Prove that $$\int_0^{\infty} \frac{\sin(2013 x)}{x(\cos x+\cosh x)}dx=\frac{\pi}{4}$$
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- The integrand is even. 2. Use Parseval's Theorem.
– Ron Gordon Feb 01 '13 at 21:12 -
@rlgordonma: good point. I'll check this way! Thanks! – user 1591719 Feb 01 '13 at 21:14
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@rlgordonma Please enlighten me: I only know Parseval for, say, $2\pi$ periodic functions and their integral over $[-\pi,\pi]$. – Julien Feb 01 '13 at 21:20
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@julien: it is also applied to functions defined on $L^2(-\infty,\infty)$. – Ron Gordon Feb 01 '13 at 21:24
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@rlgordonma Do you have a reference? The only analogue of Parseval I know of on $L^2(\mathbb{R})$ is Plancherel. – Julien Feb 01 '13 at 21:27
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@julien: here's an example of using Parseval in evaluating a tough integral: http://math.stackexchange.com/questions/288049/how-to-integrate-int-0-infty-frac-sin-x-cosh-x-cos-x-cdot-frac/288095#288095 – Ron Gordon Feb 01 '13 at 21:29
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2And 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 01 '13 at 21:29
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2@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 01 '13 at 21:33
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1@Chris'ssister: Nice question, as always. +1 – Mikasa Feb 07 '13 at 11:35
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@JoelReyesNoche: It's not a contest question. – user 1591719 Feb 22 '13 at 15:34
3 Answers
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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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1Yeah, it seems that the complex analysis makes miracles. Very nice. (+1) – user 1591719 Feb 25 '13 at 20:44
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@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
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@Chris'ssisterandpals: Not to be picky, but my name is Rob :-) thanks for the bounty, arrrgh! – robjohn Feb 25 '13 at 21:13
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@Chris'ssisterandpals: I am trying this integral in Mathematica, and it is taking a long time. – robjohn Feb 25 '13 at 21:35
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yeah. I tried to compute it with Mathematica, but I got no result. – user 1591719 Feb 25 '13 at 21:37
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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 $m=2013$
Martin Gales
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How do you prove that $\sum_{k=1}^\infty(-1)^{k-1}=\frac12$ although the partial sum $s_n:=\sum_{k=1}^n(-1)^{k-1}=(1-(-1)^n)/2$ does not converge? – HorizonsMaths Feb 18 '13 at 00:33
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2@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 06:14
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@MartinGales I still don't see why the sum should be $1/2$. Are you using some "advanced calculus" to prove it? All I can see is that $s_{2n}=0$ and $s_{2n+1}=1$, and with my current knowledge of calculus, the latter tells me that this sum does not exist. – HorizonsMaths Feb 21 '13 at 08:53
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@MartinGales: an infinite alternating sum like that is a red flag that something has gone wrong in a problem like this. Have another look, especially why you can only work the case $m=1$. The result is clearly independent of $m$ - that's how all of these problems with the year stuck in as a parameter work. – Ron Gordon Feb 22 '13 at 12:44
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1@rlgordonma How do you show that the integral is independent of $m$? – Martin Gales Feb 23 '13 at 08:37
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2@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. – HorizonsMaths Feb 23 '13 at 18:58
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@MartinGales: it is a lot harder to prove than I thought, but I do believe it. Sort of like a religion at this point until I solve the problem, or see it solve to my satisfaction. BTW I got the same sum representation as you. Although your result is right, you do need to argue away from that horrendous alternating sum. – Ron Gordon Feb 23 '13 at 21:31
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1@rlgordonma I(0)=0& I(-x)=-I(x). So, the integral is not as independent of parameter as hoped. – Ishan Banerjee Feb 24 '13 at 13:43
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2@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
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1@Mercy I repeat what I have said above: $\sum_{k=1}^\infty(-1)^{k-1}$ does not converge to $\frac{1}{2}$, it is exactly equal to $\frac{1}{2}$. Like 1+1 does not converge to 2, it is equal to 2. – Martin Gales Feb 25 '13 at 12:00
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Yup the limit as k goes to infinity of (-1)^(k-1) absolutely does no convege. It oscillates between 1 and -1. – EhBabay Feb 25 '13 at 20:04
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1@MartinGales: for a series to converge, its terms must tend to $0$. Cesàro summation is a method to evaluate series which do not converge. From the Wikipedia article: "The significance of Cesàro summation is that a series which does not converge may still have a well-defined Cesàro sum." – robjohn Feb 25 '13 at 20:46
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@MartinGales: The integral is not independent of $m$. If $m$ is odd, the integral is $\frac\pi4$. If $m$ is even, this is not so. – robjohn Feb 26 '13 at 02:44
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\sin\pars{2013 x}\,\dd x \over x\bracks{\cos\pars{x} + \cosh\pars{ x}}}={\pi \over 4}:\ {\large ?}}$
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}
Felix Marin
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