How to calculate $\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cos x\, \text{d}x}$.

Question

Calculate:

$$\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cos x\, \text{d}x}$$

My attempt: Let $$ A=\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cos x\, \text{d}x},B=\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\sin x\, \text{d}x} $$ then $$ A+B=\frac{\pi ^2\ln 2}{4}-\frac{7}{8}\zeta \left( 3 \right) +\underset{I}{\underbrace{\int_0^{\frac{\pi}{2}}{x^2\cot x\ln \left( \sin 2x \right)}\, \text{d}x}} $$ $$ A-B=\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cot x\, \text{d}x} $$ Define $$ J\left( a,b \right) =\int_0^{\frac{\pi}{2}}{\frac{\sin \left( 2ax \right)}{\sin ^b\left( 2x \right)}}\, \text{d}x $$ I‘ve been stuck here for a long time, I can't figure out $I$ by the derivative of $J(a,b)$.

Maybe I'm doing the wrong way. If so, how can I figure it out? THX!

Answer 1

$$\color{blue}{\int_0^{\pi /2} {{x^2}\cot x\ln (\cos x)\mathrm{d}x} = - \frac{{{\pi ^4}}}{{720}} + \frac{{\ln ^42}}{{24}} - \frac{{{\pi ^2}\ln ^22}}{6} + \text{Li}_4\left(\frac{1}{2}\right)}$$

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It is easy to show that (using Fourier expansion of $\ln(\cos x)$ for instance): $$\int_0^{\pi/2} x^2 \cot x \ \mathrm{d}x = \frac{\pi^2 \ln 2}{4} - \frac{7\zeta(3)}{8}$$ Hence it suffices to consider $$I = \int_0^{\pi /2} {x^2}\cot x\ln (2\cos x)\mathrm{d}x $$ this alternative form will be proved convenient.

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Now consider the twins: $$P = \int_0^1 {\frac{{{{\ln }^2}x\ln (1 + x)}}{{1 - x}}\mathrm{d}x} \qquad Q = \int_0^1 {\frac{{{{\ln }^2}x\ln (1 - x)}}{{1 + x}}\mathrm{d}x} $$ We will see that $I$ arises from integrating $\ln^2 z \ln(1+z) / (1-z) $ around the semicircle contour above real axis. Indeed $$ P + \int_{ - 1}^0 {\frac{{{{\ln }^2}x\ln (1 + x)}}{{1 - x}}\mathrm{d}x} + \int_0^\pi {i{e^{ix}}\frac{{{{\ln }^2}({e^{ix}})\ln (1 + {e^{ix}})}}{{1 - {e^{ix}}}}\mathrm{d}x} = 0 $$ Hence $$\begin{aligned} P + \int_0^1 {\frac{{{{(\ln x + \pi i)}^2}\ln (1 - x)}}{{1 + x}}\mathrm{d}x} &= i\int_0^\pi {{x^2}\frac{{{e^{ix}}}}{{1 - {e^{ix}}}}\ln (1 + {e^{ix}})\mathrm{d}x} \\ &= - 4\int_0^{\pi/2} {{x^2}(\cot x + i)\left[ {\ln (2\cos x) + ix } \right]\mathrm{d}x} \end{aligned}$$ Discard imaginary part: $$P + Q - {\pi ^2}\int_0^1 {\frac{{\ln (1 - x)}}{{1 + x}}\mathrm{d}x} = - 4\int_0^{\pi /2} {\left[ {{x^2}\cot x\ln (2\cos x) - {x^3}} \right]\mathrm{d}x} = -4I + \frac{\pi^4}{16}$$

