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I am working on the integral $$I=\int_{-\pi/4}^{\pi/4}\frac{x}{\sin x}\mathrm{d}x=2\int_0^{\pi/4}\frac{x}{\sin x}\mathrm{d}x$$ Which I am fairly confident has a closed form, as $$\int_{0}^{\pi/2}\frac{x}{\sin x}\mathrm{d}x=2G$$ Where $G$ is Catalan's constant.

Preforming a tangent half angle substitution, we have that $$I=4\int_0^{\sqrt{2}-1}\frac{\arctan x}{x}\mathrm{d}x$$ Then using $$\arctan x=\sum_{n\geq0}(-1)^n\frac{x^{2n+1}}{2n+1}$$ We have $$I=4\sum_{n\geq0}\frac{(-1)^n}{(2n+1)^2}(\sqrt{2}-1)^{2n+1}$$ Which is painfully similar to $G$. I do not know how to deal with that extra $(\sqrt{2}-1)^{2n+1}$ bit though...

In another post of mine I showed that $$I=\pi\sum_{n\geq1} n\log\bigg(\frac{4n+1}{4n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}$$ And similarly I showed that $$\sum_{n\geq1}n\log\bigg(\frac{2n+1}{2n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}=\frac{4G}\pi$$ So I have to questions. How do I find an exact value for $I$? And are the last two series representations correct? Thanks.

Major Edit:

Okay so I found a closed form for the integral. Wolfy gave me $$\int\frac{x}{\sin x}\mathrm{d}x=i\bigg(\text{Li}_2(-e^{ix})-\text{Li}_2(e^{ix})\bigg)+x\log\frac{1-e^{ix}}{1+e^{ix}}$$ I guess that Wolfy didn't want to do the algebra, so I did it by hand. It took me like $10$ minutes, but I am pretty sure that $$I=-\frac34G+\frac{\pi^2}4\bigg(\frac{13}{24}-i\bigg)-\frac{i\pi}4\log(1+\sqrt{2})+\frac{i-1}{32\sqrt{2}}\bigg[\psi^{(1)}\bigg(\frac{5}{8}\bigg)-\psi^{(1)}\bigg(\frac{1}{8}\bigg)\bigg]+\frac{i+1}{32\sqrt{2}}\bigg[\psi^{(1)}\bigg(\frac{3}{8}\bigg)-\psi^{(1)}\bigg(\frac{7}{8}\bigg)\bigg]$$ Where $\psi^{(1)}$ is the first derivative of the di-gamma function.

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    $\begingroup$ I'm not sure why you are confident that your integral has a closed form. Not that Wolfram Alpha is the last word on integration, but it is usually pretty good at recognizing constants, and in this case it does not. $\endgroup$ – DeficientMathDude Dec 21 '18 at 2:49
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    $\begingroup$ @DeficientMathDude - You just contradicted yourself. $\endgroup$ – user150203 Dec 21 '18 at 3:31
  • $\begingroup$ Wolframalpha numerical value for integral = 1.62706 and 2*G = 1.83193118 $\endgroup$ – user150203 Dec 21 '18 at 3:39
  • $\begingroup$ You can write $\frac{x}{\sin(x)}$ as $x\csc(x)$. See where I'm going with this? $\endgroup$ – Aniruddh Venkatesan Dec 21 '18 at 3:42
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    $\begingroup$ @DeficientMathDude see the edit I just made $\endgroup$ – clathratus Dec 21 '18 at 3:50
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Let $$I = \int_{-\pi/4}^{\pi/4} x \, \text{cosec} \, x \, dx = 2 \int_0^{\pi/4} x \, \text{cosec} \, x \, dx.$$ After integrating by parts we have \begin{align} I &= -\frac{\pi}{2} \ln (1 + \sqrt{2}) + 2 \int_0^{\pi/4} \ln (\text{cosec} \, x + \cot x) \, dx\\ &= -\frac{\pi}{2} \ln (1 + \sqrt{2}) + 2 \int_0^{\pi/4} \ln (1 + \cos x) \, dx - 2 \int_0^{\pi/4} \ln (\sin x) \, dx. \end{align}

