How to evaluate $\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\:dx$

Question

How can i evaluate $$\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\:dx$$ I started like this $$\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\:dx=\frac{x^2\ln \left(\sin \left(x\right)\right)}{2}|^{\frac{\pi }{2}}_0-\frac{1}{2}\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\:dx$$ but this way doesnt turn things any simpler, i also tried using the substitution $t=\tan \left(\frac{x}{2}\right)$ and got this, $$4\int _0^{1}\arctan \left(t\right)\ln \left(\frac{2t}{1+t^2}\right)\:\frac{1}{1+t^2}\:dt$$ $$=4\ln \left(2\right)\int _0^{1}\frac{\arctan \left(t\right)}{1+t^2}\:dt+4\int _0^{1}\frac{\arctan \left(t\right)\ln \left(t\right)}{1+t^2}\:dt-4\int _0^{1}\frac{\arctan \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt$$ That first integral is very simple but the rest look very difficult, could you help me evaluate this one?

Answer 1

$$\int_0^{\pi/2}x\ln(\sin x)dx=\int_0^{\pi/2}x\left(-\ln2-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}\right)dx$$

$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\int_0^{\pi/2}x\cos(2nx)dx$$

$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\left(\frac{\cos(n\pi)}{4n^2}+\frac{\pi\sin(n\pi)}{4n}-\frac{1}{4n^2}\right)$$

$$=-\frac{\pi^2}{8}\ln2-\sum_{n=1}^\infty\frac{1}{n}\left(\frac{(-1)^n}{4n^2}+\frac{0}{4n}-\frac{1}{4n^2}\right)$$

$$=-\frac{\pi^2}{8}\ln2-\frac14\text{Li}_3(-1)+\frac14\zeta(3)$$

$$=-\frac{\pi^2}{8}\ln2+\frac{7}{16}\zeta(3)$$

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Bonus: With subbing $x\to \pi/2-x$ we have

$$\int_0^{\pi/2}x\ln(\cos x)dx=\int_0^{\pi/2}(\pi/2-x)\ln(\sin x)dx$$

$$=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin x)dx-\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=\frac{\pi}{2}\left(-\frac{\pi}{2}\ln2\right)-\left(-\frac{\pi^2}{8}\ln2+\frac{7}{16}\zeta(3)\right)$$ $$=-\frac{\pi^2}{8}\ln(2)-\frac7{16}\zeta(3)$$

Or we can use the Fourier series of $\ \ln(\cos x)=-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$.

Also by subtracting the two integrals gives

$$\int_0^{\pi/2}x\ln(\tan x)dx=\frac78\zeta(3)$$

Or we can use the Fourier series of $\ \ln(\tan x)=-2\sum_{n=1}^\infty\frac{\cos((4n-2)x)}{2n-1}.$

Answer 2

Define on $[0;\infty[$ the function $R$ by,

for all $x\in [0;\infty[$, $\displaystyle \text{R}(x)=\int_0^x \dfrac{\ln t}{1+t^2}\,dt=\int_0^1 \dfrac{x\ln(tx)}{1+t^2x^2}\,dt$.

