Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$

Question

In order to compute, in an elementary way,

$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$

(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$)

i need to show, in a simple way, that:

$\displaystyle \int_0^1 \dfrac{\arctan x \log x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$

$G$ is the Catalan's constant.

Answer 1

I finally get a solution (i swear i didn't know it when i have posted the question)

Define for $x\in [0,1]$ the function $F$:

$\displaystyle F(x)=\int_0^x \dfrac{\ln t}{1+t}dt$

Notice that $F(1)=-\dfrac{\pi^2}{12}$

(use Taylor's development)

and, after performing the change of variable $y=\dfrac{t}{x}$,

$\displaystyle F(x)=\int_0^1 \dfrac{x\ln(xy)}{1+xy}dy$

Since that:

$\Big[F(x)\arctan x\Big]_0^1=-\dfrac{\pi^3}{48}$

then,

$\displaystyle -\dfrac{\pi^3}{48}=\int_0^1 \dfrac{F(x)}{1+x^2}dx+\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(xy)}{(1+xy)(1+x^2)}dxdy$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(x)}{(1+xy)(1+x^2)}dxdy+\int_0^1\int_0^1 \dfrac{x\ln(y)}{(1+xy)(1+x^2)}dxdy$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\left[\dfrac{\ln x\ln(1+xy)}{1+x^2}\right]_{y=0}^{y=1} dx+ \displaystyle \int_0^1 \left[-\dfrac{\ln y\ln(1+xy)}{1+y^2}+\dfrac{\ln y\ln(1+x^2)}{2(1+y^2)}+\dfrac{y\ln y\arctan x}{1+y^2}\right]_{x=0}^{x=1}dy$

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx= \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\dfrac{\ln y\ln(1+y)}{1+y^2}dy+\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln y}{1+y^2}dy+ \dfrac{\pi}{4}\times \int_0^1 \dfrac{y\ln y}{1+y^2}dy$

Using Taylor's development,

$\displaystyle \int_0^1 \dfrac{y\ln y}{1+y^2}dy=-\dfrac{\pi^2}{48}$

And it's well known that, $\displaystyle -G=\int_0^1\dfrac{\ln y}{1+y^2}dy$

Therefore,

$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=-\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{192}$

And finally,

$\displaystyle \int_0^1 \dfrac{\arctan x \ln x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$

(I hope there is no mistake, this proof is too wonderful to be true )

NB:

Added, July 2, 2019.

The above computation is the result of "reverse engineering". I was searching for a way to express $\pi^3$ as in integral. If you introduce the function, for $x\in [0;1]$, \begin{align}\displaystyle F(x)&=\int_0^x \dfrac{\ln t}{1+t}dt\\ &=\int_0^1 \dfrac{x\ln(tx)}{1+tx}dt \end{align} Observe that, \begin{align}\frac{\partial F(x)}{\partial x}&=\dfrac{\ln x}{1+x}\\ F(1)&=-\frac{\pi^2}{12} \end{align}

Then, \begin{align}-\frac{\pi^3}{48}&=\Big[F(x)\arctan x\Big]_0^1\\ \end{align} And, \begin{align}\frac{\partial F(x)}{\partial x}\arctan x=\frac{\arctan x\ln x}{1+x}\end{align}

Thus, one can apply integration by parts, \begin{align}\int_0^1 \frac{\arctan x\ln x}{1+x}\,dx&=\int_0^1 \frac{\partial F(x)}{\partial x}\arctan x\,dx\end{align} and so on,