The second of these integrals is perhaps (?) reasonably well known (for an evaluation, see here). The result is: $$\int_0^{\pi/4} \ln (\sin x) \, dx = -\frac{1}{2} \mathbf{G} - \frac{\pi}{4} \ln 2,$$ where $\mathbf{G}$ is Catalan's constant. Thus

$$I = -\frac{\pi}{2} \ln (1 + \sqrt{2}) + \mathbf{G} + \frac{\pi}{2} \ln 2 + 2I_1.$$

For the first of the integrals we will make use of the following formula, a proof of which can be found here $$\ln (1 + \cos x) = 2 \sum_{n = 1}^\infty (-1)^{n + 1} \frac{\cos (nx)}{n} - \ln 2.$$

So \begin{align} I_1 &= \int_0^{\pi/4} \ln (1 + \cos x) \, dx\\ &= 2 \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \int_0^{\pi/4} \cos (nx) \, dx - \ln 2 \int_0^{\pi/4} dx\\ &= -\frac{\pi}{4} \ln 2 + 2 \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n^2} \sin \left (\frac{n \pi}{4} \right )\\ &= -\frac{\pi}{4} \ln 2 + 2 \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{odd}}}^\infty \frac{(-1)^{n + 1}}{n^2} \sin \left (\frac{n \pi}{4} \right ) + + 2 \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{even}}}^\infty \frac{(-1)^{n + 1}}{n^2} \sin \left (\frac{n \pi}{4} \right )\\ &= -\frac{\pi}{4} \ln 2 + 2 \sum_{n = 1}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2n - 1) \right ] - \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2} \sin \left (\frac{n \pi}{2} \right )\\ &= -\frac{\pi}{4} \ln 2 + 2 S_1 - \frac{1}{2} S_2. \end{align}

For the second of these sums, \begin{align} S_2 &= \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{odd}}}^\infty \frac{1}{n^2} \sin \left (\frac{\pi n}{2} \right ) + \sum_{\stackrel{{\Large{n = 1}}}{n \in \text{even}}} \frac{1}{n^2} \sin \left (\frac{\pi n}{2} \right )\\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\ &= \mathbf{G}, \end{align} after a shift of the index $n \mapsto 2n + 1$ in the odd sum has been made while the even sum is identically equal to zero.

For the first of the sums, as it converges absolutely we can split it up as folows: $$S_1 = \sum_{\stackrel{{\Large{n = 1}}}{n \in 1,5,9,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ] + \sum_{\stackrel{{\Large{n = 1}}}{n \in 2,6,10,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ] + \sum_{\stackrel{{\Large{n = 1}}}{n \in 3,7,11,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ] + \sum_{\stackrel{{\Large{n = 1}}}{n \in 4,8,12,\ldots}}^\infty \frac{1}{(2n - 1)^2} \sin \left [\frac{\pi}{4} (2k - 1) \right ].$$

Shifting the indices as follows: $n \mapsto 4n - 3, n \mapsto 4n - 2, n \mapsto 4n - 1, n \mapsto 4n$ leads to \begin{align} S_1 &= \frac{1}{\sqrt{2}} \left [\sum_{n = 1}^\infty \frac{1}{(8n - 7)^2} + \sum_{n = 1}^\infty \frac{1}{(8n - 5)^2} - \sum_{n = 1}^\infty \frac{1}{(8n - 3)^2} - \sum_{n = 1}^\infty \frac{1}{(8n - 1)^2} \right ]\\ &= \frac{1}{\sqrt{2}} \left [\sum_{n = 0}^\infty \frac{1}{(8n + 1)^2} + \sum_{n = 0}^\infty \frac{1}{(8n + 3)^2} - \sum_{n = 0}^\infty \frac{1}{(8n + 5)^2} - \sum_{n = 0}^\infty \frac{1}{(8n + 7)^2} \right ]\\ &= \frac{1}{64 \sqrt{2}} \left [\sum_{n = 0}^\infty \frac{1}{(n + 1/8)^2} + \sum_{n = 0}^\infty \frac{1}{(n + 3/8)^2} - \sum_{n = 0}^\infty \frac{1}{(n + 5/8)^2} - \sum_{n = 0}^\infty \frac{1}{(n + 7/8)^2} \right ]\\ &= \frac{1}{64 \sqrt{2}} \left [\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right ], \end{align} where we have made use of the series representation for the polygamma function of order one (as known as the trigamma function). So the value for $I_1$ is: $$I_1 = -\frac{\pi}{4} \ln 2 - \frac{1}{2} \mathbf{G} + \frac{1}{32 \sqrt{2}} \left [\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right ],$$ leading to a final result of $$\int_{-\pi/4}^{\pi/4} x \, \text{cosec} \, x \, dx = -\frac{\pi}{2} \ln (1 + \sqrt{2}) + \frac{1}{16 \sqrt{2}} \left [\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right ].$$


