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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 '13 at 21:12
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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 '13 at 21:29
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@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. –  1015 Feb 1 '13 at 21:33
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@Chris'ssister: Nice question, as always. +1 –  Babak S. Feb 7 '13 at 11:35
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Why the coefficient of 2013? Is this a contest question? –  Joel Reyes Noche Feb 22 '13 at 12:53

3 Answers 3

up vote 16 down vote accepted
+100

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 '13 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 '13 at 21:05
    
@Chris'ssisterandpals: Not to be picky, but my name is Rob :-) thanks for the bounty, arrrgh! –  robjohn Feb 25 '13 at 21:13
    
OK! Welcome. :-) –  Chris's sis Feb 25 '13 at 21:24
    
@Chris'ssisterandpals: I am trying this integral in Mathematica, and it is taking a long time. –  robjohn Feb 25 '13 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 $\frac{1}{1+x}=1-x+x^2-x^3+...$ –  Martin Gales Feb 20 '13 at 15:31
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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 '13 at 6:14
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@Mercy Look at Cesàro summation on Wikipedia. –  Martin Gales Feb 23 '13 at 8:35
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@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 '13 at 18:58
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@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 '13 at 22:37

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\sin\pars{2013 x}\,\dd x \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}={\pi \over 4}:\ {\large ?}}$

Note that \begin{align}&\color{#c00000}{\cos\pars{x} + \cosh\pars{x}} =\cos\pars{x} + \cos\pars{x\ic} \\[3mm]&=\cos\pars{{x + x\ic \over 2} + {x - x\ic \over 2}} +\cos\pars{{x + x\ic \over 2} - {x - x\ic \over 2}} =2\cos\pars{x + x\ic \over 2}\cos\pars{x - x\ic \over 2} \\[3mm]&=2\cos\pars{\mu x}\cos\pars{\mu^{*}x} \qquad\mbox{where}\qquad\mu\equiv{1 + \ic \over 2} = \expo{\pi\ic/4} \end{align}

Then, \begin{align}&\color{#c00000}{% \int_{0}^{\infty}{\sin\pars{m x}\,\dd x \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}} ={1 \over 4}\int_{-\infty}^{\infty} {\sin\pars{mx} \over x\cos\pars{\mu x}\cos\pars{\mu^{*}x}}\,\dd x\end{align} With the identity $\ds{{\sin\pars{\xi} \over \xi} = \int_{0}^{1}\cos\pars{k\xi}\,\dd k =\Re\int_{0}^{1}\expo{\ic\verts{k}\xi}\,\dd k}$ we'll have: $$ \color{#c00000}{% \int_{0}^{\infty}{\sin\pars{m x}\,\dd x \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}} ={1 \over 4}\,m\,\Re\int_{0}^{1}\color{#00f}{\int_{-\infty}^{\infty} {\expo{\ic\verts{mk}x}\,\dd x \over \cos\pars{\mu x}\cos\pars{\mu^{*}x}}}\,\dd k \tag{1} $$

Zeros of $\ds{\cos\pars{\mu x}}$ are given by $\ds{\pars{~\mbox{with}\ n \in {\mathbb Z}~}}$: $$ x_{n} = \pars{n + \half}\,{\pi \over \mu} =\pars{2n + 1}\,\pars{1 - \ic}\,{\pi \over 2}\,,\qquad \Im\pars{x_{n}} > 0\quad\mbox{if}\quad n=-1,-2,-3,\ldots $$ Similarly, zeros of $\ds{\cos\pars{\mu^{*}x}}$ are given by $\ds{\pars{~\mbox{with}\ n \in {\mathbb Z}~}}$: $$ x_{n}^{*} = \pars{n + \half}\,{\pi \over \mu^{*}} =\pars{2n + 1}\,\pars{1 + \ic}\,{\pi \over 2}\,,\qquad \Im\pars{x_{n}^{*}} > 0\quad\mbox{if}\quad n=0,1,2,\ldots $$ $\large\begin{array}{|c|}\hline \mbox{Note that}\quad \ds{x_{-n - 1} = -x_{n}}\\ \hline\end{array}$.

