3
$\begingroup$

Evaluate the integral $$ \int_0^{\infty} \frac{\cosh(ax)}{\cosh(x)}\,dx, $$ where $|a|<1$.

Consider the closed loop integral of $\displaystyle\frac{e^{az}}{\cosh(z)}$ where the contour $C$ is $y=0, y=\pi, x=-R$, and $x=R$.

So far I have found that $$\displaystyle\int_C \frac{e^{az}}{\cosh(z)}\,dz=2\pi e^{\frac{\pi i a}{2}}$$ by the residue theorem. Not sure how to compare this to the first part where we are integrating from $0$ to infinity.

$\endgroup$
  • $\begingroup$ Hint: Use Jordan's lemma and the even-ness of cosh. $\endgroup$ – Batman Nov 28 '15 at 2:34
  • 1
    $\begingroup$ I could write an entire novel as an answer to this question. $\endgroup$ – Lucian Nov 28 '15 at 6:01
2
$\begingroup$

I give to you a different approach, not using residue calculus. In case you only wanted a solution using complex analysis, tell me and I will delete my answer.

We write $$ \begin{aligned} \frac{\cosh ax}{\cosh x}&=\frac{e^{(a-1)x}+e^{(-(a+1)x}}{1+e^{-2x}}\\ &=\sum_{k=0}^{+\infty}(-1)^k\bigl(e^{(a-1-2k)x}+e^{-(a+1+2k)x}\bigr) \end{aligned} $$ and then integrate term-wise, to find that $$ \int_0^{+\infty}\frac{\cosh ax}{\cosh x}\,dx=\sum_{k=0}^{+\infty}(-1)^k\Bigl(\frac{1}{1-a+2k}+\frac{1}{1+a+2k}\Bigr). $$ Now, as it happens there is a nice series expansion of the secant function looking very much like this (see for example the series expansion here), $$ \sec z=2\sum_{k=0}^{+\infty}(-1)^k\Bigl(\frac{1}{\pi-2z+2k\pi}+\frac{1}{\pi+2z+2k\pi}\Bigr). $$ In particular, with $z=a\pi/2$, $$ \sec\Bigl(\frac{a\pi}{2}\Bigr)=\frac{2}{\pi}\sum_{k=0}^{+\infty}(-1)^k\Bigl(\frac{1}{1+-a+2k}+\frac{1}{1+a+2k}\Bigr). $$

Hence $$ \int_0^{+\infty}\frac{\cosh ax}{\cosh x}\,dx=\frac{\pi}{2}\sec\Bigl(\frac{a\pi}{2}\Bigr). $$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Since $$ \cosh(z)=0\iff e^z=-e^{-z}\iff e^{2z}=-1\iff z=\left(k+\tfrac12\right)\pi i,k\in\mathbb{Z} $$ the singularities of $\frac{\cosh(az)}{\cosh(z)}$ are at $z=\left(k+\tfrac12\right)\pi i$ for $k\in\mathbb{Z}$.

The residue of $\frac1{\cosh(z)}$ at $z=\left(k+\tfrac12\right)\pi i$ is $$ \frac1{\sinh\left(\left(k+\tfrac12\right)\pi i\right)} =\frac{-i}{\sin\left(\left(k+\tfrac12\right)\pi\right)}=(-1)^{k+1}i $$ Therefore, the residue of $\frac{\cosh(az)}{\cosh(z)}e^{i\delta z}$ at this point is $(-1)^{k+1}i\cos\left(a\left(k+\frac12\right)\pi\right)e^{-\delta\left(k+\frac12\right)\pi}$.

Since $\frac{\cosh(az)}{\cosh(z)}e^{i\delta z}$ vanishes exponentially in the upper halfplane as $\left|z\right|\to\infty$, its integral over $\mathbb{R}$ is $2\pi i$ times the sum of the residues in the upper halfplane. $$ \begin{align} \int_{-\infty}^\infty\frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x &=\lim_{\delta\to0}\int_{-\infty}^\infty\frac{\cosh(ax)}{\cosh(x)}e^{i\delta x}\,\mathrm{d}x\\ &=2\pi\lim_{\delta\to0}\sum_{k=0}^\infty(-1)^k\cos\!\left(a\left(k+\tfrac12\right)\pi\right)e^{-\delta\left(k+\frac12\right)\pi}\\ &=2\pi\lim_{\delta\to0}\operatorname{Re}\left(\sum_{k=0}^\infty(-1)^ke^{(ia-\delta)\left(k+\frac12\right)\pi}\right)\\ &=2\pi\lim_{\delta\to0}\operatorname{Re}\left(e^{(ia-\delta)\pi/2}\sum_{k=0}^\infty(-1)^ke^{(ia-\delta)k\pi}\right)\\ &=2\pi\lim_{\delta\to0}\operatorname{Re}\left(\frac{e^{(ia-\delta)\pi/2}}{1+e^{(ia-\delta)\pi}}\right)\\ &=2\pi\operatorname{Re}\left(\frac{e^{ia\pi/2}}{1+e^{ia\pi}}\right)\\[6pt] &=\pi\sec\left(\frac{a\pi}2\right) \end{align} $$ Therefore, since the integrand is even, $$ \int_0^\infty\frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x=\frac\pi2\sec\left(\frac{a\pi}2\right) $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $\frac{\cosh(az)}{\cosh(z)}\sim e^{(a-1)|x|}$ for large $|x|$ and $e^{i\delta z}\sim e^{-\delta y}$ for large $y$ $\endgroup$ – robjohn Nov 28 '15 at 23:04
  • $\begingroup$ I just got an upvote on this answer. I guess I have answered this question before. At least the method used was different. $\endgroup$ – robjohn Nov 30 '15 at 14:20
2
$\begingroup$

