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I have this integral, $$I_n=\displaystyle \int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \qquad n\in \mathbb{Z}^+.$$ We have the results $$ \begin{align} I_1 & = 2C, \\ I_2 &= \pi\log 2, \\ I_4 & = -\frac{\pi^3}{12} + 2\pi\log 2 + \frac{\pi^3}{3}\log 2-\frac{3\pi}{2}\zeta(3), \end{align} $$ where $C$ is Catalan's constant. Can we prove any of these results, or make any progress on $I_3$, or the general case?

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  • $\begingroup$ I think your best chance is a recurrence relationship of $I(n)$ in terms of $I(n-2)$ and/or $I(n-1)$. $\endgroup$ – Lucian May 8 '14 at 13:12
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    $\begingroup$ $\displaystyle{\large I_{4}}$ must be $\displaystyle{\large\color{#c00000}{-\,{\pi^{3} \over 12}} + 2\pi\ln\left(2\right) + {\pi^{3} \over 3}\,\ln\left(2\right) - {3\pi \over 2}\,\zeta\left(3\right)}$. $\endgroup$ – Felix Marin Jul 28 '14 at 2:55
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 16:59
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Integrating by parts, we have $$ \int \frac{x^2}{\sin^2 x} \, dx= -x^2 \cot x +\int 2x \cot x \, dx\\= -x^2 \cot x + 2x \ln \sin x - \int 2 \ln \sin x \, dx $$ Evaluating this between $0$ and $\pi/2$, we find that the boundary terms vanish (by taking the appropriate limits), so we are left with the well-known integral $$ -2 \int_0^{\pi/2} \ln \sin x \, dx =\pi \ln 2 $$

Edit: I have found a way to do $I_1$. Integrating by parts, $$ \int \frac{x}{\sin x} \, dx= x \ln \tan \frac{x}{2} - \int \ln \tan \frac{x}{2} \, dx $$ Evaluating between $0$ and $\pi/2$ yields $$ I_1 = -2 \int_0^{\pi/4} \ln \tan x \, dx\\ = -2 \int_{-\infty}^{0} x \frac{e^x}{1+e^{2x}} \, dx\\ = 2 \sum_{k \geq 0} (-1)^k\int_{0}^{\infty} x e^{-(2k+1)x} \, dx \\ = 2 \sum_{k \geq 0} (-1)^k \frac{1}{(2k+1)^2} \\ = 2C $$ as was to be proved.

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Integrating by parts 3 times,

$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= - \frac{x^{4}}{3} \cot(x) \left(\csc^{2} (x) +2 \right) \Bigg|^{\pi/2}_{0} + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \csc^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{2}{3}x^{3} \cot^{2}(x) \Bigg|^{\pi/2}_{0} + 2 \int_{0}^{\pi /2} x^{2} \cot^{2}(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx + 2 \int_{0}^{\pi /2} x^{2} \cot^{2} (x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx -2x^{2} \Big( x + \cot(x) \Big) \Bigg|^{\pi/2}_{0} +4 \int_{0}^{\pi /2} x\Big(x+ \cot(x) \Big) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{4} + 4 \int_{0}^{\pi /2} x^{2} \ dx + 4 \int_{0}^{\pi /2} x \cot(x) \ dx \\ &= \frac{8}{3} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx - \frac{\pi^{3}}{12} + 4 \int_{0}^{\pi /2} x \cot(x) \ dx . \end{align}$$

In general, $$ \int_{a}^{b} f(x) \cot(x) \ dx = 2 \sum_{n=1}^{\infty} \int_{a}^{b} f(x) \sin (2nx) \ dx .$$

See here.

So $$ \begin{align} \int_{0}^{\pi /2} x^{3} \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x^{3} \sin (2nx) \ dx \\ &= 2 \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1} \pi^{3}}{16n} - \frac{(-1)^{n-1} 3\pi}{8n^{3}} \right) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{3 \pi}{4} \eta(3) \\ &= \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3). \end{align}$$

And

$$ \begin{align} \int^{\pi /2}_{0} x \cot(x) \ dx &= 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x \sin(2nx) \ dx \\ &= -\frac{\pi}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \\ &= \frac{\pi \ln 2}{2} . \end{align}$$

Therefore,

$$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= \frac{8}{3} \left( \frac{\pi^{3}}{8} \ln (2) - \frac{9 \pi }{16} \zeta(3) \right) - \frac{\pi^{3}}{12} + 4 \left(\frac{\pi \ln 2}{2} \right) \\ &= - \frac{\pi^{3}}{12} + 2 \pi \ln(2) + \frac{\pi^{3}}{3} \ln (2) - \frac{3 \pi}{2} \zeta(3) . \end{align}$$