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Thus it suffices to find $P$ and $Q$. If you're familiar with Euler sum or polylogarithm, these logarithm integrals of weight $4$ are standard. But nevertheless I will delineate how they're obtained. $$\begin{aligned} P &= \int_0^1 {\frac{{{{\ln }^2}x\ln (1 - {x^2})}}{{1 - x}}\mathrm{d}x} - \int_0^1 {\frac{{{{\ln }^2}x\ln (1 - x)}}{{1 - x}}\mathrm{d}x} \\ &= \int_0^1 {\left( {\frac{x}{{1 - {x^2}}} + \frac{1}{{1 - {x^2}}}} \right){{\ln }^2}x\ln (1 - {x^2})\mathrm{d}x} - \int_0^1 {\frac{{{{\ln }^2}x\ln (1 - x)}}{{1 - x}}\mathrm{d}x} \\ &= - \frac{7}{8}\int_0^1 {\frac{{{{\ln }^2}x\ln (1 - x)}}{{1 - x}}\mathrm{d}x} + \int_0^1 {\frac{{{{\ln }^2}x\ln (1 - {x^2})}}{{1 - {x^2}}}\mathrm{d}x} \\ &= - \frac{7}{8}\int_0^1 {\frac{{{{\ln }^2}x\ln (1 - x)}}{{1 - x}}\mathrm{d}x} + \frac{1}{8}\int_0^1 {{x^{ - 1/2}}\frac{{{{\ln }^2}x\ln (1 - x)}}{{1 - x}}\mathrm{d}x} \end{aligned}$$ Thus the value of $P$ can be calculated from the partial derivatives of $$\int_0^1 x^{a-1} (1-x)^{b-1} \mathrm{d}x = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ and the result is $$P = -\frac{19\pi^4}{720} + \frac{7}{2}\ln 2 \zeta(3)$$

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$Q$ does not succumb under such line of attack, indeed: $$Q = \frac{\pi^4}{90} + \frac{\pi^2 \ln^2 2}{6} - \frac{\ln^4 2}{6} - 4 \text{Li}_4 \left(\frac{1}{2}\right) $$

Denote $$A= \int_0^1 \frac{\ln x \ln^2(1-x)}{x} \mathrm{d}x \qquad B= \int_0^1 \frac{\ln x \ln^2(1+x)}{x} \mathrm{d}x \qquad C= \int_0^1 \frac{\ln x \ln(1+x) \ln(1-x)}{x} \mathrm{d}x $$ Due to the Euler sum $\sum H_n / n^3 = \pi^4 / 72$ (or via beta function as above), $A = -\pi^4 / 180$. Integration by parts on $C$ gives $$2C = P-Q$$ Also $$A + B + 2C = \int_0^1 {\frac{{\ln x{{\ln }^2}(1 - {x^2})}}{x}\mathrm{d}x} = \frac{1}{4}\int_0^1 {\frac{{\ln x{{\ln }^2}(1 - x)}}{x}\mathrm{d}x} = \frac{A}{4}$$ The reason for introducing these integrals is that $B$ is tamer than $Q$, its indefinite integral is a simple combination of polylogarithms up to order $4$, from which you can calculate: $$B = \frac{\pi^4}{24}+ \frac{\pi^2 \ln^2 2}{6} - \frac{\ln^4 2}{6} - 4\text{Li}_4(\frac{1}{2}) - \frac{7}{2}\ln 2 \zeta(3)$$

From these you can find the value of $Q$, hence finally $I$.

Answer 2

In the book, Almost Impossible Integrals, Sums and Series, page $247$ Eq $3.288$ we have

$$\cot x\ln(\cos x)=\sum_{n=1}^\infty(-1)^n\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac1n\right)\sin(2nx)$$

$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\sin(2nx),\quad 0<x<\frac{\pi}{2}$$

Thus,

$$\int_0^{\pi/2}x^2\cot x\ln(\cos x)dx=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\int_0^{\pi/2}x^2\sin(2nx)dx\right)$$

$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\frac{\cos(n\pi)}{4n^3}-\frac{3\zeta(2)\cos(n\pi)}{4n}-\frac{1}{4n^3}\right)$$

$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\frac{(-1)^n}{4n^3}-\frac{3\zeta(2)(-1)^n}{4n}-\frac{1}{4n^3}\right)$$

$$=\frac14\int_0^1\frac{1-t}{t(1+t)}\left(\sum_{n=1}^\infty\frac{t^n}{n^3}-\frac{3\zeta(2)t^n}{n}-\frac{(-t)^n}{n^3}\right)dt$$

$$=\frac14\int_0^1\left(\frac1t-\frac2{1+t}\right)\left(\text{Li}_3(t)+3\zeta(2)\ln(1-t)-\text{Li}_3(-t)\right)dt$$