Converting the trigamma functions to Clausen functions of order two

Inspired by the answer given by Zacky in terms of the Clausen function of order two, here I will show how to convert my answer in terms of the 4 trigamma functions into 2 Clausen functions of order 2.

The relation between the Clausen function of order two and the trigamma function is given by (a proof of this can be found here) $$\text{Cl}_2 \left (\frac{q \pi}{p} \right ) = \frac{1}{(2p)^{2m} (2m - 1)!} \sum_{n = 1}^p \sin \left (\frac{qn\pi}{p} \right ) \left [\psi^{(1)} \left (\frac{n}{2p} \right ) + (-1)^q \psi^{(1)} \left (\frac{n + p}{2p} \right ) \right ].$$ Setting $m = 1, q = 1, p = 4$ gives \begin{align} \text{Cl}_2 \left (\frac{\pi}{4} \right ) &= \frac{1}{64} \left [\frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{1}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) \right \} + \psi^{(1)} \left (\frac{1}{4} \right ) - \psi^{(1)} \left (\frac{3}{4} \right ) \right.\\ & \qquad \left. + \frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right \} \right ], \qquad (*) \end{align} and setting $m = 1, q = 3, p = 4$ gives \begin{align} \text{Cl}_2 \left (\frac{3\pi}{4} \right ) &= \frac{1}{64} \left [\frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{1}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) \right \} - \psi^{(1)} \left (\frac{1}{4} \right ) + \psi^{(1)} \left (\frac{3}{4} \right ) \right.\\ & \qquad \left. + \frac{1}{\sqrt{2}} \left \{\psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) \right \} \right ]. \qquad (**) \end{align} On adding ($*$) to ($**$) we see that $$\psi^{(1)} \left (\frac{1}{8} \right ) + \psi^{(1)} \left (\frac{3}{8} \right ) - \psi^{(1)} \left (\frac{5}{8} \right ) - \psi^{(1)} \left (\frac{7}{8} \right ) = 32 \sqrt{2} \left [\text{Cl}_2 \left (\frac{\pi}{4} \right ) + \text{Cl}_2 \left (\frac{3\pi}{4} \right ) \right ],$$ giving $$\int_{-\pi/4}^{\pi/4} \frac{x}{\sin x} \, dx = -\frac{\pi}{2} \ln (1 + \sqrt{2}) + 2 \, \text{Cl}_2 \left (\frac{\pi}{4} \right ) + 2 \, \text{Cl}_2 \left (\frac{3\pi}{4} \right ).$$

