# Evaluating $\int_0^1 \frac{\ln x\ln(1+x^2)\arctan x}{x}dx$

I came accross the integral $$J=\int_0^1 \frac{\ln x\ln(1+x^2)\arctan x}{x}dx$$

Probably $$J=2\beta(4)-\frac{35}{64}\pi\zeta(3)$$

$$\beta(4)=-\frac{1}{6}\int_0^1 \frac{\ln^3 x}{1+x^2}dx$$

I think i can compute it using also Integrating $\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2} dx$ using real methods

Is it possible to compute it using (generalised) harmonic series?

Edit:

\begin{align*} K1&=\int_0^1 \frac{\ln(1-x)\ln^2 x}{1+x^2}\,dx\\ C_1&=\int_0^1 \frac{\ln x}{1-x}dx,C_2=\int_0^1 \frac{\ln^2 x}{1+x^2}dx,C_3=\int_0^1 \frac{\ln^2 x}{1-x}dx,C_4=\int_0^1 \frac{\ln x}{1+x^2}dx\\ C_5&=\int_0^1 \frac{\ln^2 x}{1+x}dx,C_6=\int_0^1 \frac{\ln^3 x}{1+x^2}dx\\ K_1=&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^2 t}{1+t^2}dt-C_2\right)\ln(1-x)\right]_0^1+\int_0^1 \frac{1}{1-x}\left(\int_0^x \frac{\ln^2 t}{1+t^2}dt-C_2\right)dx\\ &=\int_0^1 \frac{1}{1-x}\left(\int_0^x \frac{\ln^2 t}{1+t^2}dt-C_2\right)dx\\ &=\int_0^1 \int_0^1 \left(\frac{t^2x\ln^2(tx)}{(1+t^2x^2)(1+t^2)}-\frac{\ln^2(tx)}{(1+t^2x^2)(1+t^2)}+\frac{\ln^2 t}{(1-x)(1+t^2)}-\frac{C_2}{1-x}\right)dtdx+\\ &2\left(\int_0^1 \frac{\ln t}{1-x}\,dx\right)\left(\int_0^1 \frac{\ln t}{1+t^2}\,dt\right)+\left(\int_0^1 \frac{1}{1+t^2}\,dt\right)\left(\int_0^1 \frac{\ln^2 x}{1-x}\,dx\right)\\ &=2C_1C_4+\frac{\pi}{4}C_3+\\& \int_0^1 \int_0^1 \left(\frac{t^2x\ln^2(tx)}{(1+t^2x^2)(1+t^2)}-\frac{\ln^2(tx)}{(1+t^2x^2)(1+t^2)}+\frac{\ln^2 t}{(1-x)(1+t^2)}-\frac{C_2}{1-x}\right)dtdx\\ &=2C_1C_4+\frac{\pi C_3}{4}+\int_0^1 \left(\frac{1}{1+t^2}\int_0^t \frac{u\ln^2 u}{1+u^2}du-\frac{1}{t(1+t^2)}\int_0^t\frac{\ln^2 u}{1+u^2}du\right)dt+\\ &\int_0^1 \frac{1}{1-x}\left(\int_0^1 \frac{\ln^2 t}{1+t^2}dt-C2\right)dx\\ &=2C_1C_4+\frac{\pi}{4}C_3+\int_0^1 \left(\frac{1}{1+t^2}\int_0^t \frac{u\ln^2 u}{1+u^2}du-\frac{1}{t(1+t^2)}\int_0^t\frac{\ln^2 u}{1+u^2}du\right)dt\\ &=2C_1C_4+\frac{\pi C_3}{4}+\frac{\pi}{4}\left(\int_0^1 \frac{u\ln^2 u}{1+u^2}du\right)-\int_0^1 \frac{t\ln^2 t\arctan t}{1+t^2}dt+\frac{\ln 2}{2}\int_0^1 \frac{\ln^2 u}{1+u^2}du+\\ &\int_0^1 \frac{\ln^3 t}{1+t^2}dt-\frac{1}{2}\int_0^1 \frac{\ln^2 t\ln(1+t^2)}{1+t^2}dt\\ &=2C_1C4+\frac{\pi C_3}{4}+\frac{\pi C_5}{32}-\int_0^1 \frac{t\ln^2 t\arctan t}{1+t^2}dt+\frac{C_2\ln 2}{2}+C_6-\\&\frac{1}{2}\int_0^1 \frac{\ln^2 t\ln(1+t^2)}{1+t^2}dt\\ &=\frac{\pi^2\text{G}}{3}+\frac{35\pi\zeta(3)}{64}+\frac{\pi^3\ln 2}{32}-6\beta(4)-\int_0^1 \frac{t\ln^2 t\arctan t}{1+t^2}dt-\\&\frac{1}{2}\int_0^1 \frac{\ln^2 t\ln(1+t^2)}{1+t^2}dt\\ \end{align*} Moreover, \begin{align*}\int_0^1 \frac{t\ln^2 t\arctan t}{1+t^2}dt&\overset{\text{IBP}}=\frac{1}{2}\Big[\ln^2 t\ln(1+t^2)\arctan t\Big]-\\&\frac{1}{2}\int_0^1 \ln(1+t^2)\left(\frac{\ln^2 t}{1+t^2}+\frac{2\arctan t\ln t}{t}\right)dt\\ &=-\frac{1}{2}\int_0^1 \ln(1+t^2)\left(\frac{\ln^2 t}{1+t^2}+\frac{2\arctan t\ln t}{t}\right)dt\\ \int_0^1 \frac{\ln t\ln(1+t^2)\arctan t}{t}dt&=-\int_0^1 \frac{t\ln t\arctan t}{1+t^2}dt-\frac{1}{2}\int_0^1\frac{\ln(1+t^2)\ln^2 t}{1+t^2}dt \end{align*} Therefore, $$\displaystyle \boxed{J=K_1-\frac{\pi^2\text{G}}{3}-\frac{35\pi\zeta(3)}{64}-\frac{\pi^3\ln 2}{32}+6\beta(4)}$$

