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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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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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  • $\begingroup$ 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 $\int_{0}^{\frac{\pi}{2}}xln(sin(x))dx-\frac{7}{8}\zeta(3)=\int_{0}^{1}\frac{tanh^{-1}(x)ln(x)}{x(1-x^{2})}dx$. I do not know how I managed to see that in the first place:) $\endgroup$
    – Cody
    Jul 29 '11 at 16:52
  • $\begingroup$ @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.) $\endgroup$ Jul 30 '11 at 2:41
  • $\begingroup$ Wow, you're a genius Andrew. Thank you. By the way, I registered my account. Sorry for any inconveniences. $\endgroup$
    – Cody
    Jul 30 '11 at 10:42
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Unfortunately I have no clever ideas to evaluate your integral. Instead I will follow your suggested line of attack and write $$\tanh^{-1} (x) = \frac{1}{2} \ln \left (\frac{1 + x}{1 - x} \right ).$$

Doing so the integral becomes \begin{align*} I &= \frac{1}{2} \int_0^1 \ln \left (\frac{1 + x}{1 - x} \right ) \frac{\ln x}{x(1 - x^2)} \, dx\\ &= \frac{1}{2} \int_0^1 \ln \left (\frac{1 - x^2}{(1 - x)^2} \right ) \frac{\ln x}{x(1 - x^2)} \, dx\\ &= \frac{1}{2} \int_0^1 \frac{\ln (1 - x^2) \ln x}{x(1 - x^2)} \, dx - \int_0^1 \frac{\ln (1 - x) \ln x}{x (1 - x^2)} \, dx\\ &= \frac{1}{2} I_1 - I_2. \end{align*}


The integral $I_1$

Enforcing the substitution $x \mapsto \sqrt{x}$ gives $$I_1 = \frac{1}{4} \int_0^1 \frac{\ln (1 - x) \ln x}{x(1 - x)} \, dx.$$ Making use of the generating function for the harmonic numbers $H_n$, namely $$\sum_{n = 1}^\infty H_n x^n = - \frac{\ln (1 - x)}{1 - x},$$ we have $$I_1 = -\frac{1}{4} \sum_{n = 1}^\infty \int_0^1 x^{n - 1} \ln x \, dx.$$

Noting that $$\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2},$$ a result that can be established using integration by parts, the integral becomes $$I_1 = \frac{1}{4} \sum_{n = 1}^\infty \frac{H_n}{n^2}.$$

The resulting sum, known as an Euler sum can be readily found (see here for example). Its value is $$\sum_{n = 1}^\infty \frac{H_n}{n^2} = 2 \zeta (3).$$ Thus $$I_1 = \int_0^1 \frac{\ln (1 - x^2) \ln x}{x(1 - x^2)} \, dx = \frac{1}{2} \zeta (3).$$


The integral $I_2$

From a partial fraction decomposition of $$\frac{1}{x(1 - x^2)} = \frac{1}{x} + \frac{1}{2(1 - x)} + \frac{1}{2(1 + x)},$$ the integral for $I_2$ can be written as \begin{align*} I_2 &= \int_0^1 \frac{\ln (1 - x) \ln x}{x} \, dx + \frac{1}{2} \int_0^1 \frac{x \ln x \ln (1 - x)}{1 - x} \, dx + \frac{1}{2} \int_0^1 \frac{x \ln x \ln (1 - x)}{1 + x} \, dx\\ &= I_\alpha + \frac{1}{2} I_\beta + \frac{1}{2} I_\gamma. \end{align*}

The first two integrals are relatively easy the find, the third is more difficult.

For the first \begin{align*} I_\alpha &= \int_0^1 \frac{\ln (1 - x) \ln x}{x} \, dx\\ &= -\text{Li}_2 (x) \ln x \Big{|}^1_0 + \int_0^1 \frac{\text{Li}_2 (x)}{x} \, dx\\ &= \int_0^1 \frac{\text{Li}_2 (x)}{x} \, dx\\ &= \text{Li}_3 (x) \Big{|}_0^1 = \text{Li}_3 (1) = \zeta (3). \end{align*}

For the second \begin{align*} I_\beta &= \int_0^1 \frac{x \ln x \ln (1- x)}{1- x} \, dx, \end{align*} we again make use of the generating function for the harmonic numbers. Doing so we have $$I_\beta = - \sum_{n = 1}^\infty H_n \int_0^1 x^{n + 1} \ln x \, dx.$$