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  • $\begingroup$ Nice answer! We might be able to reduce those $4$ values of the trigamma function using: $$\psi_1\left(x\right)+\psi_1\left(x+\frac12\right)=4\psi_1\left(2x\right)$$ I mean: $$\psi_1\left(\frac18\right)+\left(\frac58\right) = 4\psi_1\left(\frac14\right)$$ And same with: $$\psi_1\left(\frac38\right)+\left(\frac78\right) = 4\psi_1\left(\frac34\right)$$ And the values of $\psi_1\left(\frac14\right)$ and $\psi_1\left(\frac34\right)$ are known from here: mathworld.wolfram.com/TrigammaFunction.html $\endgroup$ – Zacky Dec 21 '18 at 12:06
  • $\begingroup$ Perfect beautiful wonderful amazing thank you for a really great answer :) $\endgroup$ – clathratus Dec 21 '18 at 19:04
  • $\begingroup$ Also: does the closed form I found simplify to the answer you give? $\endgroup$ – clathratus Dec 21 '18 at 19:56
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For the purpose of an alternative method, mostly relying on Clausen function. We have: $$I=\int_{-\pi/4}^{\pi/4}\frac{x}{\sin x} dx=2\int_0^{\pi/4}x\left(\ln\left(\tan\frac{x}{2}\right)\right)' dx=2x\ln\left(\tan\frac{x}{2}\right)\bigg|_0^\frac{\pi}{4}-2\int_0^\frac{\pi}{4}\ln\left(\tan\frac{x}{2}\right)dx=$$ $$=\frac{\pi}{2}\ln(\sqrt 2-1)-2\int_0^\frac{\pi}{4}\left(\ln\left(2\sin\frac{x}{2}\right)-\ln\left(2\cos\frac{x}{2}\right)\right)dx=$$$$=\frac{\pi}{2}\ln(\sqrt 2-1)+2\text{Cl}_2\left(\frac{\pi}{4}\right)+2\text{Cl}_2\left(\frac{3\pi}{4}\right)$$ The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.

An interesting question might be: For what values of $\phi$ does the following integral have an elementary answer (Catalan's constant included)? $\ \displaystyle{I(\phi)=\int_0^\phi \frac{x}{\sin x}dx}$. So far I only know about $I\left(\frac{\pi}{6}\right)$.

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    $\begingroup$ I would be happy with the 2 order two Clausen functions as an answer. Far more compact compared to the answer I gave in terms of 4 trigamma functions. And besides, are the elementary functions such as $e^x$ and $\sin x$ not themselves just series in disguise? $\endgroup$ – omegadot Dec 21 '18 at 12:01
  • $\begingroup$ You are definetly right. In my mind, by elementary I reffer of something that is learnt in school (which maybe is not quite the best description of something elementary). Maybe Clausen and Trigamma function are studied somewhere, but I only heard of them from the internet. $\endgroup$ – Zacky Dec 21 '18 at 12:12
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    $\begingroup$ I love special functions. Thanks Zacky $\endgroup$ – clathratus Dec 21 '18 at 18:58
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    $\begingroup$ $I\left(\frac{\pi}{2}\right)=2G$ , $G$ is the Catalan constant. $\endgroup$ – FDP Dec 21 '18 at 20:48
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Too long for comments.

$$I=4\sum_{n\geq0}\frac{(-1)^n}{(2n+1)^2}(\sqrt{2}-1)^{2n+1}=4 \left(\sqrt{2}-1\right) \, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2};2 \sqrt{2}-3\right)$$

On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could the integral, I got $$I=-\frac{\pi}{4} \, \log \left(3+2 \sqrt{2}\right)+\frac 1 {16 \sqrt 2} \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi ^{(1)}\left(\frac{3}{8}\right)-\psi ^{(1)}\left(\frac{5}{8}\right)-\psi ^{(1)}\left(\frac{7}{8}\right) \right)$$

What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.

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  • $\begingroup$ Yes $\mathbf{G}$ does indeed appear twice before canceling out as can be seen in the answer I provide. $\endgroup$ – omegadot Dec 21 '18 at 10:58
  • $\begingroup$ @omegadot. Using another way, it did appear four times in complex expressions and ran away ! By the way, your answer is real nice ! $\to +1$. Cheers. $\endgroup$ – Claude Leibovici Dec 21 '18 at 11:22
  • $\begingroup$ So does the closed form I found simplify to this? $\endgroup$ – clathratus Dec 21 '18 at 19:02
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{-\pi/4}^{\pi/4}{x \over \sin\pars{x}}\,\dd x,\ \pars{~\mbox{OP already shows that}\ I = \bbox[10px,#ffd]{\!\!\!\!\! 4\!\int_{0}^{\root{2} - 1}\!\!{\arctan\pars{x} \over x}\,\dd x}~} \end{align}


