# Use residues to evaluate $\int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x$, where $|a|<1$

Use residues to evaluate $$\int_0^\infty \frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x$$ where $|a|<1$.

Try considering the integral of the form $$\int_C \frac{\exp(az)}{\cosh(z)}\,\mathrm dz,$$ where $C$ is the contour given by $y=0,\, y=\pi,\, x=-R,\, x=R$.

• This answer (with a rotation in the complex plane and the path for the numerator or, simpler, setting $a:=ib$ first) should help. Nov 21, 2012 at 12:44
• @RaymondManzoni: the problem I see with the present problem, and with the link you gave, is that we have the hyperbolic cosine also in the numerator, and not only cosine. This messes things greatly, at least for me, though I think this may have been a typo by the OP since he's been adviced to use that exponential function in the numerator... Nov 21, 2012 at 14:46
• @DonAntonio: Hmmm... I don't think so. The numerator may be $e^{az}$ as here (or $e^{iaz}$ as in the other link) but we are supposing $|a|<1$ and are dividing by $\cosh(z)$. For $|z|\gg 1$ the norm of this fraction will be majored by $\dfrac {e^{|a|z}}{e^z}\sim e^{-(1-|a|)z}$ (the denominator is giving us convergence) so that the two vertical integrals will disappear at the limit (while the two horizontal ones are identical up to a global coefficient and the limit is taken as a P.V.). Or did I forget something? Nov 21, 2012 at 17:27
• @RaymondManzoni , but how do you pass to $\,\cosh az\,$ from $\,e^{az}\,$? I know the vertical integrals vanish in the limit and both horizontal ones are the same up to a constant ($\,e^{(a-1)}\,$ , if I remember correctly), yet I can't see the byperbolic cosine appearing afterwards... Nov 21, 2012 at 17:34
• @DonAntonio: At the end we should get $\displaystyle P.V. \int_{-\infty}^\infty \frac {e^{az}}{\cosh(z)} dz=\frac 12\int_0^\infty \frac{\cosh(az)}{\cosh(z)}dz$ (after separing properly the negative and positive part from $0$ to $R$ changing variable for the negative part and taking the limit as a principal value). Nov 21, 2012 at 17:40

This doesn't use residues until we use $$\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}=\pi\csc(\pi z)$$ which can be proven using residues.

We just expand things in powers of $e^x$: \begin{align} &\int_0^\infty\frac{\cosh(ax)}{\cosh(x)}\,\mathrm{d}x\\ &=\int_0^\infty e^{(a-1)x}\frac{1+e^{-2ax}}{1+e^{-2x}}\,\mathrm{d}x\\ &=\int_0^\infty\left(e^{(a-1)x}-e^{(a-3)x}+e^{(a-5)x}-\dots\right)\,\mathrm{d}x\\ &+\int_0^\infty\left(e^{(-a-1)x}-e^{(-a-3)x}+e^{(-a-5)x}-\dots\right)\,\mathrm{d}x\\ &=\frac1{1-a}-\frac1{3-a}+\frac1{5-a}-\dots\\ &+\frac1{1+a}-\frac1{3+a}+\frac1{5+a}-\dots\\ &=\frac1{a+1}-\frac1{a+3}+\frac1{a+5}-\dots\\ &-\frac1{a-1}+\frac1{a-3}-\frac1{a-5}-\dots\\ &=\frac12\left(\dots+\frac1{\frac{a+1}2-2}-\frac1{\frac{a+1}2-1}+\frac1{\frac{a+1}2}-\frac1{\frac{a+1}2+1}+\frac1{\frac{a+1}2+2}-\dots\right)\\ &=\frac12\pi\csc\left(\pi\frac{a+1}2\right)\\ &=\frac\pi2\sec\left(\frac\pi2a\right) \end{align}