NB: I assume that: \begin{align*} C_1&=\int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}\\ C_2&=\int_0^1 \frac{\ln^2 x}{1+x^2}dx=\frac{\pi^3}{16}\\ C_3&=\int_0^1 \frac{\ln^2 x}{1-x}dx=2\zeta(3)\\ C_4&=\int_0^1 \frac{\ln x}{1+x^2}dx=-\text{G}\\ C_5&=\int_0^1 \frac{\ln^2 x}{1+x}dx=\frac{3}{2}\zeta(3)\\ C_6&=\int_0^1 \frac{\ln^3 x}{1+x^2}dx=-6\beta(4)\\ \end{align*}

• The integral J can be calculated using the well-known series expansion of $\ln(1+x^2)\arctan x$ then using the generating function of $H_n/n^3$ but tedious calculations are involved. Commented Jul 30, 2020 at 6:34
• Here is the series expansion math.stackexchange.com/questions/3240779/… Commented Jul 30, 2020 at 6:38
• @Pisco: Definition of $\beta(4)$ was incorrect
– FDP
Commented Jul 30, 2020 at 6:49

Using integration by parts, we get \begin{align*} J &= \int_0^1 \frac{\log(x)\log(1+x^2)\arctan (x)}{x}dx \\ &= \frac{\log^2(x) \log(1+x^2)\arctan(x)}{2}\Big|_0^1 - \frac{1}{2}\int_0^1 \frac{\log^2(x)\log(1+x^2)}{1+x^2}dx - \int_0^1 \frac{x \log^2(x) \arctan(x)}{1+x^2}dx \\ &= - \frac{1}{2}\int_0^1 \frac{\log^2(x)\log(1+x^2)}{1+x^2}dx - \int_0^1 \frac{x \log^2(x) \arctan(x)}{1+x^2}dx \quad \color{blue}{\cdots (1)} \end{align*} Let $$I_1 = \int_0^1 \frac{x \log^2(x) \arctan(x)}{1+x^2}dx$$ and $$I_2=\int_0^1 \frac{\log^2(x)\log(1+x^2)}{1+x^2}dx$$. We can make use of the following well known series expansions: \begin{align*} \frac{\arctan (x)}{1+x^2} &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n x^{2n+1} , \quad |x|< 1\\ \frac{\log(1+x^2)}{1+x^2} &= \sum_{n=1}^\infty (-1)^{n+1} H_n x^{2n} , \quad |x|<1 \end{align*} where $$\tilde{H}_n = \sum_{i=0}^n \frac{1}{2i+1}$$. This gives us \begin{align*} I_1 &= \sum_{n=0}^\infty (-1)^n \tilde{H}_n\int_0^1 x^{2n+2} \log^2(x)\; dx \\ &= 2\sum_{n=0}^\infty \frac{(-1)^n \tilde{H}_n}{(2n+3)^3} \end{align*} and \begin{align*} I_2 &= \sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{2n}\log^2(x)\; dx\\ &= 2\sum_{n=1}^\infty \frac{(-1)^{n+1} H_n}{(2n+1)^3} \end{align*} Therefore, we have $$J = -\sum_{n=0}^\infty \frac{(-1)^n (H_{n+1} + 2\tilde{H}_n)}{(2n+3)^3} = -2\sum_{n=1}^\infty \frac{(-1)^{n+1} H_{2n}}{(2n+1)^3} \quad \color{blue}{\cdots (2)}$$ Note that Ali Shather's series expansion could also have been used to obtain equation (2).

An easy way to evaluate the Euler sum is to use the method of residues (see, for e.g. "Euler Sums and Contour Integral Representations" by Philippe Flajolet and Bruno Salvy). We'll integrate the function $$f(z) = \pi \csc(\pi z) \frac{\gamma+\psi_0(-2z+1)}{(-2z+1)^3}$$ around the positively oriented square, $$C_N$$, with vertices $$\pm \left(N+\frac{1}{4} \right)\pm \left(N+\frac{1}{4} \right)i$$. It is easy to see that $$\lim_{N\to \infty}\int_{C_N}f(z)\; dz = 0$$ Hence, the sum of all residues of $$f(z)$$ at its poles is equal to $$0$$.

