Prove that $\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$

Prove that $$\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$$

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1. The integrand is even. 2. Use Parseval's Theorem. –  Ron Gordon Feb 1 at 21:12
And yes, it is more properly called Plancherel's Theorem. But many people (like my Optics colleagues) just call it Parseval's theorem all the same. –  Ron Gordon Feb 1 at 21:29
@rlgordonma Thank you for the link. Ah! Plancherel, this I know. I see, Wikipedia confirms that it is often called Parseval in other domains of science and engineering fields. –  julien Feb 1 at 21:33
@Chris'ssister: Nice question, as always. +1 –  B.S. Feb 7 at 11:35
Why the coefficient of 2013? Is this a contest question? –  Joel Reyes Noche Feb 22 at 12:53

Because the integrand is even, contour integration yields \begin{align} &\int_0^\infty\frac{\sin(mx)}{x(\cos(x)+\cosh(x))}\mathrm{d}x\\ &=\frac12\int_{-\infty}^\infty\frac{\sin(mx)}{x(\cos(x)+\cosh(x))}\mathrm{d}x\\ &=\frac12\int_{-\infty-i}^{\infty-i}\frac{\sin(mx)}{x(\cos(x)+\cosh(x))}\mathrm{d}x\\ &=\frac1{4i}\int_{\gamma^+}\frac{e^{imz}}{z(\cos(z)+\cosh(z))}\mathrm{d}z -\frac1{4i}\int_{\gamma^-}\frac{e^{-imz}}{z(\cos(z)+\cosh(z))}\mathrm{d}z\\ &=\frac{2\pi i}{4i}\frac12 +2\frac{2\pi i}{4i}\sum_{k=0}^\infty(-1)^{k+1}\frac{\mathrm{sech}\left(\frac{2k+1}{2}\pi\right)}{\frac{2k+1}{2}\pi}\frac{\cos\left(m\frac{2k+1}{2}\pi\right)}{\exp\left(m\frac{2k+1}{2}\pi\right)}\\ &=\frac\pi4+\pi\sum_{k=0}^\infty(-1)^{k+1}\frac{\mathrm{sech}\left(\frac{2k+1}{2}\pi\right)}{\frac{2k+1}{2}\pi}\frac{\cos\left(m\frac{2k+1}{2}\pi\right)}{\exp\left(m\frac{2k+1}{2}\pi\right)}\\ &=\frac\pi4\qquad\text{for odd m} \end{align} where $\gamma^+$ goes from $-R-i$ to $+R-i$ then circles counterclockwise back in the upper half plane along $|z+i|=R$ and $\gamma^+$ goes from $-R-i$ to $+R-i$ then circles clockwise back in the lower half plane along $|z+i|=R$.

The residue of $f(z)=\dfrac{e^{imz}}{z(\cos(z)+\cosh(z))}$ at $z=0$ is $\frac12$.

Let $\alpha=\frac{1+i}{2}$. As noted in this answer, $f$ has singularities at $\pm(2k+1)\pi\alpha$ and $\pm(2k+1)\pi\overline{\alpha}$.

The sum of the residues in the upper half plane at $(2k+1)\pi\alpha$ and $-(2k+1)\pi\overline{\alpha}$ is the same as the sum of the residues in the lower half plane at $(2k+1)\pi\overline{\alpha}$ and $-(2k+1)\pi\alpha$. Both are equal to $$(-1)^{k+1}\frac{\mathrm{sech}\left(\frac{2k+1}{2}\pi\right)}{\frac{2k+1}{2}\pi}\frac{\cos\left(m\frac{2k+1}{2}\pi\right)}{\exp\left(m\frac{2k+1}{2}\pi\right)}$$ Note that for odd $m$, these are all $0$; $x=m\frac{2k+1}{2}\pi\equiv\frac\pi2\pmod{\pi}\Rightarrow\cos(x)=0$. This is not so for even $m$. Thus, $$\int_0^\infty\frac{\sin(2013x)}{x(\cos(x)+\cosh(x))}\mathrm{d}x=\frac\pi4$$

