}\,\,\, &\ {1 \over 2\pi}\int_{0}^{\infty} {\expo{-\bracks{1/2 - \alpha/\pars{4\pi}}x}\,\,\, - \expo{-\bracks{1/2 + \alpha/\pars{4\pi}}x} \over 1 - \expo{-x}} \,\dd x \\[5mm] = &\ {1 \over 2\pi}\int_{0}^{\infty} {1 - \expo{-\bracks{1/2 + \alpha/\pars{4\pi}}x} \over 1 - \expo{-x}}\,\dd x \\[2mm] - &\ {1 \over 2\pi}\int_{0}^{\infty} {1 - \expo{-\bracks{1/2 - \alpha/\pars{4\pi}}x} \over 1 - \expo{-x}}\,\dd x \\[5mm] = & {1 \over 2\pi} \bracks{\Psi\pars{{1 \over 2} + {\alpha \over 4\pi}} - \Psi\pars{{1 \over 2} - {\alpha \over 4\pi}}} \\[5mm] = &\ {1 \over 2\pi}\,\pi \cot\pars{\pi\bracks{{1 \over 2} - {\alpha \over 4\pi}}} = {1 \over 2}\tan\pars{\alpha \over 4}\label{2}\tag{2} \end{align}
With (\ref{1}) and (\ref{2}): \begin{align} &\bbox[5px,#ffd] {\int_{-\infty}^{\infty}{\expo{2x} \over \cosh\pars{\pi x}} \,\dd x} = {1 \over 2}\tan\pars{\pi + 2 \over 4} + {1 \over 2}\tan\pars{\pi - 2 \over 4} \\[5mm] = &\ {1 \over 2}\,{1 + \tan\pars{1/2} \over 1 - \tan\pars{1/2}} + {1 \over 2}\,{1 - \tan\pars{1/2} \over 1 + \tan\pars{1/2}} = {1 + \tan^{2}\pars{1/2} \over 1 - \tan^{2}\pars{1/2}} \\[5mm] = &\ {1 \over \cos^{2}\pars{1/2} - \sin^{2}\pars{1/2}} = {1 \over \cos\pars{2\bracks{1/2}}} \\[5mm] = & \bbx{\sec\pars{1}} \approx 1.8508 \\ & \end{align}