# Interesting integral some how related to $\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$

I hope I used the correct tag (definite integral).

I ran across an integral that is rather tough and I am wondering if anyone could give me a shove in the right direction.

$\displaystyle\int_{0}^{1}\frac{\tanh^{-1}(x)\ln(x)}{x(1-x^{2})}\text{ d}x=\frac{-7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$

This solution is almost exactly like the solution to $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x$, which is $\displaystyle \frac{7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$

I solved the latter integral by using the identity $\displaystyle -\ln(\sin(x))-\ln(2)=\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}$, then integrating:

$\displaystyle -\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x-\frac{{\pi}^{2}}{8}\ln(2)=\int_{0}^{\frac{\pi}{2}}x\cos(2x)\text{ d}x+\frac{\int_{0}^{\frac{\pi}{2}}\cos(4x)\text{ d}x}{2}+\frac{\int_{0}^{\frac{\pi}{2}}\cos(6x)}{3}\cdot\cdot\cdot\cdot$

But, $\displaystyle \int_{0}^{\frac{\pi}{2}}x\cdot \cos(2kx)\text{ d}x=-\left(\frac{1+(-1)^{k+1}}{(2k)^{2}}\right)$

Thus: $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1+(-1)^{k+1}}{k^{3}}-\frac{{\pi}^{2}}{8}\ln(2)$

$\displaystyle =\frac{1}{2}\displaystyle\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{3}}-\frac{{\pi}^{2}}{8}\ln(2)$

and so on. Which results in the solution I mentioned in the beginning.

Sorry for all that, but I wanted to show you what I was using in order to some how relate it to the integral I am wanting to solve. I have been trying and trying to relate the aforementioned $\displaystyle \tanh$ integral with this one. The solutions are so nearly the same, I figured there has to be a way to relate them and solve the integral. Does anyone have some ideas?. I have tried the identity $\displaystyle \tanh^{-1}(x)=\frac{1}{2}\left[\ln(1+x)-\ln(1-x)\right]$, then breaking it up:

Resulting in $\displaystyle \frac{1}{2}\int_{0}^{1}\frac{\ln(x)\ln(1+x)}{x(x^{2}-1)}\text{ d}x-\frac{1}{2}\int_{0}^{1}\frac{\ln(x)\ln(1-x)}{x(x^{2}-1)} \text{ d}x$, then I used the various series representations for $\displaystyle \ln(1+x)$, $\displaystyle \frac{1}{1-x^{2}}$, etc. I tried double integrals, but I always get stuck.

I even broke it up per partial fraction expansion, but several of the resulting integrals were still nasty.

Does anyone have some clever ideas?.

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## 1 Answer

A straightforward way could be to consider the function $$I(a,b)=\int_0^1 \frac{\left(\frac{1+x}{1-x}\right)^a x^b}{(1+x)^2} \, dx=$$ $$\Gamma (1-a) \Gamma (b+1) \, _2\tilde{F}_1(2-a,b+1;-a+b+2;-1),\quad a>1,\ b>-1,$$ where $_2\tilde{F}_1$ is the regularized hypergeometric function (see the first integral representation in the reference). To take the second derivative $\frac {\partial^2I(a,b)}{\partial a\partial b}$ and manually evaluate the limit $a\to1+0$, $b\to-1+0$, which gives the desired integral.

Updated

Here is another idea. The integral $\int_{0}^{\frac{\pi}{2}}x\log(\sin(x))dx$ can be evaluated in the same way as the Gauss integral $I=\int_{0}^{\frac{\pi}{2}}\log \sin x\,dx\;$. Namely, making change of variables $y=\pi/2-x\;$ we have $I=\int_{0}^{\frac{\pi}{2}}\log \cos x\, dx$, so $$I=\frac12 \int_{0}^{\frac{\pi}{2}}\log \frac12\sin 2x\, dx=\frac14 \int_{0}^{\pi}\log \frac12\sin x\, dx=\frac12 \int_{0}^{\frac{\pi}{2}}\log \frac12\sin x\, dx,$$ which lead to an equation on $I$ etc.

Now for the function $f(x)=\frac12\frac{\log (|x|) \log \left(\left|\frac{x+1}{1-x}\right|\right)}{2 x \left(1-x^2\right)}$ there are two changes of variables leaving in place the logarithms in $f$:

1) $y=\frac{1-x}{1+x}$,

2) $y=1/x$.

The first one can be regarded as analogue to $x\to \pi/2-y$ and the second transform the integral segment to $[1,+\infty)$ which perhaps corresponds to integrating on $[\pi/2,\pi]$. May be combining whose observations would lead to the desired result. For example, the first one leads to $$\int_0^1f(x)\,dx=\int_0^1 \frac{(y+1) \log (y) \log \left(\frac{1+y}{1-y}\right)}{4 (1-y) y}\,dx,$$ and denoting $$I_1=\int_0^1f(x)\,dx=\frac{1}{16} \left(-7 \zeta (3)-\pi ^2 \log (4)\right),$$ $$I_2=\int_1^\infty f(x)\,dx=\frac{1}{16} \left(7 \zeta (3)-\pi ^2 \log (4)\right),$$ we have $I_1+I_2=-\frac{1}{4} \pi ^2 \log (2)\;$, $I_1-I_2=-\frac{7 \zeta (3)}{8}\;$.

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Thank you. I am not that familiar with hypergeometric functions. But, I will study it. I was thinking perhaps I could do some series manipulation, subs, or what not, but apparently it's not that easy. Somehow $\int_{0}^{\frac{\pi}{2}}xln(sin(x))dx-\frac{7}{8}\zeta(3)=\int_{0}^{1}\frac{tan‌​h^{-1}(x)ln(x)}{x(1-x^{2})}dx$. I do not know how I managed to see that in the first place:) – Cody Jul 29 '11 at 16:52
@Cody: users are always allowed to comment on their own questions, without reputation requirements. But in your case, since you are using an unregistered account, sometimes the software easily lose track of who you are (if you switch computers/browsers/IP addresses). Please consider registering your account. (I've merged your two unregistered account that appeared on this question.) – Willie Wong Jul 30 '11 at 2:41
Wow, you're a genius Andrew. Thank you. By the way, I registered my account. Sorry for any inconveniences. – Cody Jul 30 '11 at 10:42