Observe that $\text{R}(0)=\lim_{x\rightarrow}\text{R}(x)=0$ \begin{align} A_2&=\int_0^{\frac{\pi}{2}}t\ln(\cos t)\,dt\\ B_2&=\int_0^{\frac{\pi}{2}}t\ln(\sin t)\,dt\\ A_2+B_2&=\int_0^{\frac{\pi}{2}}t\ln\left(\frac{1}{2}\sin(2t)\right)\,dt\\ &=\int_0^{\frac{\pi}{2}}t\ln\left(\sin(2t)\right)\,dt-\frac{\pi^2\ln 2}{8}\\ &\overset{x=2t}=\frac{1}{4}\int_0^\pi x\ln(\sin x)\,dx-\frac{\pi^2\ln 2}{8}\\ &\overset{t=\pi-x}=\frac{1}{4}\int_0^\pi (\pi-x)\ln(\sin x)\,dx-\frac{\pi^2\ln 2}{8}\\ 2(A_2+B_2)&=\frac{\pi}{4}\int_0^\pi \ln(\sin x)\,dx-\frac{\pi^2\ln 2}{4}\\ A_2+B_2&=\frac{\pi}{8}\int_0^\pi \ln(\sin x)\,dx-\frac{\pi^2\ln 2}{8}\\ &=-\frac{\pi^2\ln 2}{4}\\ B2-A2&=\int_0^{\frac{\pi}{2}}t\ln(\tan t)\,dt\\ &\overset{x=\tan t}=\int_0^\infty \frac{\ln x\arctan x}{1+x^2}\,dx\\ \end{align} \begin{align} U_2&=\int_0^\infty \frac{\arctan\left(\frac{1}{x}\right)\ln x}{1+x^2}\,dx\\ V_2&=\int_0^\infty \frac{\arctan\left(x\right)\ln x}{1+x^2}\,dx\\ U_2+V_2&=\frac{\pi}{2}\int_0^\infty \frac{\ln x}{1+x^2}\,dx\\ &=0\\ U_2&\overset{\text{IBP}}=\left[R(x)\arctan\left(\frac{1}{x}\right)\right]_0^\infty +\int_0^\infty \frac{R(x)}{1+x^2}\,dx\\ &=\int_0^\infty \left(\int_0^1 \dfrac{x\ln(tx)}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx\\ &=\int_0^\infty \left(\int_0^1 \dfrac{x\ln(x)}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx+\\ &\int_0^1 \left(\int_0^\infty \dfrac{x\ln(t)}{(1+t^2x^2)(1+x^2)}\,dx\right)\,dt\\ &=V_2+\int_0^1 \frac{\ln^2 t}{t^2-1}\,dt\\ &=V_2+\int_0^1 \frac{t\ln^2 t}{1-t^2}\,dt-\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &\overset{u=t^2}=B+\frac{1}{8}\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=V_2-\frac{7}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=V_2-\frac{7}{8}\times 2\zeta(3)\\ &=V_2-\frac{7}{4}\zeta(3)\\ U_2&=-\frac{7}{8}\zeta(3)\\ V_2&=\frac{7}{8}\zeta(3)\\ B_2-A_2&=\frac{7}{8}\zeta(3)\\ A_2&=-\frac{7}{16}\zeta(3)-\frac{1}{8}\pi^2\ln 2\\ B_2&=\boxed{\frac{7}{16}\zeta(3)-\frac{1}{8}\pi^2\ln 2}\\ \end{align}

NB: I assume following results: \begin{align} \int_0^\infty \frac{\ln x}{1+x^2}\,dx&=0\\ \int_0^1 \frac{\ln^2 x}{1-x}\,dx&=2\zeta(3)\\ \int_0^\pi \ln(\sin x)\,dx&=-\pi\ln 2 \end{align}

Answer 3

An incomplete solution requiring some more interesting work:

$$I=\int_{0}^{\pi/2} x \ln (\sin x) dx= \int_{0}^{1} \ln t ~\frac{\sin^{-1} t}{\sqrt{1-t^2}} dt.$$ Using the MacLaurin series for $$\frac{\sin^{-1} t}{\sqrt{1-t^2}}= \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} t^{2n+1}$$

See: Deriving Maclaurin series for $\frac{\arcsin x}{\sqrt{1-x^2}}$. Then $$I=\sum_{n=0}^{\infty} \int_{0}^{1}\ln t ~\frac{(2n)!!}{(2n+1)!!} t^{2n+1}=-\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!}~ \int_{0}^{\infty}u~e^{-(2n+2)u}~du~~( t=e^{-u})$$ $$\implies I=-\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \frac{1}{(2n+2)^2}$$ I have numerically confirmed using Mathematica that $I$ is nothing but $$\frac{1}{16}[-\pi^2 \ln 4+7 \zeta(3)]$$ The same as obrained @Ali Shather in his very nice solution above. Can some one fill the gap here! I may come back.

