10

The following problem

\begin{align} &\int_{0}^{\pi/2} x\ln^{2}\left(\sin\left(x\right)\right)\,{\rm d}x \\[5mm] = & \ \frac{1}{2}\ln^{2}\left(2\right)\zeta\left(2\right) - \frac{19}{32}\,\zeta\left(4\right) + \frac{1}{24}\,\ln^{4}\left(2\right) + \operatorname{Li}_{4}\left(\frac{1}{2}\right) \label{1}\tag1 \end{align}

was already solved in this solution.

The question here is how to prove $(\ref{1})$ by utilizing the Fourier series of

\begin{align} &\tan\left(x\right) \ln\left(\sin\left(x\right)\right) = -\sum_{n=1}^\infty\left[% \psi\left(\frac{n+1}{2}\right) - \psi\left(\frac{n}{2}\right)-\frac1n\right]\sin(2nx) \\[5mm] = & \ -\sum_{n=1}^\infty\left(\int_{0}^{1}\frac{1 - t}{1 + t}\,\,t^{n - 1}\,\,{\rm d}t\right) \sin\left(2nx\right),\quad 0 < x <\frac{\pi}{2} \end{align}

I wonder what kind of clever manipulation we need to do to create the integrand in $(1)$. I am sure it would be an amazing solution.

Thank you in advance.

This Fourier series can be found in the book, Almost Impossible Integrals, Sums and series, page $243$, Eq$(3.281)$.

Felix Marin
  • 94,079
Ali Olaikhan
  • 27,891

4 Answers4

6

Some generalizations. Enjoy!

  • $\small \int_0^{\frac{\pi }{2}} x^3 \log ^2(2 \sin (x)) \, dx=-\frac{3}{4}\zeta(\bar 5,1)+\frac{3}{4} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{3 \zeta (3)^2}{8}+\frac{21}{32} \pi ^2 \zeta (3) \log (2)-\frac{187 \pi ^6}{26880}+\frac{1}{32} \pi ^2 \log ^4(2)-\frac{1}{32} \pi ^4 \log ^2(2)$

  • $\small \int_0^{\frac{\pi }{2}} x^3 \log^3 (2 \sin (x)) \, dx=\frac{9}{4} \zeta(\bar5,1,1)+\frac{9}{4} \pi ^2 \text{Li}_5\left(\frac{1}{2}\right)+\frac{9}{4} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{\pi ^4 \zeta (3)}{16}-\frac{759 \pi ^2 \zeta (5)}{512}-\frac{45 \zeta (7)}{512}+\frac{63}{64} \pi ^2 \zeta (3) \log ^2(2)+\frac{3}{40} \pi ^2 \log ^5(2)-\frac{1}{16} \pi ^4 \log ^3(2)$

  • $\scriptsize \int_0^{\frac{\pi }{2}} x^4 \log ^4(2 \sin (x)) \, dx=-\frac{7}{4} \pi ^3 \zeta(\bar5,1)-15 \pi \zeta(\bar7,1)-6 \pi \zeta(\bar5,1,\bar1,1)+9 \pi \log ^2(2)\zeta(\bar5,1)+18 \pi \log (2) \zeta(\bar5,1,1)-24 \pi \text{Li}_5\left(\frac{1}{2}\right) \zeta (3)-\frac{1}{6} \pi ^5 \text{Li}_4\left(\frac{1}{2}\right)+6 \pi ^3 \text{Li}_6\left(\frac{1}{2}\right)+3 \pi ^3 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+6 \pi ^3 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{3 \pi ^3 \zeta (3)^2}{8}+\frac{2835 \pi \zeta (3) \zeta (5)}{64}+\frac{1}{5} \pi \zeta (3) \log ^5(2)+\frac{13}{24} \pi ^3 \zeta (3) \log ^3(2)-\frac{31}{8} \pi \zeta (5) \log ^3(2)+\frac{9}{2} \pi \zeta (3)^2 \log ^2(2)-\frac{8}{15} \pi ^5 \zeta (3) \log (2)-\frac{65}{32} \pi ^3 \zeta (5) \log (2)+\frac{465}{16} \pi \zeta (7) \log (2)-\frac{36493 \pi ^9}{4838400}+\frac{1}{12} \pi ^3 \log ^6(2)-\frac{5}{72} \pi ^5 \log ^4(2)+\frac{25 \pi ^7 \log ^2(2)}{1008}$