Answer 2

Let we deal with a basic problem first, i.e. the computation of $$ C_{2n+1} = \int_{0}^{1}\frac{x^{2n+1}\log x}{1+x}\,dx = \int_{0}^{+\infty}\frac{t e^{-(2n+2)t}}{1+e^{-t}}\,dt\tag{1}$$ Since $\int_{0}^{+\infty}t e^{-mt}\,dt = \frac{1}{m^2}$, we have: $$ -C_{2n+1} = \frac{1}{(2n+2)^2}-\frac{1}{(2n+3)^2}+\frac{1}{(2n+4)^2}-\ldots=\frac{\psi'(n+1)-\psi'\left(n+\frac{3}{2}\right)}{4}\tag{2}$$ and: $$ I=\int_{0}^{1}\frac{\arctan(x)\log(x)}{1+x}\,dx = -\sum_{n\geq 0}\frac{(-1)^n C_{2n+1}}{2n+1}=-\sum_{m\geq 0}\sum_{n\geq 0}\frac{(-1)^{n+m}}{(2n+1)(2n+m+2)^2}\tag{3}$$ By reindexing the last double series, $$ I = -\sum_{s=0}^{+\infty}\sum_{p=0}^{s}\frac{(-1)^s}{(2p+1)(p+s+2)^2}=-\sum_{p=0}^{+\infty}\sum_{s\geq p}\frac{(-1)^s}{(p+s+2)^2(2p+1)}\tag{4}$$ hence, in terms of the Hurwitz zeta function: $$ I = -\sum_{p\geq 0}\frac{(-1)^p}{4(p+1)}\left(\zeta\left(2,p+1\right)-\zeta\left(2,p+\frac{3}{2}\right)\right)\tag{5}$$ or, by using the inverse Laplace transform: $$ I = -\int_{0}^{+\infty}\frac{s e^{s/2}\log(1+e^{-s})}{4(1+e^{s/2})}\,ds =-\int_{0}^{+\infty}\frac{s e^s \log(1+e^{-2s})}{1+e^s}\,ds\tag{6}$$ where the last integral is a bit more manageable than the initial one (we made the arctangent function disappear). The constants $K,\log 2$ and $$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32} \tag{7}$$ (see here for the last identity) should simply appear by integration by parts.

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With a suitable change of variable and differentiation under the integral sign, we may probably also exploit the integral remainder term in the second Binet's formula for $\log\Gamma$.

Answer 3

Hint:

set $x=e^{-y}$ we have \begin{align} & \int_{0}^{1}{\frac{{{\tan }^{-1}}x\,\,\ln x}{1+x}}\,dx=\int_{0}^{\infty }{\,\frac{-y\,{{e}^{-y}}{{\tan }^{-1}}({{e}^{-y}})\,}{1+{{e}^{-y}}}}\,dy \\ \\ & {-{e}^{-y}}{{\tan }^{-1}}({{e}^{-y}})=-{e}^{-y}\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{2n-1}{{e}^{-(2n-1)y}}}=\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{2n-1}{{e}^{-2n\,y}}} \\ \\ & \frac{1}{1+{{e}^{-y}}}=\sum\limits_{n=0}^{\infty }{{{(-1)}^{n}}{{e}^{-ny}}} \\ \end{align}

Answer 4

This is a long solution but I hope you find it useful.

First lets consider the integral: \begin{align*} I&=\int_0^1\frac{\ln x\arctan x}{x(1+x)}\ dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1\frac{x^{2n}\ln x}{1+x}\ dx\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{\partial}{\partial{n}}\int_0^1\frac{x^{2n}}{1+x}\ dx\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{\partial}{\partial{n}}\left(H_n-H_{2n}+\ln2\right)\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}-\zeta(2)\right)\\ &=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}\right)-\frac{\pi^3}{48}\tag{1} \end{align*}

on the other hand, \begin{align*} I=\int_0^1\frac{\ln x\arctan x}{x(1+x)}\ dx=\int_0^1\frac{\ln x\arctan x}{x}\ dx-\int_0^1\frac{\ln x\arctan x}{1+x}\ dx\tag{2} \end{align*} where \begin{align*} \int_0^1\frac{\ln x\arctan x}{x}\ dx&=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln x\ dx=-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=-\frac{\pi^3}{32} \end{align*}

