Interesting integral $\int_{0}^{1}\frac{\tanh^{-1}x\ln x}{x(1-x^{2})}{ d}x$

Question

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

$$\int_{0}^{1}\frac{\tanh^{-1}x\ln x}{x(1-x^{2})}{ d}x=-\frac{7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$$ This solution is almost exactly like the solution to $\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x=\frac{7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$

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?.

Answer 1

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 $I_1+I_2=-\frac{1}{4} \pi ^2 \log (2)\;$, $I_1-I_2=-\frac{7 \zeta (3)}{8}\;$.

Answer 2

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*}

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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 $$I_1 = \int_0^1 \frac{\ln (1 - x^2) \ln x}{x(1 - x^2)} \, dx = \frac{1}{2} \zeta (3).$$

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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 $$I_\beta = \zeta (3) - (2 - \zeta (2)) = \zeta (3) + \zeta (2) - 2.$$

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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*}

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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.

Answer 3

Split the integral as follows $$\int_{0}^{1}\frac{\tanh^{-1}x\ln x}{x(1-x^{2})}dx \overset{x\to\frac{1-x}{1+x}} =\int_{0}^{1}\frac{\tanh^{-1}x\ln x}{2x} + \frac{\tanh^{-1}x\ln x}{1-x}\ { d}x $$ whete \begin{align} & \int_{0}^{1}\frac{\tanh^{-1}x\ln x}{2x}\overset{ibp} =-\frac14\int_0^1 \frac{\ln^2x}{1-x^2}dx=-\frac7{16}\zeta(3)\\ \\ &\int_{0}^{1}\frac{\tanh^{-1}x\ln x}{1-x}\ { d}x \\ =& \int_{0}^{1}\tanh^{-1}x \ d\left[\text{Li}_2(1-x)\right] \overset{ibp} =-\int_0^1 \frac{\text{Li}_2(1-x)}{1-x^2}dx \\ =&\int_0^1 \frac1{1-x^2}\bigg( \int_0^1 \frac{(1-x)\ln t}{1-(1-x)t}dt\bigg)dx\\ =& \int_0^1 \int_0^1 \frac{\ln t}{(1+x)(1-t+tx )}dx\ dt\\ =& \int_0^1 \frac{\ln t\ln(2-2t)}{1-2t}dt \overset{y=1-2t}=\frac12\int_{-1}^1 \frac{\ln\frac{1-y}2\ln(1+y)}ydy\\ =& -\frac12\ln2 \int_{-1}^1 \frac{\ln(1+y)}ydy = -\frac{\pi^2}8\ln2 \\ \end{align}