Then, \begin{align} I & = \bbox[10px,#ffd]{4\int_{0}^{\root{2} - 1}{\arctan\pars{x} \over x}\,\dd x} = 4\,\Im\int_{0}^{\root{2} - 1}{\ln\pars{1 + \ic x} \over x}\,\dd x \\[5mm] & \stackrel{{\large x\ =\ \ic t} \atop {\large t\ =\ -\ic x}}{=}\,\,\, 4\,\Im\int_{0}^{-\pars{\root{2} - 1}\ic} {\ln\pars{1 - t} \over t}\,\dd t \\[5mm] & = -4\,\Im\int_{0}^{-\pars{\root{2} - 1}\ic} \mrm{Li}_{2}'\pars{t}\,\dd t = -4\,\Im\mrm{Li}_{2}\pars{-\bracks{\root{2} - 1}\ic} \\[5mm] & = \bbx{4\,\Im\mrm{Li}_{2}\pars{\bracks{\root{2} - 1}\ic}} \approx 1.6271 \end{align}

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  • $\begingroup$ Your final expression $\Im\mrm{Li}_2([\sqrt 2 - 1]i)$ is missing a factor of $-4$ infront otherwise it does not approx to $1.6271$. $\endgroup$ – mrtaurho Dec 21 '18 at 16:41
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    $\begingroup$ @mrtaurho Thanks. Indeed it's $\displaystyle\color{red}{+}4$ because I already switched the $\displaystyle\mathrm{Li}_{2} $ argument sign. Note that $\displaystyle\overline{\mathrm{Li}_{2}\left(\,{z}\,\right)} = \mathrm{Li}_{2}\left(\,{\overline{z}}\,\right)$. $\endgroup$ – Felix Marin Dec 21 '18 at 17:17
  • $\begingroup$ I appreciate the quick and easy approach. Thanks $\endgroup$ – clathratus Dec 21 '18 at 19:03
  • $\begingroup$ @clathratus You're welcome !!!. $\endgroup$ – Felix Marin Dec 22 '18 at 16:22
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I tried to solve the problem with the a minimum of previous knowledge, by just doing manipulations with series for some time, and then at a certain point, identify known functions. In comparison with the other solutions presented here my method seems to start one level lower, i.e. it is more elementary.

I have found that this method can be applied also to similar related problems discussed here, for instance to Integral $T_n=\int_{0}^{\pi/2}x^{n}\ln(1+\tan x)\,dx$.

Series expansion

The integral in question is

$$f = \int_{-\frac{\pi}{4}}^{+\frac{\pi}{4}} \frac{x}{\sin(x)}\,dx\tag{1}$$

The starting point (and the main idea) is an expansion of $\csc(x)$ into a geometric series, the rest is more or less a sequel of it.

$$\frac{x}{\sin(x)} = \frac{2 i x}{\left( e^{i x} -e^{-ix}\right)}= \frac{2 i x e^{-ix} }{1-e^{-2 i x}}=2 i x \sum_{k=0}^\infty e^{-(2k+1) i x }\tag{2}$$

The indefinite integral is then, interchanging the order of integral and sum

$$f_a = \int \frac{x}{\sin(x)}\,dx = \int 2 i x \sum_{k=0}^\infty e^{-(2k+1) i x }\,dx= 2 i \sum_{k=0}^\infty \int x e^{-(2k+1) i x }\,dx\\= 2 i \sum_{k=0}^\infty \left(e^{-i (2 k+1) x} \left(\frac{1}{(2 k+1)^2}+\frac{i x}{2 k+1}\right)\right)= A_a + B_a\tag{3}$$

Here the subscript $a$ indicates antiderivative.

From here we can proceed in two directions, stick to the series and try to simplify it and identify known function in the end, or alternatively, try to identify known functions early. Both approaches have their virtues and lead to insights to the structures involved.

Late identification

Turning to the definite integral, i.e. taking $a$ in the limits from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ we have

$$A = 2 i \sum_{k=0}^\infty \frac{1}{(2k+1)^2}\left( e^{-i (2 k+1) \frac{\pi}{4}} - e^{i (2 k+1) \frac{\pi}{4}} \right)= 4 \sum_{k=0}^\infty \frac{\sin(\frac{\pi}{4}(2k+1)}{(2k+1)^2}\tag{4}$$

Now the different values of the $\sin$ appearing here are these four

$$\{ \sin(\frac{\pi}{4}),\sin(\frac{3\pi}{4}),\sin(\frac{5\pi}{4}),\sin(\frac{7\pi}{4})\}= \frac{1}{\sqrt{2}}\{1,1,-1,-1 \} $$

with cyclic repetition.