By parts, it can be shown that $$\int_0^1 x^{n + 1} \ln x \, dx = -\frac{1}{(n + 2)^2},$$ thus $$I_\beta = \sum_{n = 1}^\infty \frac{H_n}{(n + 2)^2}.$$

To find the resulting sum we begin by shifting the summation index $n \mapsto n - 1$. Thus $$I_\beta = \sum_{n = 2}^\infty \frac{H_{n - 1}}{(n + 1)^2}.$$ Now from properties for the harmonic numbers $$H_n = H_{n - 1} + \frac{1}{n}. \tag1$$ Thus \begin{align*} I_\beta &= \sum_{n = 2}^\infty \frac{H_n}{(n + 1)^2} - \sum_{n = 2}^\infty \frac{1}{n(n + 1)^2}\\ &= \sum_{n = 1}^\infty \frac{H_n}{(n + 1)^2} - \sum_{n = 1}^\infty \frac{1}{n(n + 1)^2}. \end{align*}

In the first sum, shifting the summation index again by $n \mapsto n - 1$ and applying property (1) resulting in \begin{align*} \sum_{n = 1}^\infty \frac{H_n}{(n + 1)^2} &= \sum_{n = 2}^\infty \frac{H_n}{n^2} - \sum_{n = 2}^\infty \frac{1}{n^3} = \sum_{n = 1}^\infty \frac{H_n}{n^2} - \sum_{n = 1}^\infty \frac{1}{n^3} = 2 \zeta (3) - \zeta (3) = \zeta (3). \end{align*}

For the second sum, as $$\frac{1}{n(n + 1)^2} = \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{(n + 1)^2},$$ we have \begin{align*} \sum_{n = 1}^\infty \frac{1}{n(n + 1)^2} &= \sum_{n = 1}^\infty \left (\frac{1}{n} - \frac{1}{n + 1} \right ) - \sum_{n = 1}^\infty \frac{1}{(n + 1)^2}. \end{align*} The first sum telescopes to $1$. For the second sum, it is $$\sum_{n = 1}^\infty \frac{1}{(n + 1)^2} = \sum_{n = 2}^\infty \frac{1}{n^2} = \sum_{n = 1}^\infty \frac{1}{n^2} - 1 = \zeta (2) - 1.$$ So $$\sum_{n = 1}^\infty \frac{1}{n(n + 1)^2} = 1 - (\zeta (2) - 1) = 2 - \zeta (2),$$ giving $$I_\beta = \zeta (3) - (2 - \zeta (2)) = \zeta (3) + \zeta (2) - 2.$$


The integral $I_\gamma$

Now for the difficult one. To evaluate it we will use double infinite sums. Since $$\frac{\ln (1 - x)}{1 + x} = -\sum_{k = 0}^\infty \sum_{n = 0}^\infty \frac{(-1)^k}{n + 1} x ^{x + k + 1},$$ the integral can be rewritten as $$I_\gamma = - \sum_{k = 0}^\infty \sum_{n = 0}^\infty \frac{(-1)^k}{n + 1} \int_0^1 x^{n + k + 2} \ln x \, dx.$$ As $$\int_0^1 x^{n + k + 2} \ln x \, dx = - \frac{1}{(n + k + 3)^2},$$ we can write $$I_\gamma = \sum_{k = 0}^\infty \sum_{n = 1}^\infty \frac{(-1)^k}{(n + 1)(n + k + 3)^2}.$$

