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How to prove

$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$ and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$ (Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{\sin x} dx = 7~\zeta(3) ,\text { equivalently }\ \int_{0}^{1}\frac{x - x^{2} }{ \sin(\pi x)}dx = 7\frac{\zeta (3)}{\pi^{3}},$$ has been solved here.)

Wolfgang
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  • I suggest you to You may use series like $$\frac{1}{\sin^2(\pi z)} = \sum_{k \in \mathbb{Z}} \frac{1}{(z-k)^2}$$and similar series for $\frac{1}{\sin^3(\pi z)}$. – Christophe Leuridan Apr 26 '22 at 21:29
  • @ChristopheLeuridan I'm sure this comes from an application of the residue theorem - what exactly does one do? Presumably a cotangent box contour, but with which function? – FShrike Apr 26 '22 at 21:49
  • I forgot a constant. The right formula is $$\frac{\pi^2}{sin^2(\pi z)} = \sum_{k \in \mathbb{Z}} \frac{1}{(z-k)^2}.$$ You prove that the difference of the two sides of the equality defines an entire function (all singularities are removable), which is $1$-periodic, so it suffices to check that the function is bounded on the band $0 \le \Re(z) \le 1$. – Christophe Leuridan Apr 26 '22 at 21:55
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    To answer the second minor question. $$\int_{0}^{\pi} \frac{x^3 (\pi-x)^3}{\sin^3 x}, dx=126\pi^2 \zeta (3) -\left(1395+\frac{279\pi^2}{2}\right)\zeta(5) +\frac{5715}{4}\zeta(7)$$ – KStarGamer Apr 27 '22 at 01:03
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    You inspired me an interesting problem. https://math.stackexchange.com/questions/4439605/approximation-of-int-0-pi-bigx-pi-x-csc-x-bigk-dx-quad-forall-k?noredirect=1#comment9295976_4439605 – Claude Leibovici May 01 '22 at 05:15

4 Answers4

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Integrate by parts twice

\begin{align} \int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =& \>2 \int_0^{ \frac\pi2}\frac{x^2(\pi-x)^2}{\sin^2 x} dx \\ =&\>4 \int_0^{\frac\pi2}(\pi^2x-3\pi x^2+2x^3)\cot x\> dx\\ =&\> 4\int_0^{\frac\pi2} (6\pi x-6x^2 -\pi^2)\ln(2\sin x)dx \end{align} Then, utilize the known results $\int_0^{\frac\pi2} \ln(2\sin x)dx=0$, $\int_0^\frac{\pi}{2}x\ln(2\sin x)dx= \frac7{16}\zeta(3)$ and $\int_0^{\frac\pi2} x^2\ln(2\sin x)\,dx= \frac{3\pi}{16}\zeta(3)$ to arrive at \begin{align} \int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), \end{align}

Quanto
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Let's integrate the function $$\frac{z^{2}(\pi-z)^{2}}{\sin^{2}(z)}$$ around a tall rectangular contour with vertices at $z=0$, $z= \pi$, $z= \pi + iR$, and $z=iR$.

If we let $R \to \infty$, the integral vanishes along the top of the contour.

Since there are no singularities inside the contour, we have $$\int_{0}^{\pi} \frac{x^{2}(\pi-x)^{2}}{\sin^{2}(x)} \, \mathrm dx + \int_{0}^{\infty} \frac{(\pi + i t)^{2}(-it)^{2}}{\sinh^{2}(t)} \, i \, \mathrm dt - \int_{0}^{\infty}\frac{(it)^{2}(\pi-it)^{2}}{\sinh^{2}(t)} \, i \, \mathrm dt =0.$$

Equating the real parts on both sides of the equation, we get $$ \begin{align} \int_{0}^{\pi} \frac{x^{2}(\pi-x)^{2}}{\sin^{2}(x)} \, \mathrm dx &=4 \pi \int_{0}^{\infty} \frac{t^{3}}{\sinh^{2}(t)} \, \mathrm dt \\ &= 16 \pi \int_{0}^{\infty}\frac{ t^{3}e^{-2t}}{(1-e^{-2t})^{2}} \, \mathrm dt \\ &= 16 \pi \int_{0}^{\infty} t^{3} \sum_{n=1}^{\infty} n e^{-2tn} \, \mathrm dt \\ &= 16 \pi \sum_{n=1}^{\infty} n \int_{0}^{\infty} t^{3} e^{-2tn} \, \mathrm dt \\ &= 16 \pi \sum_{n=1}^{\infty} n \, \frac{\Gamma(4)}{(2n)^{4}} \\ &= 6 \pi \sum_{n=1}^{\infty} \frac{1}{n^{3}} \\ &= 6 \pi \zeta(3). \end{align}$$

Random Variable
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2

A more general solution:

I’ll state the following propositions that can be trivially proven using induction, sums of geometric series, second derivatives and the use of $\tan^2 x=\sec^2 x-1$.