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For $n\in\mathbb{N}$ we have:

$$\int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^ndx=\sum\limits_{j=0}^{n-1}2^{n-j}\begin{bmatrix}n\\j+1\end{bmatrix}\sum\limits_{v=0}^j\binom{j}{v}(-n)^{j-v+1}\sum\limits_{l=1}^{\big\lfloor\frac{n+1}{2}\big\rfloor}\frac{(-1)^l\Big(\tfrac{\pi}{2}\Big)^{n-2l+1}}{(n-2l+1)!}f_n(2l-v),$$

with the Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$ defined by $\displaystyle\sum\limits_{k=0}^n\begin{bmatrix}n\\k\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k),$

with $f_{2m-1}(s):=(-1)^{m-1}\beta(s),$ where $\beta(s)$ := Dirichlet $\beta$ function,

and $f_{2m}(s):=(-1)^{m-1}2^{-s}\eta(s),$ where $\eta(s)$ := Dirichlet $\eta$ function,

for $m\in\mathbb{N}$, with the analytical extensions $(s\in\mathbb{C})$

$$\begin{align} \beta(1-s)&=\bigg(\dfrac{2}{\pi}\bigg)^s\sin\bigg(\dfrac{\pi s}{2}\bigg)~\Gamma(s)~\beta(s) \\\\ \eta(1-s)&=\dfrac{2^s-1}{1-2^{s-1}}~\pi^{-s}\cos\bigg(\dfrac{\pi s}{2}\bigg)~\Gamma(s)~\eta(s) \end{align}$$

and with the simplifications $(k\in\mathbb{N}_0)$

$$\begin{align} \beta(-2k-1)~&=~0 \\\\ \beta(-2k)~&=~-\frac{2^{4k+1}}{2k+1}~B_{2k+1}\bigg(\frac{1}{4}\bigg) \\\\ \beta(2k+1)~&=~(-1)^{k-1}~\frac{(2\pi)^{2k+1}}{2(2k+1)!}~B_{2k+1}\bigg(\frac{1}{4}\bigg) \\\\ \eta(1-k)~&=~\frac{2^k-1}{k}~B_k \\\\ \eta(2k)~&=~(-1)^{k-1}~\frac{2^{2k-1}-1}{(2k)!}~B_{2k}~\pi^{2k} \\\\ \eta(2k+1)~&=~\bigg(1-\frac{1}{2^{2k}}\bigg)\zeta(2k+1) \end{align}$$


Examples include:

$$\begin{align} \int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^1dx~&=~2\beta(2)\approx1.83 \\\\ \int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^2dx~&=~\pi\ln2\approx2.178 \\\\ \int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^3dx~&=~-6\beta(4)+\bigg(\frac{3}{4}\pi^2+6\bigg)\beta(2)-\frac{3}{8}\pi^2\approx2.64 \\\\ \int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^4dx~&=~-\frac{3}{2}\pi\zeta(3)+\bigg(\frac{\pi^3}{3}+2\pi\bigg)\ln2-\frac{\pi^3}{12}\approx3.27 \\\\ \int\limits_0^{\pi/2}\bigg(\frac{x}{\sin x}\bigg)^5dx~&=~90\beta(6)-\bigg(\frac{45}{4}\pi^2+100\bigg)\beta(4)+\bigg(\frac{45}{192}\pi^4+\frac{25}{4}\pi^2+10\bigg)\beta(2)- \\ &-\bigg(\frac{55}{384}\pi^4+\frac{5}{8}\pi^2\bigg)\approx4.135 \end{align}$$

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  • $\begingroup$ Fantastic and underrated answer. $\endgroup$ – Brevan Ellefsen Aug 14 '17 at 2:04
  • $\begingroup$ @BrevanEllefsen : Very kind of you, thanks ! (But: for a better rating I should explain more of course ... :-) ) $\endgroup$ – user90369 Aug 14 '17 at 8:27
  • $\begingroup$ How do we know your initial claim? $\endgroup$ – clathratus Mar 3 at 2:33
  • $\begingroup$ @clathratus : I have to explain more, of course, but it's a lot, I need too much time for it. And my answer came 2 years too late. Therefore I wrote only the answer for interested people, not the whole explanation. If you like to do busywork, you should start with $\sin x = (e^{ix}-e^{-ix})/(i2)$, the rest comes on its own. By the way: You can also look at here for similar ideas. $\endgroup$ – user90369 Mar 3 at 16:11
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Using a CAS, I obtained $$I_3=\frac{1}{256} \left(192 \left(8+\pi ^2\right) C-96 \pi ^2-\psi ^{(3)}\left(\frac{1}{4}\right)+\psi ^{(3)}\left(\frac{3}{4}\right)\right)$$ $$I_6=\frac{1}{320} \pi \left(40 \pi ^2 (-12 \zeta (3)-1+20 \log (2))+240 (15 (\zeta (5)-\zeta (3))+\log (16))+\pi ^4 (32 \log (2)-11)\right)$$

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  • $\begingroup$ Are you will to share the name of the CAS? My old Maple give solutions only in terms of $\mathrm{polylog}(n,\pm i)$ and Wolfram Alpha shows only numerical values? $\endgroup$ – gammatester May 8 '14 at 12:28
  • $\begingroup$ It is a modiied version of Maxima. I must say that I got a lot of trouble with $I_5$. $\endgroup$ – Claude Leibovici May 8 '14 at 12:31
  • $\begingroup$ @ClaudeLeibovici : The maxima which I've installed (v5.24.0) can't integrate even when n=1. Is that modified version available online? $\endgroup$ – gar May 9 '14 at 10:38
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    $\begingroup$ @gar. No, it is a version we have modified in my research team many years ago precisely for integration of very nasty functions we worked on at that time. If life is not too short, it is part of my plans to clean it up and propose it to whoever could be interested. $\endgroup$ – Claude Leibovici May 9 '14 at 10:51
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    $\begingroup$ @ClaudeLeibovici : Oh, I see. That would be a great thing to do! Thanks for your time and work. $\endgroup$ – gar May 9 '14 at 10:56

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