$$=\frac14\underbrace{\int_0^1\frac{\text{Li}_3(t)-\text{Li}_3(-t)}{t}dt}_{\mathcal{I}_1}-\frac12\underbrace{\int_0^1\frac{\text{Li}_3(t)-\text{Li}_3(-t)}{1+t}dt}_{\mathcal{I}_2}$$ $$+\frac34\zeta(2)\underbrace{\int_0^1\frac{\ln(1-t)}{t}dt}_{\mathcal{I}_3}-\frac32\zeta(2)\underbrace{\int_0^1\frac{\ln(1-t)}{1+t}dt}_{\mathcal{I}_4}$$

$$\mathcal{I}_1=\text{Li}_4(1)-\text{Li}_4(-1)=\zeta(4)+\frac78\zeta(4)=\boxed{\frac{15}{8}\zeta(4)}$$

By integration by parts we have

$$\mathcal{I}_2=\frac74\ln(2)\zeta(3)-\int_0^1\frac{\ln(1+t)\text{Li}_2(t)}{t}dt+\int_0^1\frac{\ln(1+t)\text{Li}_2(-t)}{t}dt$$

$$=\frac74\ln(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 t^{n-1}\text{Li}_2(t)dt-\frac12\text{Li}_2^2(-t)|_0^1$$

$$=\frac74\ln(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n} \left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\frac5{16}\zeta(4)$$ $$=\frac74\ln(2)\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac5{16}\zeta(4)$$ substitute

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$

we get

$$\mathcal{I}_2=\boxed{-2\operatorname{Li_4}\left(\frac12\right)-\frac{25}{16}\zeta(4)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$

$$\mathcal{I}_3=-\text{Li}_2(1)=\boxed{-\zeta(2)}$$

$$\mathcal{I}_4=\int_0^1\frac{\ln(1-t)}{1+t}dt=\int_0^1\frac{\ln x}{2-x}dx=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln xdx$$ $$=-\sum_{n=1}^\infty\frac{1}{n^22^n}=-\text{Li}_2\left(\frac12\right)=\boxed{\frac12\ln^22-\frac12\zeta(2)}$$

Combine all boxed results we finally get

$$\int_0^{\pi/2}x^2\cot x\ln(\cos x)dx=\text{Li}_4\left(\frac12\right)-\frac18\zeta(4)-\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2)$$

Answer 3

By Fourier series $$ \log\cos x=\sum_{n\geq 1}\frac{1-\cos(2nx)}{n}(-1)^n \tag{A}$$ $$ \int_{0}^{\pi/2}x^2\cot(x)\log\cos(x)\,dx\\ = \int_{0}^{\pi/2}\left(\frac{\pi^2}{12}+\sum_{n\geq 1}\frac{(-1)^n \cos(2nx)}{n^2}\right)\sum_{n\geq 1}\frac{\cos(x)-\cos(2nx)\cos(x)}{n\sin(x)}(-1)^n\,dx\tag{B}$$ so the problem boils down to the evaluation of some alternating Euler sums.

Answer 4

Let $$ I = \sum_{n=1}^{\infty} \frac{2^{2n} H_n}{n^3 \binom{2n}{n}} $$ from here we have

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

replace $x$ by $\sqrt{x}$ we get $$\sum_{n=1}^\infty\frac{2^{2n}x^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$

multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=2\int_0^1\frac{\arcsin^2(\sqrt{x})\ln(1-x)}{x}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx$$

Substitute $\tan x = t$, then the integral becomes $$ I = 4 \int_0^{\infty} \frac{\arctan^2(t) \ln(1 + t^2)}{t(1 + t^2)} \, dt = 4A , $$ where $$ A = \int_0^{\infty} \frac{\arctan^2(t) \ln(1 + t^2)}{t(1 + t^2)} \, dt . $$

Now observe: $$ \ln(1 + ix) = \frac{1}{2} \ln(1 + x^2) + i \arctan(x) . $$ Cubing both sides and taking the real part: $$ \text{Re}\{\ln^3(1 + ix)\} = \frac{1}{8} \ln^3(1 + x^2) - \frac{3}{2} \ln(1 + x^2) \arctan^2(x) . $$

Hence, $$ A = -\frac{2}{3} \, \text{Re} \int_0^{\infty} \frac{\ln^3(1 + ix)}{x(1 + x^2)} \, dx + \frac{1}{12} \int_0^{\infty} \frac{\ln^3(1 + x^2)}{x(1 + x^2)} \, dx . $$