Infiniticism
  • 9,030
  • 1
    (+1) They are definitely enjoyable, and more enjoyable by the art of mathematics. – user97357329 Aug 22 '20 at 13:53
  • 1
    @AliShather These are known as alternating multiple zeta values (MZV). For example, $$\zeta(\bar{5},1,1) = \sum_{n_1 > n_2 > n_3 \geq 1} \frac{(-1)^{n_1}}{n_1^5 n_2 n_3}$$ the bar indicates which summation is alternating. The number $5+1+1 = 7$ is called weight of this MZV. – pisco Aug 22 '20 at 14:58
  • @pisco thank you for clarifying thats what i wanted to know. – Ali Olaikhan Aug 22 '20 at 15:03
5

To solve this integral use the following integral which can be proved using contour integration.

$$\int_{0}^{\frac π 2}(\cos^{p-1}x )\cos(ax)dx=\frac{π}{2^p}\frac{\Gamma(p)}{\Gamma(\frac {p+a+1}{2})\Gamma(\frac {p-a+1}{2})}$$

On differentiating with respect to $a$, we get

$$\int_{0}^{\frac π 2}x(\cos^{p-1}x )\sin(ax)dx=\frac{π}{2^p}\Gamma(p)\frac{\psi(\frac {p+a+1}{2})-\psi(\frac {p-a+1}{2})}{\Gamma(\frac {p+a+1}{2})\Gamma(\frac {p-a+1}{2})}$$

where $p>0$ and $ -(p+1)<a<(p+1)$ and $\Gamma(x)$ and $\psi(x)$ are the gamma and digamma functions.Let $p_1=(\frac {p+a+1}{2})$ and $p_2=(\frac {p-a+1}{2})$.On differentiating the above equation.

$$\frac{\partial }{\partial a}I(a,p)=\frac{π}{2^{p+1}}\frac{\Gamma(p)}{\Gamma(p_1)\Gamma(p_2)}[-(\psi(p_1)-\psi(p_2))^2+\psi'(p_1)+\psi'(p_2)]$$$$=\int_{0}^{\frac π 2}x^2(cos^{p-1}x )\cos(ax)dx$$

On putting $a=0$, we get

$$J_p=\int_{0}^{\frac π 2}x^2(\cos^{p-1}x )dx=\frac{π}{2^{p+1}}\frac{\Gamma(p)}{\Gamma^2(\frac{p+1}{2})}\psi'(\frac{p+1}{2})$$

Again differentiating with respect to parameter $p$ ,we get

$$\frac d {dp}J_p=\int_{0}^{\frac π 2}x^2(\cos^{p-1}x)\log(\cos x)dx=\frac{π}{2^{p+1}}\frac{\Gamma(p)}{\Gamma^2(\frac{p+1}{2})}\psi'(\frac{p+1}{2})[-\log2+\psi(p)-\psi(\frac{p+1}{2})+\frac {\psi''(\frac{p+1}{2})}{\psi'(\frac{p+1}{2})}$$

From the above integral we can also calculate below integral by letting $p=1$.

$$\int_{0}^{\frac π 2}x^2\log(\cos x)dx=\frac{π^3}{24}[-\log2+\frac{\psi''(1)}{\psi'(1)}]=-\frac{π^3}{24}\log2-\fracπ2 \zeta(3)$$

Now, $$J''_p=\int_{0}^{\frac π 2}x^2(\cos^{p-1}x)\log^2(\cos x)dx=\frac{π}{2^{p+1}}\frac{\Gamma(p)}{\Gamma^2(\frac{p+1}{2})}\psi'(\frac{p+1}{2}) \left ( \left ( [-\log2+\psi(p)-\psi(\frac{p+1}{2})+\frac {\psi''(\frac{p+1}{2})}{\psi'(\frac{p+1}{2})} \right )^2+\psi'(p)-\frac12\psi'(\frac{p+1}{2})+\frac12\frac d {dp}\frac {\psi''(\frac{p+1}{2})}{\psi'(\frac{p+1}{2})} \right ) $$

On letting $p=1$ ,we get

$$\int_{0}^{\fracπ2}x^2\log^2(\cos x)dx=\fracπ {1440}[11π^4+60π^2\log^2 2+720\zeta(3)\log2]$$

Now you can get your desired integral using all of the above integrals but this is lenghty task as it involves some other integrals which will create polylogarithms and the term containing $ \zeta(3)$ will cancel.