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Yeah, it seems that the complex analysis makes miracles. Very nice. (+1) –  Chris's sis Feb 25 at 20:44
@Chris'ssisterandpals: I first looked to see if the function was even so that I could extend the domain of integration to $(-\infty,+\infty)$ in order to use contour integration. –  robjohn Feb 25 at 21:05
@Chris'ssisterandpals: Not to be picky, but my name is Rob :-) thanks for the bounty, arrrgh! –  robjohn Feb 25 at 21:13
OK! Welcome. :-) –  Chris's sis Feb 25 at 21:24
@Chris'ssisterandpals: I am trying this integral in Mathematica, and it is taking a long time. –  robjohn Feb 25 at 21:35

Consider a general case:

$$I(m)=\int_0^{\infty} \frac{\sin(m x)}{x(\cos x+\cosh x)}dx$$ where $m$ is a positive integer.

At first, let's try with m=1

$$I(1)=\int_0^{\infty} \frac{\sin x}{x(\cos x+\cosh x)}dx$$ As a first step we note that

$$\frac{\sin x}{\cos x+\cosh x}=-2\sum_{k=1}^{\infty}(-1)^ke^{-xk}\sin(xk)$$ This relationship can be derived from the following geometric series sum:
$$\sum_{k=1}^{\infty}(-1)^ke^{-xk}e^{ixk}=-\frac{e^{-x}e^{ix}}{1+e^{-x}e^{ix}}$$ Imaginary part of this gives the relationship above. Thus, the integral:

$$I(1)=-2\sum_{k=1}^{\infty}(-1)^k\int_{0}^{\infty}e^{-xk}\sin(xk)\frac{dx}{x}$$ But

$$\int_{0}^{\infty}e^{-xk}\sin(xk)\frac{dx}{x}=\frac{\pi}{4}$$ This result follows from the well known integral

$$\int_{0}^{\infty}e^{-xa}\sin(xk)dx=\frac{k}{a^2+k^2}$$ if we integrate the last with respect to $a$ from $a=k$ to $a=\infty$. We get:

$$I(1)=-2\sum_{k=1}^{\infty}(-1)^k\frac{\pi}{4}=\frac{\pi}{2}\sum_{k=1}^{\infty}(-1)^{k-1}$$ The sum

$$\sum_{k=1}^{\infty}(-1)^{k-1}=1-1+1-1+...=\frac{1}{2}$$ was already known in the times of Leibniz.

Finally:

$$I(1)=\frac{\pi}{4}$$ But even in the next case of $m=2$ i have no clue, how to evaluate the integral. Numerical calculations suggest that the result holds for every $m$ including $m=2013$

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@Mercy $\lim_{x\to1-0}(1-x+x^2-x^3+...)=\lim_{x\to1-0}\frac{1}{1+x}=\frac{1}{2}$. I am not saying that the sum converges to $\frac{1}{2}$. I am saying that the sum is equal to $\frac{1}{2}$. –  Martin Gales Feb 21 at 6:14
@Mercy Look at Cesàro summation on Wikipedia. –  Martin Gales Feb 23 at 8:35
@MartinGales I think you are mixing things up, it is clear that the series $\sum_{k=1}^\infty(-1)^{k-1}$ does not converge, and I'm surprised you still believe it does. In fact, a necessary condition for the convergence of a series $\sum_{k=1}^\infty a_k$ is that $a_k \to 0$ which obviously is not satisfied. –  Mercy Feb 23 at 18:58
@rlgordonma I(0)=0& I(-x)=-I(x). So, the integral is not as independent of parameter as hoped. –  Ishan Banerjee Feb 24 at 13:43
@MartinGales I'm affraid I have to agree with Mercy, I don't see why the series $\sum_{k=1}^\infty(-1)^{k-1}$ should be convergent since the limit $\lim_{k \to \infty}(-1)^{k-1}$ is not zero. –  albmiz-mth Feb 24 at 22:37