Answer 4

Assuming that you could enjoy polylogarithms, the antiderivative does exist (have a look here) $$I=\int x\log \left(\sin \left(x\right)\right)\:dx$$ Using the bounds, the results are

• at $\frac \pi 2$, $\frac{1}{48} \left(9 \zeta (3)+i \pi ^3-6 \pi ^2 \log (2)\right)$
• at $0$, $\frac{1}{48} \left(i \pi ^3-12 \zeta (3) \right)$

and, then, the result.

Answer 5

This solution is based on Cauchy's integral theorem.

Integrate \begin{equation*} f(z) = \log(z)\dfrac{\log(1-z^2)}{z} \end{equation*} where \begin{equation*} \log(z)=\ln|z|+i\arg(z), \qquad -\pi<\arg(z)<\pi, \end{equation*} over the boundary $\gamma$ of the unit circle in the first quadrant. Let $\gamma = \gamma_1+\gamma_2+\gamma_3$. Here \begin{alignat*}{1} \gamma_1(x)&=x,\, 0\le x\le 1\\ \gamma_2(t)&=e^{it}, \, 0 \le t \le {\pi}/2\\ \gamma_3(y)&=iy,\, y \mbox{ from } 1 \mbox{ to } 0. \end{alignat*} From Cauchy's integral theorem we get \begin{gather*} 0 = \int_{\gamma_1}f(z)\,\mathrm{d}z + \int_{\gamma_2}f(z)\,\mathrm{d}z + \int_{\gamma_3}f(z)\,\mathrm{d}z=\\[2ex] \int_{0}^{1}\ln(x)\dfrac{\log(1-x^2)}{x} \,\mathrm{d}x+ \int_{0}^{\pi/2}\log(e^{it})\dfrac{\log(1-e^{i2t})}{e^{it}}ie^{it} \,\mathrm{d}t- \int_{0}^{1}\log(iy)\dfrac{\log(1+y^2)}{iy}i\, \mathrm{d}y=\\[2ex] \int_{0}^{1}\ln(x)\dfrac{\log(1-x^2)}{x} \,\mathrm{d}x+ \int_{0}^{\pi/2}i^2t\left(\ln(2\sin(t))+i\arg\left(1-e^{i2t}\right)\right)\,\mathrm{d}t-\\[2ex] \int_{0}^{1}\left(\ln(y)+i\dfrac{\pi}{2}\right)\dfrac{\log(1+y^2)}{y}\, \mathrm{d}y. \end{gather*} We extract the real part of every integral. \begin{gather*} 0 = -\int_{0}^{1}\left(\sum_{k=1}^{\infty}\dfrac{x^{2k-1}\ln(x)}{k}\right)\, \mathrm{d}x - \int_{0}^{\pi/2}t\ln(2)\, \mathrm{d}t -\int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t-\\[2ex] -\int_{0}^{1}\left(\sum_{k=1}^{\infty}(-1)^{k-1}\dfrac{y^{2k-1}\ln(y)}{k}\right)\, \mathrm{d}y =\\[2ex] \sum_{k=1}^{\infty}\dfrac{1}{4k^3}-\dfrac{\pi^2}{8}\ln(2) -\int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t +\sum_{k=1}^{\infty}\dfrac{(-1)^{k-1}}{4k^3}=\\[2ex] \dfrac{1}{4}\zeta(3) -\dfrac{\pi^2}{8}\ln(2) -\int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t + \dfrac{3}{16}\zeta(3). \end{gather*} Consequently \begin{equation*} \int_{0}^{\pi/2}t\ln(\sin(t))\, \mathrm{d}t = \dfrac{7}{16}\zeta(3)-\dfrac{\pi^2}{8}\ln(2). \end{equation*}

Answer 6

Using the trapezioidal rule like numerical integration:

$$\displaystyle{\int \limits _{1}^{\frac \pi2}x\ln(\sin (x))\,dx\approx \frac{1}{2}h\left(f(1)+f\left(\frac\pi2\right)\right)=-0.03672410\\h=\frac \pi2-1}$$

I think that this is a short way to find the value of the $ \int \limits _{1}^{\frac \pi2}x\ln(\sin (x))\,dx$.