These are some generalisations of log-sine integrals:

$$\int_0^zx^m\log\left(2\sin\frac{x}{2}\right)\mathrm{d}x=\frac{z^{m+1}}{m+1}\log\left(2\sin\frac{z}{2}\right)-\frac{z^{m+1}}{(m+1)^2}+\frac{2z^{m+1}}{m+1}\sum_{k=1}^{\infty}\frac{\zeta(2k)}{m+2k+1}\left(\frac{z}{2\pi}\right)^{2k} \;\;\;\;\; (|z|<2\pi;m\in\mathbb{N}).$$

$$\int_0^{2\pi}\left(2\sin\frac12 \theta\right)^{\lambda}e^{i\nu\theta}\mathrm{d}\theta=2\pi e^{i\nu\pi}\frac{\Gamma(1+\lambda)}{\Gamma(1+\frac12\lambda+\nu)\Gamma(1+\frac12\lambda-\nu)}.$$

$$2^p\int_0^\pi x^n\log^p(\sin (x))\mathrm{d}x=\pi^n\left(\frac{\pi}{n+1}\frac{\mathrm{d}^p}{\mathrm{d}m^p}\left(\left.\frac{\binom{2m}{m}}{4^m}\right)\right|_{m=0}-\sum_{k=1}^{\infty}\frac{\partial^p}{\partial m^p}\left(\left.\frac{\binom{2m}{m+k}}{4^m}\right)\right|_{m=0}\sum_{j=1}^{\lfloor\frac{n}{2}\rfloor}\frac{n!(-1)^{j+k}}{(n+1-2j)!(2\pi)^{2j-1}k^{2j}}\right).$$

$$\frac{1}{2\pi}\int_0^{2\pi}\left[\log\left(2\sin \frac{x}{2}\right)\right]^n\mathrm{d}x=\frac{(-1)^nn!}{2}a_n\ \ \ \ \ (n\in N_0)\\[2ex] \text{coefficients $a_n$ are given by:}\\[2ex] \frac{2^{-2z}\Gamma\left(\frac12-z\right)}{\sqrt{\pi}\ \Gamma(1-z)}=\sum_{n=0}^{\infty}a_nz^n$$

Infiniticism
  • 9,030
Paras
  • 1,472
5

From here we have

$$\frac23\arcsin^4x=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}=\sum_{n=1}^\infty\frac{H_{n}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}-\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^4{2n\choose n}}$$

Set $x=1$ we get

$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=\sum_{n=1}^\infty\frac{4^nH_{n}^{(2)}}{n^2{2n\choose n}}-\frac{15}{4}\zeta(4)\tag1$$

In this question we showed $$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}+12\ln^2(2)\zeta(2)\tag2$$

Adding $(1)$ and $(2)$ yields

$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{4}\zeta(4)-\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}$$

By using the Fourier series of $\tan x\ln(\sin x)$, we showed in this solution:

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$

substitute this result we get

$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\frac{19}{4}\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\tag3$$

Now we use the well-known series expansion of $\arcsin^2 x$:

$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{4^n x^{2n}}{n^2{2n\choose n}}$$

Multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{2n-1}\ln xdx=\frac{1}{4n^2}$ we get

$$\frac18\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=-\int_0^1\frac{\ln x\arcsin^2(x)}{x}dx$$

$$\overset{IBP}{=}\int_0^1\frac{\ln^2x\arcsin(x)}{\sqrt{1-x^2}}dx\overset{x=\sin\theta}{=}\int_0^{\pi/2}x\ln^2(\sin x)dx\tag4$$

From $(3)$ and $(4)$ we obtain

$$\int_0^{\pi/2} x\ln^2(\sin x)dx=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$

