Evaluate $\int_0^{\pi/2}x\cot{(x)}\ln^4\cot\frac{x}{2}\,\mathrm dx$

Question

How to evaluate the following integral ?: $$ \int_{0}^{\pi/2}x\cot\left(\, x\,\right)\ln^{4}\left[\,\cot\left(\,{x \over 2}\,\right)\,\right]\,{\rm d}x $$ It seems that evaluate to $$ {\pi \over 16}\left[\, 5\pi^{4}\ln\left(\, 2\,\right) - 6\pi^{2}\zeta\left(\, 3\,\right) -{93 \over 4}\,\zeta\left(\, 5\,\right) \,\right] $$ Exactly ?.

Answer 1

How to evaluate the following integral ?

Using M.N.C.E.'s hints, rewrite his first integral in terms of $\displaystyle\int_0^1\frac{\ln^5x}{1+x^2}dx$ using integration by parts with regard to $\dfrac{dx}x=d\big(\ln x\big)$, then expand $\dfrac1{1+x^2}$ into its binomial series, and switch the order of summation and integration. This will yield a very familiar series. Similar tricks apply for the second one as well, only first rewrite $\ln^4x$ as $\bigg[\dfrac{d^4}{dn^4}x^n\bigg]_{\large n=0}\quad$ and then switch the order of integration and differentiation. Once again, you will encounter a very familiar series.

Answer 2

This is a partial answer and I need more work. Denote the original integral by $I$. From M.N.C.E's idea, define $$ I_1=\int_0^1\frac{\arctan x\ln^4x}{x}dx, I_2(a)=\int_0^1\frac{x\arctan(a x)\ln^4x}{1+x^2}dx $$ and then $I_2(0)=0$ and $I=2I_1-4I_2(1)$. Now \begin{eqnarray} I_1&=&-\frac15\int_0^1\frac{1}{1+x^2}\ln^5xdx\\ &=&-\frac15\int_0^1\sum_{n=0}^\infty(-1)^nx^{2n}\ln^5xdx\\ &=&\frac{120}5\sum_{n=0}^\infty(-1)^n\frac{1}{(2n+1)^6}\\ &=&\frac{120}5\left(\sum_{n=0}^\infty\frac{1}{(4n+1)^6}-\sum_{n=0}^\infty\frac{1}{(4n+3)^6}\right)\\ &=&\frac{3}{512}(\zeta(6,\frac14)-\zeta(6,\frac34)). \end{eqnarray} From this, we have $$ \zeta(6,\frac14)+\zeta(6,\frac24)+\zeta(6,\frac34)=(4^6-1)\zeta(6)=\frac{13\pi^6}{3},\zeta(6,\frac24)=\frac{\pi^6}{15} $$ or $$ \zeta(6,\frac14)=\frac{64\pi^6}{15}-\zeta(6,\frac34). $$ Thus $$ I_1= \frac{3}{512}(\frac{64\pi^6}{15}-2\zeta(6,\frac34)).$$ But \begin{eqnarray} I_2'(a)&=&\int_0^1\frac{x^2\ln^4x}{(1+a^2x^2)(1+x^2)}dx\\ &=&\frac1{1-a^2}\int_0^1\left(\frac{1}{1+a^2x^2}-\frac{1}{1+x^2}\right)\ln^4xdx\\ &=&\frac1{1-a^2}\int_0^1\sum_{n=0}^\infty(-1)^n(a^{2n}-1)x^{2n}\ln^4xdx\\ &=&\frac{24}{1-a^2}\sum_{n=0}^\infty(-1)^n(a^{2n}-1)\frac{1}{(2n+1)^5}\\ &=&-24\sum_{n=0}^\infty(-1)^n(1+a^2+(a^2)^2+\cdots+(a^2)^{n-1})\frac{1}{(2n+1)^5} \end{eqnarray} and hence \begin{eqnarray} I_2(1)&=&-24\int_0^1\sum_{n=0}^\infty(-1)^n(1+a^2+(a^2)^2+\cdots+(a^2)^{n-1})\frac{1}{(2n+1)^5}da\\ &=&-24\sum_{n=0}^\infty(-1)^n\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)\frac{1}{(2n+1)^5}\\ &=&-24\sum_{n=0}^\infty(-1)^n(H_{2n+1}-\frac{1}{2}H_{n}-\frac{1}{2n+1})\frac{1}{(2n+1)^5}\\ &=&-24(\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^5}-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^5}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^6})\\ &=&-24(\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^5}-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^5})+\frac{3}{512}(\zeta(6,\frac14)-\zeta(6,\frac34))\\ &=&-24(\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^5}-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^5})+\frac{3}{512}(\frac{64\pi^6}{15}-2\zeta(6,\frac34)) \end{eqnarray}

Answer 3

Let $$J = \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx} \qquad K = \int_0^1 {\frac{x{\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx}$$ Then by M.N.C.E.'s comment, $$\tag{1}I = \int_0^{\pi /2} {x\cot x{{\ln }^4}\left( {\cot \frac{x}{2}} \right)dx} = 2J - 4K$$

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Here is a symmetry of the integrand that we can exploit: $$\begin{aligned} K &= \int_0^1 {\frac{{x\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx} \\&= \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx} - \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{{x(1 + {x^2})}}dx} \\&= J - \int_1^\infty {\frac{{x\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}x}}{{1 + {x^2}}}dx} \\& = J - \int_1^\infty {\frac{1}{x}\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}xdx} + \int_1^\infty {\frac{1}{{x(1 + {x^2})}}\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}xdx} \\& = J - J + \frac{\pi }{2}\int_1^\infty {\frac{{{{\ln }^4}x}}{{x(1 + {x^2})}}dx} - \int_1^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} \\&= \frac{\pi }{2}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx} - \int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} + \int_0^1 {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} \\& = \frac{\pi }{2}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx} - \int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} + \int_0^1 {\frac{{{{\ln }^4}x\arctan x}}{x}dx} - \int_0^1 {\frac{{x{{\ln }^4}x\arctan x}}{{1 + {x^2}}}dx} \\& = \frac{\pi }{4}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx} - \frac{1}{2}\int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} + \frac{J}{2} \end{aligned}$$ The fact that exponent $4$ is even is paramount here. Plugging into $(1)$, the $J$ miraculously cancelled: $$I = 2\underbrace{\int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}}_{L} - \pi \underbrace{\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx}}_{45\zeta(5)/64} $$

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The crux of the problem is, indeed, evaulating the remaining integral. Note that $$ L = \int_0^{\frac{\pi }{2}} {x\cot x{{\ln }^4}(\tan x)dx} $$ and we have the following formula:

For $-2<p<1$, $$\int_0^{\frac{\pi }{2}} {x{{\tan }^p}xdx} = \frac{\pi }{4}\csc \frac{{p\pi }}{2}\left[ {\psi ( - \frac{p}{2} + 1) - 2\psi ( - p + 1) - \gamma } \right]$$

Hence the value of $L$ follows from it by differentiating four times and set $p=-1$: $$L = -\frac{3 \pi ^3 \zeta (3)}{16}-\frac{3 \pi \zeta (5)}{8}+\frac{5}{32} \pi ^5 \ln 2$$

Finally, we obtain $$I = 2L - \pi \frac{45\zeta(5)}{64} = \color{blue}{-\frac{3 \pi ^3 \zeta (3)}{8}-\frac{93 \pi \zeta (5)}{64}+\frac{5}{16} \pi ^5 \ln 2}$$