By a partial fraction decomposition we have $$\frac{1}{(n + 1)(n + k + 3)^2} = \frac{1}{(k + 2)^2(n + 1)} - \frac{1}{(k + 2)^2 (k + n + 3)} - \frac{1}{(k + 2)(k + n + 3)^2}.$$ Now for the inner infinite sum over $n$ we have \begin{align*} \sum_{n = 0}^\infty \frac{1}{(n + 1)(n + k + 3)^2} &= \sum_{n = 0}^\infty \left [\frac{1}{(k + 2)^2(n + 1)} - \frac{1}{(k + 2)^2 (k + n + 3)} - \frac{1}{(k + 2)(k + n + 3)^2} \right ]\\ &= \frac{1}{(k + 2)^2} \sum_{n = 1}^{k + 2} \frac{1}{n} - \frac{1}{k + 2} \sum_{n = k + 3}^\infty \frac{1}{n^2}\\ &= \frac{1}{(k + 2)^2} H_{k + 2} - \frac{1}{k + 2} \sum_{n = 1}^\infty \frac{1}{n^2} + \frac{1}{k + 2} \sum_{n = 1}^{k + 2} \frac{1}{n^2}\\ &= \frac{H_{k + 2}}{(k + 2)^2} - \frac{\zeta (2)}{k + 2} + \frac{H^{(2)}_{k + 2}}{k + 2}. \end{align*} Here $H^{(a)}_n$ denotes the Generalised Harmonic numbers.

Thus $$I_\gamma = \sum_{k = 0}^\infty \frac{(-1)^k H_{k + 2}}{(k + 2)^2} - \zeta (2) \sum_{k = 0}^\infty \frac{(-1)^k}{k + 2} + \sum_{k = 0}^\infty \frac{(-1)^k H^{(2)}_{k + 2}}{k + 2}.$$

For the first sum, let $k \mapsto k - 2$, then $$S_1 = \sum_{k = 2}^\infty \frac{(-1)^k H_k}{k^2} = 1 + \sum_{k = 1}^\infty \frac{(-1)^k H_k}{k^2} = 1 + A(1,2).$$

Here $$A(p,q) = \sum_{k = 1}^\infty \frac{(-1)^{k + 1} H^{(p)}_k}{k^q},$$ correspond to the alternating Euler sums whose first few values can be found here. Since $A(1,2) = \frac{5}{8} \zeta (3)$ we have $$S_1 = 1 - \frac{5}{8} \zeta (3).$$

For the second sum $$S_2 = \sum_{k = 0}^\infty \frac{(-1)^k}{k + 2} = -\sum_{k = 1}^\infty \frac{(-1)^{k + 1}}{k} + 1 = -\ln (2) + 1.$$

For the third sum, let $k \mapsto k - 2$, then $$S_3 = \sum_{k = 2}^\infty \frac{(-1)^k H^{(2)}_k}{k} = 1 - \sum_{k = 1}^\infty \frac{(-1)^{k + 1} H^{(2)}_k}{k} = 1 - A(2,1).$$ And as $A(2,1) = \zeta (3) - \frac{1}{2} \zeta (2) \ln (2)$, we have $$S_3 = 1 + \frac{1}{2} \zeta (2) \ln (2) - \zeta (3).$$

So finally \begin{align*} I_\gamma &= \left (1 - \frac{5}{8} \zeta (3) \right ) - \zeta (2) \left (1 - \ln (2) \right ) + \left (1 + \frac{1}{2} \zeta (2) \ln (2) - \zeta (3) \right )\\ &= -\frac{13}{8} \zeta (3) + 2 - \zeta (2) + \frac{3}{2} \zeta (2) \ln (2). \end{align*}


So on combining all the results found we have \begin{align*} I_2 &= I_\alpha + \frac{1}{2} I_\beta + \frac{1}{2} I_\gamma\\ &= \zeta (3) + \frac{1}{2} \left (\zeta (3) + \zeta (2) - 2 \right ) + \frac{1}{2} \left (-\frac{13}{8} \zeta (3) + 2 - \zeta (2) + \frac{3}{2} \zeta (2) \ln (2) \right )\\ &= \frac{11}{16} \zeta (3) + \frac{3}{4} \zeta (2) \ln (2). \end{align*} And \begin{align*} I &= \frac{1}{2} I_1 - I_2\\ &= \frac{1}{4} \zeta (3) - \left (\frac{11}{16} \zeta (3) + \frac{3}{4} \zeta (2) \ln (2) \right )\\ &= -\frac{7}{16} \zeta (3) - \frac{3}{4} \zeta (2) \ln (2), \end{align*} or $$\int_0^1 \frac{\tanh^{-1} (x) \ln (x)}{x(1 - x^2)} \, dx = -\frac{7}{16} \zeta (3) - \frac{\pi^2}{8} \ln (2),$$ as required.

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  • $\begingroup$ Lots of stuff here. Good job. $\endgroup$
    – edmz
    Jan 11 '18 at 13:00

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