Denoting the desired integral by $$I=\int_0^{\infty}\!\!\mathrm{d}x\,\frac{\cosh(ax)}{\cosh x},$$ we may extend the integral to $-\infty<x<\infty$, since the integrand is a symmetric function, i.e. \begin{align*}I&=\frac{1}{2}\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\frac{\cosh(ax)}{\cosh x}\\&=\frac{1}{4}\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\left(e^{ax}+e^{-ax}\right)\,\mathrm{sech}\,x\\&=\frac{f(a)+f(-a)}{4},\qquad(A)\end{align*} where we have defined $$\displaystyle f(z):=\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,e^{zx}\,\mathrm{sech}\,x.$$ Note, as $x\to\infty$, the hyperbolic secant behaves as $e^{-x}$, hence the integrand $\sim e^{-(1-z)x}$. The integral converges iff $z<1$. Likewise, as $x\to-\infty$, the integrand goes like $e^{(z+1)x}$, hence convergence of the integral requires $z>-1$. Thus $f(z)$ is well-defined for all $|z|<1$, which is also the permitted range for the parameter $a$. That said, let’s return to the evaluation of $f(z)$. $$f(z)=2\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\frac{e^{zx}}{e^x+e^{-x}}=2\int_{-\infty}^{\infty}\!\!\mathrm{d}x\,\frac{e^{(z-1)x}}{1+e^{-2x}}.$$ Substituting $$\displaystyle 1+e^{-2x}=\frac{1}{u}\Rightarrow\mathrm{d}x=\frac{\mathrm{d}u}{2u^2}e^{2x}=\frac{\mathrm{d}u}{2u(1-u)},$$ and changing the limits of integration appropriately, we arrive at \begin{align*}f(z)&=\int_0^1\!\!\mathrm{d}u\,\frac{u^{\frac{z-1}{2}}}{(1-u)^{\frac{z+1}{2}}}\\&=\int_0^1\!\!\mathrm{d}u~u^{\frac{1+z}{2}-1}(1-u)^{\frac{1-z}{2}-1}\\&=\mathrm{B}\!\left(\frac{1+z}{2},\frac{1-z}{2}\right),\end{align*} where $\mathrm{B}(x,y)$ is the beta function (https://en.wikipedia.org/wiki/Beta_function). Expressing the beta function in terms of the gamma function (https://en.wikipedia.org/wiki/Gamma_function), via $$\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},$$ and noting that $\Gamma(1)=1$, we can further simplify our result: \begin{align*}f(z)&=\Gamma\left(\frac{1+z}{2}\right)\Gamma\left(\frac{1-z}{2}\right)\\&=\Gamma\left(\frac{1+z}{2}\right)\Gamma\left(1-\frac{1+z}{2}\right)\\&=\frac{\pi}{\sin\left(\frac{\pi}{2}+\frac{\pi z}{2}\right)}=\frac{\pi}{\cos\left(\frac{\pi z}{2}\right)}.\end{align*} In the third step above we used Euler’s well know reflection formula (https://en.wikipedia.org/wiki/Reflection_formula) $$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}.$$ Having evaluated $f(z)$ in closed form, we are merely a substitution away from the desired result (cf Eq. (A)). Thus, $$\boxed{\int_0^{\infty}\!\!\mathrm{d}x\,\frac{\cosh(ax)}{\cosh x}=\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}}$$

Cheers!

As the original query that asked to use residues has not been answered completely I will contribute some ideas.

Suppose $a$ is a rational number $p/q$ where $p<q$ and $p-q$ is odd. Use a rectangular contour that consists of four segments: $\Gamma_0$ along the real axis from $-R$ to $R$, $\Gamma_1$ parallel to the imaginary axis to $R + \pi i q$, $\Gamma_2$ parallel to the real axis but in the opposite direction to $-R + \pi i q$ and finally, $\Gamma_3$ parallel to the imaginary axis to $-R$ on the real axis.

Now set $$f(z) = \frac{e^{az}+e^{-az}}{e^z+e^{-z}}$$ so that we are looking for $$\frac{1}{2} \int_{-\infty}^\infty f(z) dz$$ and integrate $f(z)$ along $\Gamma_0 - \Gamma_1 - \Gamma_2 - \Gamma_3$. Examine each segment in turn as $R$ goes to infinity. Clearly the integral along $\Gamma_0$ is simply the integral we are looking for. The contributions of $\Gamma_1$ and $\Gamma_3$ vanish in the limit. Along $\Gamma_2$ we have $x= t + \pi i q$, getting $$\int_\infty^{-\infty} \frac{e^{\frac{p}{q}t + \pi i p}+e^{-\frac{p}{q}t - \pi i p}}{e^{t+ \pi i q}+e^{-t- \pi i q}} dt = - (-1)^{p-q} \int_{-\infty}^\infty \frac{e^{\frac{p}{q}t}+e^{-\frac{p}{q}t}}{e^{t}+e^{-t}} dt =\int_{-\infty}^\infty \frac{e^{\frac{p}{q}t}+e^{-\frac{p}{q}t}}{e^{t}+e^{-t}} dt.$$ The last equality is because $p-q$ is odd.