Infiniticism
  • 9,030
Ali Olaikhan
  • 27,891
0

We know that $$\ln \left( \frac{1+e^{2\mathrm{i}x}}{2} \right) =\ln \left( \cos x \right) +\mathrm{i}x\\\ln \left( \frac{1+e^{2\mathrm{i}x}}{2} \right) =\ln \left( \cos x \right) +\mathrm{i}x\\\mathrm{Im}\left[ \ln ^3\left( \frac{1+e^{2\mathrm{i}x}}{2} \right) \right] =3x\ln ^2\left( \cos x \right) -x^3\\\mathrm{Im}\left[ \ln ^3\left( \frac{1+e^{2\mathrm{i}x}}{2} \right) \right] =3x\ln ^2\left( \cos x \right) -x^3\\$$ \begin{align*} &\int_0^{\frac{\pi}{2}}{x\ln ^2\left( \cos x \right)}\mathrm{d}x \\ =&\frac{1}{3}\int_0^{\frac{\pi}{2}}{x^3}\mathrm{d}x+\frac{1}{3}\mathrm{Im}\int_0^{\frac{\pi}{2}}{\ln ^3\left( \frac{1+e^{2\mathrm{i}x}}{2} \right)}\mathrm{d}x \\ =&\frac{\pi ^4}{192}+\frac{1}{6}\mathrm{Im}\int_0^{\pi}{\ln ^3\left( \frac{1+e^{\mathrm{i}x}}{2} \right)}\mathrm{d}x \\ =&\frac{\pi ^4}{192}-\frac{1}{6}\mathrm{Re}\int_C{\frac{\ln ^3\left( \frac{1+z}{2} \right)}{z}}\mathrm{d}z \end{align*}\\begin{align*} &\int_0^{\frac{\pi}{2}}{x\ln ^2\left( \cos x \right)}\mathrm{d}x \\ =&\frac{1}{3}\int_0^{\frac{\pi}{2}}{x^3}\mathrm{d}x+\frac{1}{3}\mathrm{Im}\int_0^{\frac{\pi}{2}}{\ln ^3\left( \frac{1+e^{2\mathrm{i}x}}{2} \right)}\mathrm{d}x \\ =&\frac{\pi ^4}{192}+\frac{1}{6}\mathrm{Im}\int_0^{\pi}{\ln ^3\left( \frac{1+e^{\mathrm{i}x}}{2} \right)}\mathrm{d}x \\ =&\frac{\pi ^4}{192}-\frac{1}{6}\mathrm{Re}\int_C{\frac{\ln ^3\left( \frac{1+z}{2} \right)}{z}}\mathrm{d}z \end{align*}\ Representing the upper half unit circle, then according to the residue theorem: \begin{align*} &\mathrm{Re}\int_C{\frac{\ln ^3\left( \frac{1+z}{2} \right)}{z}}\mathrm{d}z \\ =&-\int_0^1{\frac{1}{x}\left[ \ln ^3\left( \frac{1+x}{2} \right) -\ln ^3\left( \frac{1-x}{2} \right) \right]}\mathrm{d}x \\ =&3\ln ^22\int_0^1{\frac{\ln \left( 1-x \right)}{x}}\mathrm{d}x-3\ln 2\int_0^1{\frac{\ln ^2\left( 1-x \right)}{x}}\mathrm{d}x+\int_0^1{\frac{\ln ^3\left( 1-x \right)}{x}}\mathrm{d}x \\ &-3\ln ^22\int_0^1{\frac{\ln \left( 1+x \right)}{x}}\mathrm{d}x+3\ln 2\int_0^1{\frac{\ln ^2\left( 1+x \right)}{x}}\mathrm{d}x-\int_0^1{\frac{\ln ^3\left( 1+x \right)}{x}}\mathrm{d}x \\ =&-\frac{2\pi ^2}{15}-\pi ^2\ln ^22+\frac{\ln ^42}{4}+6\mathrm{Li}_4\left( \frac{1}{2} \right) \end{align*}\begin{align*} &\mathrm{Re}\int_C{\frac{\ln ^3\left( \frac{1+z}{2} \right)}{z}}\mathrm{d}z \ =&-\int_0^1{\frac{1}{x}\left[ \ln ^3\left( \frac{1+x}{2} \right) -\ln ^3\left( \frac{1-x}{2} \right) \right]}\mathrm{d}x \ =&3\ln ^22\int_0^1{\frac{\ln \left( 1-x \right)}{x}}\mathrm{d}x-3\ln 2\int_0^1{\frac{\ln ^2\left( 1-x \right)}{x}}\mathrm{d}x+\int_0^1{\frac{\ln ^3\left( 1-x \right)}{x}}\mathrm{d}x \ &-3\ln ^22\int_0^1{\frac{\ln \left( 1+x \right)}{x}}\mathrm{d}x+3\ln 2\int_0^1{\frac{\ln ^2\left( 1+x \right)}{x}}\mathrm{d}x-\int_0^1{\frac{\ln ^3\left( 1+x \right)}{x}}\mathrm{d}x \ =&-\frac{2\pi ^2}{15}-\pi ^2\ln ^22+\frac{\ln ^42}{4}+6\mathrm{Li}_4\left( \frac{1}{2} \right) \end{align*}\ $$\int_0^{\frac{\pi}{2}}{x\ln ^2\left( \cos x \right)}\mathrm{d}x=\frac{79\pi ^4}{2880}+\frac{\pi ^2\ln ^22}{6}-\frac{\ln ^42}{24}-\mathrm{Li}_4\left( \frac{1}{2} \right) \\\int_0^{\frac{\pi}{2}}{x\ln ^2\left( \cos x \right)}\mathrm{d}x=\frac{79\pi ^4}{2880}+\frac{\pi ^2\ln ^22}{6}-\frac{\ln ^42}{24}-\mathrm{Li}_4\left( \frac{1}{2} \right) \\$$

Martin.s
  • 1