We consider the integral of analytic function $$ F(z)=\frac{\cosh(az)}{\cosh(z)} $$ on the rectangle contour of $C=C_1\cup C_2\cup C_3\cup C_4$, which are $y=0, \:x=R,\:y=\pi,\:x=-R$ respectively. By Cauchy's residue theorem, we have $$ \int_{-R}^RF(x)dx+\int_{C_2}F(z)dz+\int_{C_3}F(z)dz+\int_{C_4}F(z)dz=2\pi iRes\left(F,\frac{\pi i}{2}\right)\tag1 $$ Note that only pole in the contour is $z=\frac{\pi i}{2}$. Since $$ Res\left(F,\frac{\pi i}{2}\right)=\lim_{z\to \frac{\pi i}{2}}(z-\frac{\pi i}{2})F(z)=\lim_{z\to \frac{\pi i}{2}}\frac{\cosh(az)}{\sinh(z)}=- i\cos \frac{\pi a}{2}\tag2 $$ Moreover \begin{align} \int_{C_3}F(z)dz&=\int_R^{-R}\frac{\cosh(a(x+\pi i))}{\cosh(x+\pi i)}\:dx \\ &=\int_{-R}^{R}\frac{\cosh(ax)\cosh(a\pi i)+\sinh(ax)\sinh(a\pi i)}{\cosh(x)}\:dx \\ &=\cosh(a\pi i)\int_{-R}^{R}\frac{\cosh(ax)}{\cosh(x)}\:dx\tag{*} \end{align} (*): Since $\frac{\sinh(ax)}{\cosh(x)}$ is odd $$ \int_{-R}^{R}\frac{\sinh(ax)}{\cosh(x)}\:dx=0 $$ Also since $|a|<1$ $$ \left|\int_{C_2}F(z)dz\right|=\left|\int_{0}^{\pi}\frac{e^{a(R+iy )}+e^{-a(R+iy )}}{e^{R+iy}+e^{-(R+iy )}}idy\right|\leqslant \frac{2\pi e^{|a| R}}{e^{R}-e^{-R}}=\frac{2\pi e^{(|a|-1) R}}{1-e^{-2R}}\to0\tag3 $$ as $R\to\infty$.

Along with $$ \left|\int_{C_4}F(z)dz\right|=\left|\int_{\pi}^{0}\frac{e^{a(-R+iy )}+e^{-a(-R+iy )}}{e^{-R+iy}+e^{-(-R+iy )}}idy\right|\leqslant \frac{2\pi e^{|a| R}}{e^{R}-e^{-R}}=\frac{2\pi e^{(|a|-1) R}}{1-e^{-2R}}\to0\tag4 $$ as $R\to\infty$.

By $(1),(2),(3),(4)$, we have $$ \lim_{R\to\infty}(1+\cosh(a\pi i))\int_{-R}^{R}\frac{\cosh(ax)}{\cosh(x)}\:dx=2\pi\cos {\frac{\pi a}{2}} $$ i.e. $$ \int_{-\infty}^{\infty}\frac{\cosh(ax)}{\cosh(x)}\:dx=\frac{2\pi\cos {\frac{\pi a}{2}}}{1+\cosh(a\pi i)}=\frac{2\pi\cos {\frac{\pi a}{2}}}{2\cos^2 {\frac{\pi a}{2}}}=\pi\sec{\frac{\pi a}{2}} $$ Since $\frac{\cosh(ax)}{\cosh(x)}$ is even, we have finally $$ \int_{0}^{\infty}\frac{\cosh(ax)}{\cosh(x)}\:dx=\frac{\pi}{2}\sec{\frac{\pi a}{2}} $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.