Collecting terms with the same numerator in $A$ we have

$$A = \frac{4}{\sqrt{2}}\left(\frac{1}{1^2}+ \frac{1}{3^2} - \frac{1}{5^2} -\frac{1}{7^2}+ \frac{1}{9^2}+ \frac{1}{11^2} - \frac{1}{13^2} -\frac{1}{15^2}+ ... \right) \\= \frac{4}{\sqrt{2}}\left( \sum_{k=0}^\infty \frac{1}{(8k+1)^2}+\frac{1}{(8k+3)^2}-\frac{1}{(8k+5)^2}-\frac{1}{(8k+7)^2}\right) \\= \frac{1}{16\sqrt{2}}\left( \sum_{k=0}^\infty \frac{1}{(k+\frac{1}{8})^2}+\frac{1}{(k+\frac{3}{8})^2}-\frac{1}{(k+\frac{5}{8})^2}-\frac{1}{(k+\frac{7}{8})^2}\right) \tag{5a}$$

Proceeding similarly with $B$ we find

$$B = \pi \frac{1}{8 \sqrt{2}}\sum _{k=0}^{\infty } \left(\frac{1}{k+\frac{1}{8}}-\frac{1}{k+\frac{3}{8}}-\frac{1}{k+\frac{5}{8}}+\frac{1}{k+\frac{7}{8}}\right)\tag{5b}$$

The sums appearing are convergent and we could evaluate them numerically. But the conventional path is, of course, to identify them with known series.

All terms can be covered by this sum

$$\sum _{k=0}^{\infty } \frac{1}{(k+p)^s}=\psi ^{(s-1)}(p)\tag{6}$$

which can be expressend by the polygamma function.

This formula can be directly applied to $A$ giving

$$A = \frac{1}{64 \sqrt{2}}\left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi ^{(1)}\left(\frac{3}{8}\right)-\psi ^{(1)}\left(\frac{5}{8}\right)-\psi ^{(1)}\left(\frac{7}{8}\right)\right) \simeq 3.01152\tag{7a}$$

For $B$ there is the little difficulty that the series on the l.h.s. of $(7)$ is divergent just for the required $s=1$. But the formula gives the analytic continuation which remains finite.

Hence

$$B =\frac{\pi}{8 \sqrt{2}} \left(\psi ^{(0)}\left(\frac{1}{8}\right)-\psi ^{(0)}\left(\frac{3}{8}\right)-\psi ^{(0)}\left(\frac{5}{8}\right)+\psi ^{(0)}\left(\frac{7}{8}\right)\right)\simeq -1.38446 \tag{7b}$$

We shall see below that $B$ can be simplified appreciably.

Hence the integral can be expressed by polygamma functions as shown: $f = A + B \simeq 1.62706$.

Early identification

In $(3)$ let's get rid of the summation over odd $k$ by appliying the filter $\frac{1}{2}\left(1-(-1)^k\right)$ giving

$$A_a = 2 i \sum_{k=1}^\infty \frac{1}{2}\left(1-(-1)^k\right) \frac{e^{-i k x}}{ k^2}= i \sum_{k=1}^\infty \frac{(e^{-i x})^k}{ k^2}-i \sum_{k=1}^\infty \frac{(-e^{-i x})^k}{ k^2}= i \left(\text{Li}_2(e^{-i x})- \text{Li}_2(-e^{-i x})\right)\tag{8a}$$

$$B_a= -2 x\sum_{k=1}^\infty \frac{1}{2}\left(1-(-1)^k\right)\frac{(e^{-i x})^k}{ k} = -x \left(\text{Li}_1(e^{-i x})- \text{Li}_1(-e^{-i x})\right)\tag{8b}$$

Here

$$\text{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s}\tag{9}$$

is the polylog function.

