一个包含arctan与arctanh的积分

\(\Large\int_0^1\frac{\arctan x \,\operatorname{arctanh} x\, \ln x}{x}\mathrm{d}x=\frac{\pi^2}{16}\mathbf{G}-\frac{7\pi}{32}\zeta(3)\)

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\(\Large \mathrm{\mathbf{Proof:}}\)

Let \(n=0,1,2,\cdots\), We define \(I,I_{1,n},I_{2,n}\) and \(I_n\) as follows:
\(\begin{align*}I &= \int_0^1 \frac{\ln(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}\mathrm{d}x~~,~~I_{1,n} =\int_0^1 x^{2n}\ln(x)\ln(1-x)\; \mathrm{d}x \\ I_{2,n}&=\int_0^1 x^{2n}\ln(x)\ln(1+x)\; \mathrm{d}x~~,~~I_n = \int_0^1 x^{2n} \ln(x)\tanh^{-1}(x)\; \mathrm{d}x \end{align*}\)
Part I : Evaluation of \(I_{1,n}\), \(I_{2,n}\) and \(I_n\)
$$\begin{align*} I_{1,n} &= \int_0^{1 x}{2n}\ln(x)\ln(1-x) \mathrmx= \int_0^{1 x}{2n}\ln(x)\left(-\sum_^{\infty \frac{x}j} \right) \mathrmx\
&= -\sum_^{\infty \frac{1} \int_0}1 x^{2n+j}\ln(x) \mathrmx= \sum_^{\infty \frac{1}{j\left(2n+1+j \right)2}\
&= \frac{1}{(2n+1)2}\sum_\infty \left(\frac{1}-\frac{1}{j+2n+1} \right)-\frac{1}{(2n+1)}\sum_\infty \frac{1}{(j+2n+1)2}\ &= \frac{\gamma +\psi_0(2n+2)}{(2n+1)2}-\frac{\psi_1(2n+2)}{2n+1} \tag{1} \end{align*}$$
Similarly,
\(\begin{align*} I_{2,n}&=\int_0^1 x^{2n}\ln(x)\ln(1+x) \mathrm{d}x = \int_0^1 x^{2n}\ln(x)\left(\sum_{j=1}^\infty\frac{(-1)^{j+1}x^j}{j} \right) \mathrm{d}x\\&= \sum_{j=1}^\infty \frac{(-1)^{j+1}}{j}\int_0^1 x^{2n+j}\ln(x) \mathrm{d}x = \sum_{j=1}^\infty \frac{(-1)^{j}}{j\left(2n+1+j \right)^2} \\ &= \frac{1}{(2n+1)^2}\sum_{j=1}^\infty \frac{(-1)^j}{j}-\frac{1}{(2n+1)^2}\sum_{j=1}^\infty\frac{(-1)^j}{j+2n+1}-\frac{1}{2n+1}\sum_{j=1}^\infty \frac{(-1)^j}{(j+2n+1)^2} \\ &= -\frac{\ln(2)}{(2n+1)^2}+\frac{\psi_0\left( n+\dfrac{3}{2}\right)-\psi_0(n+1)}{2(2n+1)^2}+\frac{\psi_1(n+1)-\psi_1\left(n+\dfrac{3}{2} \right)}{4(2n+1)} \end{align*}\)
We can make some simplifications using the following identities:
\(\begin{align*} \psi_0\left(n+\frac{3}{2} \right) &= 2\psi_0(2n+2)-\psi_0(n+1)-2\ln(2) \\ \psi_1\left( n+\frac{3}{2}\right) &= 4\psi_1(2n+2)-\psi_1(n+1) \end{align*}\)
So, \(I_{2,n}\) can be written as:
\(\begin{align*} I_{2,n}&= -\frac{2\ln(2)}{(2n+1)^2}+\frac{\psi_0(2n+2)-\psi_0(n+1)}{(2n+1)^2}+\frac{2\psi_1(n+1)-4\psi_1(2n+2)}{4(2n+1)} \tag{2} \end{align*}\)
Also note that \(\displaystyle I_n=\frac{I_{2,n}-I_{1,n}}{2}\). Therefore,
$$\begin{align*}
I_n=-\frac{\ln(2)}{(2n+1)2}-\frac{\gamma+\psi_0(n+1)}{2(2n+1)2}+\frac{\psi_1(n+1)}{4(2n+1)}\tag{3}
\end{align*}$$
Part II : Expressing \(I\) in terms of Euler Sums