For $n\in\mathbb{N}$ $$\sec^{2n}(x)=\frac{(-1)^n \, 2^{2n}}{(2n-1)!} \sum_{k=1}^{\infty} (-1)^k e^{2ixk} \prod_{r=0}^{2n-2} (k-n+r+1)$$ $$\sec^{2n+1}(x)=\frac{(-1)^{n-1} \, 2^{2n+1}}{(2n)!} \sum_{k=1}^{\infty} (-1)^k e^{ix(2k-1)} \prod_{r=0}^{2n-1} (k-n+r)$$ Letting $x\mapsto \frac{\pi}{2}-x$, we recover representations of powers of $\csc x$.

To illustrate an example:

$$\csc^2 (x)= -4\sum_{k=1}^{\infty} (-1)^k k e^{2i\left(\frac{\pi}{2}-x\right)k}$$

Let’s use this to integrate $$\begin{align} \int_{0}^{\pi} x^2 (\pi-x)^2 \csc^2 (x)\,dx &= -4\sum_{k=1}^{\infty} (-1)^k k \int_{0}^{\pi} x^2(\pi^2-x)^2 e^{2i\left(\frac{\pi}{2}-x\right)k}\,dx\\&=-4\sum_{k=1}^{\infty} (-1)^k k \cdot \left(-\frac{3\pi \, (-1)^k}{2k^4}\right)\\&=6\pi \sum_{k=1}^{\infty}\frac{1}{k^3}\\&=6\pi \zeta(3)\end{align}$$

KStarGamer
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  • For that one, I have found this: Define $z_n:= (2^n-1) \zeta(n) $ and $p_n:= 64z_n-560z_{n+2} +1036 z {n+4} -225z {n+6}$. Note this stems from the polynomial $$(4x^2-1)(4x^2-9)(4x^2-25)=\prod_{k=-2}^3(2x+2k-1).$$ Then $$\int_{0}^{\pi} x^7 (\pi-x)^7 \csc^7 x, dx= \frac1{192}(-429p_9+1782\pi^2 p_7-630 \pi^4 p_5+28 \pi^6 p_3).$$ Advantage: it looks less discouraging than yours when grouping terms with same powers of $\pi$ together. (see here for a somewhat similar pattern). – Wolfgang Apr 28 '22 at 13:25
  • Aren't those two series divergent when $x$ is real? – Random Variable Apr 28 '22 at 14:10
  • @RandomVariable You can interpret it as a distributional identity. If you multiply the series by $e^{-k z}$ then the dominated convergence theorem holds, and proceed to take $z\to 0$. It does work out to compute the integrals here. – KStarGamer Apr 28 '22 at 16:00
  • I had forgotten a factor. Define $q_n:=7\cdot n!! \cdot p_n$. (The idea was to get smaller coefficients.) Then the correct closed form is $$\int_{0}^{\pi} x^7 (\pi-x)^7 \csc^7 x, dx= \frac1{192}(-429q_9+1782\pi^2 q_7-630 \pi^4 q_5+28 \pi^6 q_3).$$ – Wolfgang May 01 '22 at 19:34
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Letting $x\mapsto \frac{\pi}{2}-x$ changes $$ \begin{aligned} I=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right)^{2}\left(\frac{\pi}{2}+x\right)^{2}}{\cos ^{2} x} d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\pi^{2}}{4}-x^{2}\right)^{2} d(\tan x) \end{aligned} $$ Integration by parts gives $$ \begin{aligned} I &=4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x\left(\frac{\pi^{2}}{4}-x^{2}\right) \tan x d x=-8 \int_{0}^{\frac{\pi}{2}} x\left(\frac{\pi^{2}}{4}-x^{2}\right) d(\ln (\cos x)) \end{aligned} $$ Integration by parts again yields $$ I=2 \pi^{2} \int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x-24 \int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x $$ Using my post 1 and post 2, we can conclude that $$ I=-\pi^{3} \ln 2-24\left(-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \zeta (3)\right)=\frac{\pi}{6} \zeta(3) $$

Lai
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