3
$\begingroup$

Show that for $-1<p<1$, $$ \int_0^{\infty} \frac{\cos(px)}{\cosh x} \,dx = \frac{\pi}{2\cosh(p\pi/2)}.$$

I make L.S to $$\frac{e^{ipx}+e^{-ipx}}{e^{x}+e^{-x}}$$ and I cannot approach next step.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ You have a typo in the denominator. Cosh has real exponents. $\endgroup$ – Cye Waldman Apr 4 '17 at 18:33
  • $\begingroup$ oh it was mis-writing, but i still do not know next step. $\endgroup$ – Derite Apr 4 '17 at 18:47
2
$\begingroup$

Consider

$$f(z) = \frac{e^{i pz}}{\sinh(z)}$$

If we integrate around a contour of height $\pi $ and stretch it to infinity we get

enter image description here


By taking $ T \to \infty $

$$\color{red}{\int^{i\pi/2+\infty}_{-i\pi/2+\infty}f(x)\,dx}+\color{blue}{\int^{i\pi/2-\infty}_{i\pi/2+\infty}f(x)\,dx}+\\\color{red}{\int^{-i\pi/2-\infty}_{i\pi/2-\infty}f(x)\,dx}+ \color{blue}{\int^{-i\pi/2+\infty}_{-i\pi/2-\infty}f(x)\,dx} = 2\pi i \mathrm{Res}(f,0)$$


Consider

$$\color{blue}{\int^{-i\pi/2+\infty}_{-i\pi/2-\infty}\frac{e^{ipx}}{\sinh(x)}\,dx}$$

Let $x = -\pi i/2+y $

$$ i e^{\frac{\pi p}{2}}\int^{\infty}_{-\infty}\frac{e^{ipy}}{ \cosh(y)}\,dy$$


Similarly we have for

$$\color{blue}{\int^{i\pi/2-\infty}_{i\pi/2+\infty}\frac{e^{ipx}}{\sinh(x)}\,dx}$$

By letting $ x =\pi i/2+ y $

$$ i e^{-\frac{\pi p}{2}}\int^{\infty}_{-\infty}\frac{e^{ipy}}{ \cosh(y)}\,dy$$


The other integrals go to 0 hence

$$i \left(e^{\frac{\pi p}{2}}+e^{- \frac{\pi p}{2}} \right)\int^{\infty}_{-\infty}\frac{e^{ipy}}{ \cosh(y)}\,dy =2\pi i \mathrm{Res}(f,0)$$


Calculating the residue we have

$$\mathrm{Res}(f,0) = \lim_{z \to 0} z\frac{e^{ipz}}{\sinh(z)} = \lim_{z \to 0}\frac{e^{ipz}}{\cosh(z)} = 1$$


Using that we get

$$\int^{\infty}_{-\infty}\frac{e^{ipy}}{ \cosh(y)}\,dy =\frac{2\pi}{e^{\frac{\pi p}{2}}+e^{- \frac{\pi p}{2}}} $$

By taking the real part

$$\boxed{\int^{\infty}_{0}\frac{\cos(py)}{ \cosh(y)}\,dy =\frac{\pi}{2} \, \mathrm{sech}\left( \frac{\pi}{2} p\right)}$$

| cite | improve this answer | | | | |
$\endgroup$
1
$\begingroup$

I thought it might be instructive to present a purely real analysis approach.

Note that we can write

$$\begin{align} \int_0^\infty \frac{\cos(px)}{\cosh(x)}\,dx&=2\int_0^\infty \frac{e^{-x}\cos(px)}{1+e^{-2x}}\,dx\tag 1\\\\ &=2\sum_{n=0}^\infty (-1)^n\int_0^\infty e^{-(2n+1)x}\cos(px)\,dx\tag 2\\\\ &=2\sum_{n=0}^\infty (-1)^n\frac{2n+1}{(2n+1)^2+p^2} \tag 3\\\\ &=\frac{\pi}{2}\text{sech}\left(\frac{\pi p}{2}\right)\tag 4 \end{align}$$


NOTES:

In going from $(1)$ to $(2)$, we expanded $\frac{1}{1+e^{-2x}}$ in a geometric series and interchanged the order of the series and integral. Uniform convergence guarantees the legitimacy of interchanging the order of operations.

In going from $(2)$ to $(3)$, we carried out the integral.

In going from $(3)$ to $(4)$, we made use of the analysis in THIS ANSWER, in which I developed the partial fraction representation of $\sec(\pi a/2)$.

Then, we simply noted that for $a=ip$ we obtain

$$2\sum_{n=0}^\infty(-1)^n \frac{2n+1}{(2n+1)^2+p^2}=\frac{\pi}{2}\sec(i\pi p/2)=\frac{\pi}{2}\text{sech}(\pi p/2)$$

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

Try changing limits of the function according to value of p for cosh and cos and substitute.

| 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.