Then, \begin{align}&\color{#00f}{\int_{-\infty}^{\infty} {\expo{\ic\verts{mk}x}\,\dd x \over \cos\pars{\mu x}\cos\pars{\mu^{*}x}}} \\[3mm]&=2\pi\ic\bracks{% \sum_{n = -\infty}^{-1} {\expo{\ic\verts{mk}x_{n}} \over -\mu\sin\pars{\mu x_{n}}\cos\pars{\mu^{*}x_{n}}} + \sum_{n = 0}^{\infty} {\expo{\ic\verts{mk}x_{n}*}\over \cos\pars{\mu x_{n}^{*}}\pars{-\mu*}\sin\pars{\mu^{*}x_{n}^{*}}}} \\[3mm]&=2\pi\ic\sum_{n = 0}^{\infty}\bracks{% {\expo{-\ic\verts{mk}x_{n}} \over \mu\sin\pars{\mu x_{n}}\cos\pars{\mu^{*}x_{n}}} - {\expo{\ic\verts{mk}x_{n}*}\over \mu^{*}\sin\pars{\mu^{*}x_{n}^{*}}\cos\pars{\mu x_{n}^{*}}}} \\[3mm]&=-4\pi\,\Im \sum_{n = 0}^{\infty}{\mu^{*} \over \verts{\mu}^{2}}\, {\expo{-\ic\verts{mk}x_{n}} \over \sin\pars{\mu x_{n}}\cos\pars{\mu^{*}x_{n}}} \\[3mm]&=-4\pi\,\Im \sum_{n = 0}^{\infty}{\pars{1 - \ic}/2 \over 1/2}\, {\expo{-\ic\verts{mk}x_{n}}\over \sin\pars{\bracks{2n + 1}\pi/2}\cos\pars{\ic\bracks{2n + 1}\pi/2}} \end{align}

$$\color{#00f}{\int_{-\infty}^{\infty} {\expo{\ic\verts{mk}x}\,\dd x \over \cos\pars{\mu x}\cos\pars{\mu^{*}x}}} =-4\pi\,\Im \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\, \pars{1 - \ic}\expo{-\ic\verts{mk}x_{n}} $$

We replace this result in expression $\pars{1}$: \begin{align}&\color{#c00000}{% \int_{0}^{\infty}{\sin\pars{m x}\,\dd x \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}} =-\pi\,m\,\Im \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\, \pars{1 - \ic}\int_{0}^{1}\expo{-\ic\verts{m}x_{n}k}\,\dd k \\[3mm]&=-\pi\,m\,\Im \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\, \pars{1 - \ic}\,{\expo{-\ic\verts{m}x_{n}} - 1 \over -\ic\verts{m}x_{n}} \\[3mm]&=-\pi \sgn\pars{m}\Im \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \cosh\pars{\bracks{2n + 1}\pi/2}}\, \,\ic\,{\expo{-\ic\verts{m}\pars{2n + 1}\pi/2} \expo{-\verts{m}\pars{2n + 1}\pi/2} - 1 \over \pars{2n + 1}\pi/2} \\[3mm]&=-2\sgn\pars{m}\sum_{n = 0}^{\infty}{\pars{-1}^{n}\over \pars{2n + 1}\cosh\pars{\bracks{2n + 1}\pi/2}}\times \\[3mm]& \phantom{=-2\sgn\pars{m}\sum_{n = 0}^{\infty}} \bracks{\cos\pars{\verts{m}\pars{2n + 1}\,{\pi \over 2}} \expo{-\verts{m}\pars{2n + 1}\pi/2} - 1} \end{align}

Note that $$ 2\sum_{n = 0}^{\infty}{\pars{-1}^{n}\over \pars{2n + 1}\cosh\pars{\bracks{2n + 1}\pi/2}} = {\pi \over 4} $$

\begin{align}&\color{#00f}{\large% \int_{0}^{\infty}{\sin\pars{m x}\,\dd x \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}}= \\[3mm]&\color{#c00000}{\left\lbrace\begin{array}{lcl} {\pi \over 4}\,\sgn\pars{m} & \mbox{if} & m\ \mbox{is odd} \\[3mm]\sgn\pars{m}\braces{% {\pi \over 4} -2\pars{-1}^{\verts{m}/2}\sum_{n = 0}^{\infty}\pars{-1}^{n}\, {\expo{-\verts{m}\pars{2n + 1}\pi/2}\over \pars{2n + 1}\cosh\pars{\bracks{2n + 1}\pi/2}}} & \mbox{if} & m\ \mbox{is even} \end{array}\right.} \end{align}

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