For $s=1$ we have $\text{Li}_1(z) = - \log(1-z)$ so that

$$B_a = - x \left(- \log(1-e^{-i x}) + \log(1+e^{-i x})\right) = x \log\left (i \tan(\frac{x}{2 })\right)\tag{8c}$$

Now the definite integral is easily obtained from the antiderivatives, leading to

$$A = A_a(x=\frac{\pi}{4}) - A_a(x=-\frac{\pi}{4})\\= i \left(\text{Li}_2(e^{-i \frac{\pi}{4}})- \text{Li}_2(-e^{-i \frac{\pi}{4}})\right)-i \left(\text{Li}_2(e^{i \frac{\pi}{4}})- \text{Li}_2(-e^{i \frac{\pi}{4}})\right)\\= -i \left(\text{Li}_2(e^{i \frac{\pi}{4}})- \text{Li}_2(e^{-i \frac{\pi}{4}})\right)+ i \left(\text{Li}_2(-e^{i \frac{\pi}{4}})- \text{Li}_2(-e^{-i \frac{\pi}{4}})\right)\tag{10a}$$

$$B = x \log\left (i \tan(\frac{x}{2 })\right)|_{x\to \frac{\pi}{4}} - x \log\left (i \tan(\frac{x}{2 })\right)|_{x\to -\frac{\pi}{4}}= \frac{\pi}{2} \log \left(\tan \left(\frac{\pi }{8}\right)\right)\\= \frac{\pi}{2} \log \left( \sqrt{2}-1\right)\tag{10b}$$

Here we have used the trigonometric identity $\tan \left(\frac{\pi }{8}\right)=\sqrt{3-2 \sqrt{2}}=\sqrt{2}-1$.

Discussion

  • periods appearing
  • polylog or polygamma ?
  • other bundaries ($\frac{\pi}{q}$, $q=2,3,...)$
  • related integrals like $\int x^n \csc(x)\,dx$, $\int (e^{t x}-1) \csc(x)\,dx$, $\int \frac{x^n}{1-e^{i x}} \, dx$

    (to be completed)

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  • $\begingroup$ Nice work (+1). I have to look it over more, but shouldn't $(*)$ be $$\psi_{s-1}(p)=(-1)^s\Gamma(s)\sum_{k\geq0}\frac{1}{(k+p)^s}$$ $\endgroup$ – clathratus May 20 at 15:53
  • $\begingroup$ @ clathratus Thank you. I simply wanted to understand from the scratch what's going on here, and therefore assumed an uneducated perception. As to your remark: no, I don't think so. Mathematica tells me that $\sum _{k=0}^{\infty } \frac{1}{(k+p)^s} = \text{HurwitzZeta[s,p]}$ and for example $\text{HurwitzZeta[2,p]}=\psi ^{(1)}(p) $ $\endgroup$ – Dr. Wolfgang Hintze May 20 at 16:07
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\begin{align}I&=2\int_0^{\pi/4}\frac{x}{\sin x}\mathrm{d}x\\ &=2\Big[x\ln\left(\tan\left(\frac{x}{2}\right)\right)\Big]_0^{\pi/4}-2\int_0^{\pi/4}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\ &=\frac{\pi}{2}\ln(\sqrt{2}-1)-2\int_0^{\pi/4}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\ \end{align}

In the latter integral perform the change of variable $y=\dfrac{x}{2}$,

\begin{align}I&=\frac{\pi}{2}\ln(\sqrt{2}-1)-4\int_0^{\pi/8}\ln\left(\tan\left(x\right)\right)\,dx\\ &=\frac{\pi}{2}\ln(\sqrt{2}-1)-4\Big[\frac{1}{2}\text{i}\big(\text{Li}_2(\text{i}\tan x)-\text{Li}_2(-\text{i}\tan x)\big)+x\ln\left(\tan x\right)\Big]_0^{\pi/8}\\ &=\boxed{2\text{i}\left(\text{Li}_2\left(\text{i}\left(1-\sqrt{2}\right)\right)-\text{Li}_2\left(\text{i}\left(\sqrt{2}-1\right)\right)\right)} \end{align}

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