The residue at the negative integers is equal to: $$\mathop{\text{Res}}\limits_{z=-n} f(z) = (-1)^n \frac{\psi_0(2n+1)+\gamma}{(2n+1)^3} = (-1)^n \frac{H_{2n}}{(2n+1)^3} , \quad n=0,1,2,\cdots$$ Near $$z=\frac{1}{2}$$, we have $$f(z) =-\left(\pi + O((2z-1)^2) \right)\left(\frac{1}{(2z-1)^4} -\frac{\zeta(2)}{(2z-1)^2} - \frac{\zeta(3)}{2z-1} + O(2z-1)\right)$$ Therefore, \begin{align*} \mathop{\text{Res}}\limits_{z=\frac{1}{2}} f(z) &= \frac{\pi \zeta(3)}{2} \end{align*} Similarly, we have \begin{align*} \mathop{\text{Res}}\limits_{z=\frac{2n+1}{2}} f(z) &= \frac{(-1)^{n+1} \pi}{16 n^3} , \quad n=1,2,3,\cdots \end{align*} and \begin{align*} \mathop{\text{Res}}\limits_{z=n} f(z) &= \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3}- 3\frac{(-1)^{n+1}}{(2n-1)^4}, \quad n=1,2,3,\cdots \end{align*} The residue computations are a bit tedious hence I didn't write the full details. The list of local expansions of basic kernels given on page 6 of the above mentioned paper are quite useful for carrying out these computations. Now, adding up all the residues gives us: \begin{align*} \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} + \sum_{n=1} ^\infty \frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3} -3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4}&= 0\\ \implies \frac{\pi \zeta(3)}{2}+\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} + \frac{\pi}{16}\left(\frac{3\zeta(3)}{4} \right) + \sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{(2n+1)^3} -2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^4} &= 0 \\ \implies -2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} + \frac{35\pi \zeta(3)}{64} -2\beta(4) = 0 \\ \implies \boxed{\sum_{n=1}^\infty \frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} = -\beta(4) + \frac{35\pi \zeta(3)}{128}} \color{blue}{\cdots (3)} \end{align*} Finally, substitute equation (3) into (2) to obtain $$J=-2\sum_{n=1}^\infty \frac{(-1)^{n+1} H_{2n}}{(2n+1)^3} = 2\beta(4) - \frac{35\pi \zeta(3)}{64}$$.

• You use harmonic series but anyway you are the only one to give an answer.
– FDP
Commented Aug 17, 2020 at 9:17

Per Feynman’s trick \begin{align} I(a)= &\int_0^\infty \frac{\ln^2x\>[\frac1x\tan^{-1}(ax)-\frac12 \ln(1+a^2x^2)]}{1+x^2}dx\\ I’(a)=& \int_0^\infty \frac{\ln^2x(1-ax^2)}{(1+x^2)(1+a^2x^2)}dx=-\frac{\pi\ln^2a}{2(1-a)}\\ I(1) =& \int_0^1 I’(a)da = -\frac\pi2 \int_0^1 \frac{\ln^2a}{1-a} = -\pi\zeta(3) \end{align}

On the other hand

\begin{align} I(1)=&\>\bigg( \int_0^1 +\overset{x\to 1/x}{\int_1^\infty}\bigg)\frac{\ln^2x\>[\frac1x \tan^{-1}x-\frac12 \ln(1+x^2)]}{1+x^2}dx\\ =&\int_0^1 \frac{\ln^2x\tan^{-1}x}{x}\overset{ibp}{dx} +\frac\pi2 \int_0^1 \frac{x\ln^2x}{1+x^2} \overset{x^2\to x}{dx} -2 \int_0^1 \frac{x\ln^2x\tan^{-1}x}{1+x^2}dx\\ & + \int_0^1 \frac{\ln^3x}{1+x^2}dx -\int_0^1 \frac{\ln^2x\ln(1+x^2)}{1+x^2}dx\\ =& \frac23\int_0^1 \frac{\ln^3x}{1+x^2}dx +\frac\pi{16} \int_0^1 \frac{\ln^2x}{1+x}dx - \int_0^1 \frac{\ln^2x\>(2x\tan^{-1}x+\ln(1+x^2))}{1+x^2} \overset{ibp}{dx}\\ =& -4\beta(4)+\frac{3\pi}{32}\zeta(3)+2 \int_0^1 \frac{\ln^2x\ln(1+x^2)\tan^{-1}x}{x} \end{align}

which, after equating with $$I(1) = -\pi\zeta(3)$$ above, yields $$\int_0^1 \frac{\ln^2x\ln(1+x^2)\tan^{-1}x}{x} = 2\beta(4)-\frac{35}{64}\pi\zeta(3)$$