Define: $$ A = -\frac{2}{3} \, \text{Re}(B) + \frac{1}{12} C , $$ where $$ B = \int_0^{\infty} \frac{\ln^3(1 + ix)}{x(1 + x^2)} \, dx . $$

Let $\frac{1}{1 + ix} = t$, so we obtain: $$ B = \int_0^1 \frac{t \ln^3(t)}{(1 - t)(1 - 2t)} \, dt . $$

Using partial fractions: $$ B = \int_0^1 \frac{\ln^3(t)}{1 - 2t} \, dt - \int_0^1 \frac{\ln^3(t)}{1 - t} \, dt , $$ $$ B = 6 \zeta(4) - 3 \mathrm{Li}_4(2) . $$

From:

A. S. Olaikhan, An Introduction to the Harmonic Series and Logarithmic Integrals: For High School Students Up To Researchers, 2nd ed., self-published, Phoenix, AZ, 2023,

we have the identity: $$ \mathrm{Li}_4(2) = -\mathrm{Li}_4\left(\frac{1}{2}\right) + 2\zeta(4) + \ln^2(2)\zeta(2) - \frac{1}{24} \ln^4(2) - \frac{\pi}{6} \ln^3(2)i . $$

Also, define: $$ C = \int_0^{\infty} \frac{\ln^3(1 + x)}{x(1 + x)} \, dx . $$

Split the integral $C$ at $x = 1$: \begin{align*} C &= \int_0^1 \frac{\ln^3(1 + x)}{x} \, dx + 3 \int_0^1 \frac{\ln^2(x) \ln(1 + x)}{1 + x} \, dx \\ &{}-{} 3 \int_0^1 \frac{\ln^2(1 + x) \ln(x)}{1 + x} \, dx - \int_0^1 \frac{\ln^3(x)}{1 + x} \, dx . \end{align*}

Define: $$ \begin{aligned} I_1 &= \int_0^1 \frac{\ln^3(1 + x)}{x} \, dx , \\ I_2 &= \int_0^1 \frac{\ln^2(x) \ln(1 + x)}{1 + x} \, dx , \\ I_3 &= \int_0^1 \frac{\ln^2(1 + x) \ln(x)}{1 + x} \, dx , \\ I_4 &= \int_0^1 \frac{\ln^3(x)}{1 + x} \, dx . \end{aligned} $$

Using integration by parts on $I_3$, we get: $$ 3I_3 + I_1 = 0 \quad \Rightarrow \quad I_3 = -\frac{1}{3} I_1 . $$

So, $$ C = 2I_1 + 3I_2 - I_4 . $$

From Olaikhan again: $$ \begin{aligned} I_1 &= 6 \zeta(4) - \frac{21}{4} \ln(2) \zeta(3) + \frac{3}{2} \ln^2(2) \zeta(2) - \frac{1}{4} \ln^4(2) - 6 \mathrm{Li}_4\left(\frac{1}{2}\right) , \\ I_2 &= 4 \mathrm{Li}_4\left(\frac{1}{2}\right) - \frac{15}{4} \zeta(4) + \frac{7}{2} \ln(2) \zeta(3) - \ln^2(2)\zeta(2) + \frac{1}{6} \ln^4(2) , \\ I_4 &= -\frac{21}{4} \zeta(4) . \end{aligned} $$

Now, return to: $$ I = 4A = -\frac{8}{3} \text{Re}\{B\} + \frac{1}{3} C = -\frac{8}{3} \text{Re}\{B\} + \frac{2}{3} I_1 + I_2 - \frac{1}{3} I_4 . $$

Hence, we arrive at: $$ \sum_{n=1}^{\infty} \frac{2^{2n} H_n}{n^3 \binom{2n}{n}} = -8 \, \mathrm{Li}_4\left(\frac{1}{2}\right) + \zeta(4) + 8 \ln^2(2) \zeta(2) - \frac{1}{3} \ln^4(2) . $$ And hence, dividing both sides by by -8 $$ \int_0^{\frac{\pi}{2}} x^2 \cot x \ln(\cos x) \, dx = \operatorname{Li}_4\left(\frac{1}{2}\right) - \frac{1}{8} \zeta(4) - \ln^2(2)\zeta(2) + \frac{1}{24} \ln^4(2) $$