To conclude we need to compute the poles and residues inside our contour. The poles are at $$\rho_k = \frac{1}{2}\pi i + \pi i k$$ and the residues are $$\lim_{z\to \rho_k} \frac{(z-\rho_k) (e^{az} + e^{-az})}{e^z + e^{-z}} = \lim_{z\to \rho_k} \frac{(z-\rho_k) (a e^{az} -a e^{-az}) + e^{az} + e^{-az}}{e^z - e^{-z}}.$$ But $$\lim_{z\to \rho_k} \frac{1}{e^z - e^{-z}} = \frac{1}{i e^{\pi i k} - (-i) e^{-\pi i k}} = \frac{1}{i e^{\pi i k} + i e^{-\pi i k}} = \frac{e^{\pi i k}}{i (1+1)} = \frac{(-1)^k}{2i}$$ so that finally $$\operatorname{Res}_{z=\rho_k} f(z) = \frac{(-1)^k}{2i} \left( e^{a\rho_k} + e^{-a\rho_k}\right).$$ With $J$ being the integral we are looking for and $I$ the integral along $\Gamma_0$ we have $$J = \frac{1}{2} I = \frac{1}{4} 2 I = \frac{1}{4} 2\pi i \sum_{k=0}^{q-1} \operatorname{Res}_{z=\rho_k} f(z)$$ The conclusion is that $$J = \frac{1}{2} \pi i \sum_{k=0}^{q-1} \operatorname{Res}_{z=\rho_k} f(z) = \frac{\pi}{4} \sum_{k=0}^{q-1} (-1)^k \left( e^{a\rho_k} + e^{-a\rho_k}\right) = \frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k).$$ where we have used the fact that $1/2 + k < q$ implies that $k$ runs up to $q-1.$

Edit. Use the following bound to see that the integral along $\Gamma_1$ vanishes (set $z= R + it$ with $0\le t \le \pi q$): $$\left| \int_{\Gamma_1} f(z) dz \right| = \left| \int_0^{\pi q} \frac{e^{aR + ait} + e^{-aR -ait}}{e^{R+it} + e^{-R-it}} i dt \right| \le \int_0^{\pi q} \frac{e^{aR} + e^{-aR}}{e^{R} - e^{-R}} dt = \pi q e^{-(1-a) R} \frac{1-e^{-2aR}}{1-e^{-2R}}$$ Now certainly we have $$\lim_{R\to\infty}\frac{1-e^{-2aR}}{1-e^{-2R}} = 1$$ so that the integral is $\theta(e^{-(1-a) R})$ which goes to zero as $R$ goes to infinity. The integral along $\Gamma_3$ is done the same way.

There is some additional simplification that can be done which I'll do in a new answer because my browser is not coping well with those large formulas where speed is concerned.

We have $$\frac{\pi}{2} \sum_{k=0}^{q-1} (-1)^k \cosh(a\rho_k) = \frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{a\rho_k} + \frac{\pi}{4} \sum_{k=0}^{q-1}(-1)^k e^{-a\rho_k}$$

The first sum is $$\sum_{k=0}^{q-1} (-1)^k e^{a\rho_k} = \sum_{k=0}^{q-1} (-1)^k e^{a i \pi /2} e^{a \pi i k} = e^{a i \pi/2} \frac{1-(-e^{a \pi i})^q}{1 + e^{a \pi i}}$$ which is $$e^{a i \pi/2} \frac{1-(-1)^q e^{p \pi i}}{1 + e^{a \pi i}} = e^{a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{a \pi i}} = e^{a i \pi/2} \frac{2}{1 + e^{a \pi i}} = \frac{1}{\cos (a\pi/2)}$$

The second sum is $$\sum_{k=0}^{q-1} (-1)^k e^{-a\rho_k} = \sum_{k=0}^{q-1} (-1)^k e^{-a i \pi /2} e^{-a \pi i k} = e^{-a i \pi/2} \frac{1-(-e^{-a \pi i})^q}{1 + e^{-a \pi i}}$$ which is $$e^{-a i \pi/2} \frac{1-(-1)^q e^{-p \pi i}}{1 + e^{-a \pi i}} = e^{-a i \pi/2} \frac{1-(-1)^{p+q}}{1 + e^{-a \pi i}} = e^{-a i \pi/2} \frac{2}{1 + e^{-a \pi i}} = \frac{1}{\cos (a\pi/2)}$$

It follows that the original sum and the integral is $$J = \frac{\pi}{2} \frac{1}{\cos(a \pi/2)}.$$