\(\begin{align*} I &= \int_0^1 \frac{\ln(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}\mathrm{d}x= \int_0^1 \frac{\ln(x)\tanh^{-1}(x)}{x}\left(\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1} \right)\mathrm{d}x \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{2n}\ln(x)\tanh^{-1}(x) \mathrm{d}x= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} I_n \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\left(-\frac{\ln(2)}{(2n+1)^2}-\frac{\gamma+\psi_0(n+1)}{2(2n+1)^2}+\frac{\psi_1(n+1)}{4(2n+1)} \right) \\ &= -\ln(2)\frac{\pi^3}{32}-\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}\left( \gamma+\psi_0(n+1)\right)+\frac{1}{4}\sum_{n=0}^\infty \frac{(-1)^n \psi_1(n+1)}{(2n+1)^2} \end{align*}\)
Let us denote the euler sums by \(E_1\) and \(E_2\):
\(E_1 = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}\left( \gamma+\psi_0(n+1)\right)~~,~~E_2 = \sum_{n=0}^\infty \frac{(-1)^n \psi_1(n+1)}{(2n+1)^2}\)
Part III : Evaluation of \(E_1\)
We use the FS-contour method. Let \(\displaystyle f(z)=\frac{\pi\csc(\pi z) \left(\gamma+\psi_0(-z) \right)}{(2z+1)^3}\). Then the sum of all residues of \(f(z)\) is zero.
The sum of the residues at the negative integers is equal to:
$$\sum_{\infty}\textf(z) = \sum\infty \frac{(-1) \left(\gamma+\psi_0(n) \right)}{(2n-1)3}= E_1 $$
At \(z=-\dfrac{1}{2}\), the residue is
\(\text{Res}_{z=-1/2}f(z) = \frac{\pi^3}{8}\ln(2)+\frac{7\pi}{8}\zeta(3)\)
The sum of the residues at the positive integers is:
\(\sum_{n=0}^{\infty}\text{Res}_{z=n}f(z) = \sum_{n=0}^\infty \left(-6\frac{(-1)^n}{(2n+1)^4}+\frac{(-1)^n H_n}{(2n+1)^3} \right)= -6\beta(4)+E_1\)
Therefore,
$$E_1+\frac{\pi3}{8}\ln(2)+\frac{7\pi}{8}\zeta(3)+E_1-6\beta(4) = 0 \implies E_1 = \boxed{3\beta(4)-\dfrac{7\pi}{16}\zeta(3)-\dfrac{\pi3}{16}\ln(2)} $$
Part IV : Evaluation of \(E_2\)
This time we use FS contour method to the function \(\displaystyle g(z)=\frac{\pi\csc(\pi z)\psi_1(-z)}{(2z+1)^2}\).
The sum of the residues at the negative integers is:
$$\sum_}\infty \textg(z) =-\sum^{\infty \frac{(-1)}\psi_1(n)}{(2n-1)^2} = -E_2 $$
The residue at \(z=-\dfrac{1}{2}\) is :
\(\text{Res}_{z=-1/2}g(z)=-\frac{7\pi}{2}\zeta(3)\)
The sum of the residues at the positive integers is:
\(\begin{align*} \sum_{n=0}^\infty \text{Res}_{z=n}g(z) &= \sum_{n=0}^\infty \left(12\frac{(-1)^n}{(2n+1)^4}+\frac{\pi^2 (-1)^n}{2(2n+1)^2} -\frac{(-1)^n\psi_1(n+1)}{(2n+1)^2}\right) \\ &= 12\beta(4)+\frac{\pi^2}{2}\mathbf{G}-E_2 \end{align*}\)
The sum of all the residues is zero. Therefore,
\(-E_2-\frac{7\pi}{2}\zeta(3)+12\beta(4)+\frac{\pi^2}{2}\mathbf{G}-E_2 = 0 \implies E_2 = \boxed{6\beta(4)-\dfrac{7\pi}{4}\zeta(3)+\dfrac{\pi^2}{4}\mathbf{G}}\)
Part V : The Final Answer
\(\begin{align*} I &= -\frac{\pi^3 \ln(2)}{32}-\frac{E_1}{2}+\frac{E_2}{4} \\ &= -\frac{\pi^3 \ln(2)}{32}-\frac{1}{2}\left(3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\ln(2) \right)+\frac{1}{4}\left( 6\beta(4)-\frac{7\pi}{4}\zeta(3)+\frac{\pi^2}{4}\mathbf{G}\right) \\ &=\Large\boxed{\color{blue}{\dfrac{\pi^2}{16}\mathbf{G}-\dfrac{7\pi\zeta(3)}{32}}} \end{align*}\)