we can conclude from $(1)$ and $(2)$ that \begin{align*} \int_0^1\frac{\ln x\arctan x}{1+x}\ dx&=-\frac{\pi^3}{96}-\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}\right)\\ &=-\frac{\pi^3}{96}-\frac12\left(2S_1-S_2\right)\tag{3} \end{align*} \begin{align} S_1&=\sum_{n=0}^\infty\frac{(-1)^nH_{2n}^{(2)}}{2n+1}\\ &=\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}^{(2)}}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &\boxed{=\Im\sum_{n=1}^\infty\frac{i^nH_n^{(2)}}{n}-\frac{\pi^3}{32}=S_1} \end{align} \begin{align} S_2&=\sum_{n=0}^\infty\frac{(-1)^nH_{n}^{(2)}}{2n+1}\\ &=\sum_{n=0}^\infty(-1)^nH_n^{(2)}\int_0^1x^{2n}\ dx\\ &=\int_0^1\sum_{n=0}^\infty H_n^{(2)}(-x^2)^n\\ &=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx \quad \text{ apply IBP}\\ &=-\frac{\pi^3}{48}+2\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx\tag{#}\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1x^{2n}\ dx\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\times\frac{\pi^3}{32}\\ &\boxed{=-4\Im\sum_{n=1}^\infty\frac{i^nH_n}{n^2}+\frac{5\pi^3}{48}=S_2} \end{align} note that in line $\text{(#)}$, we used $\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}\ $ (see here).

Plugging $S_1$ and $S_2$ in $(3)$, we get $$\int_0^1\frac{\arctan x\ln x}{1+x}\ dx=\frac{7\pi^3}{96}-\Im\left(\sum_{n=1}^\infty\frac{i^nH_n^{(2)}}{n}+2\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\right)$$

using the generating functions: $$\sum_{n=1}^\infty\frac{x^nH_n^{(2)}}{n}=\operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\ln(1-x)\operatorname{Li}_2(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)$$

$$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln^2x\ln(1-x)+\zeta(3)$$ then \begin{align} \sum_{n=1}^\infty\frac{x^nH_n^{(2)}}{n}+2\sum_{n=1}^\infty\frac{x^nH_n}{n^2}&=3\operatorname{Li}_3(x)+\ln(1-x)\{\operatorname{Li}_2(1-x)+\ln x\ln(1-x)-\zeta(2)\}\\ &=3\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x) \end{align} where in the last line, we used the reflection identity. taking $x=i$ , we get \begin{align} \Im\left(\sum_{n=1}^\infty\frac{i^nH_n^{(2)}}{n}+2\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\right)&=\Im\left(3\operatorname{Li}_3(i)-\ln(1-i)\operatorname{Li}_2(i)\right)\\ &=\frac{17\pi^3}{192}-\frac12G\ln2 \end{align} which follows \begin{align} \int_0^1\frac{\arctan x\ln x}{1+x}\ dx&=\frac{7\pi^3}{96}-\left(\frac{17\pi^3}{192}-\frac12G\ln2\right)\\ &=\frac12G\ln2-\frac{\pi^3}{64} \end{align}

Answer 5

Different approach:

start with applying integration by parts

$$I=\int_0^1\frac{\tan^{-1}(x)\ln(x)}{1+x}dx\\=\left|(\operatorname{Li}_2(-x)+\ln(x)\ln(1+x))\tan^{-1}(x)\right|_0^1-\int_0^1\frac{\operatorname{Li}_2(-x)+\ln(x)\ln(1+x)}{1+x^2}dx$$

$$=-\frac{\pi^3}{48}-\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx-\color{blue}{\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}dx}\tag1$$

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From $$\operatorname{Li}_2(x)=-\int_0^1\frac{x\ln(y)}{1-xy}dy$$

it follows that

$$\int_0^1\frac{\operatorname{Li}_2(-x)}{1+x^2}dx=\int_0^1\frac1{1+x^2}\left(\int_0^1\frac{x\ln(y)}{1+xy}dy\right)dx$$

$$=\int_0^1\ln(y)\left(\int_0^1\frac{x}{(1+x^2)(1+yx)}dx\right)dy$$

$$=\int_0^1\ln(y)\left(\frac{\pi}{4}\frac{y}{1+y^2}-\frac{\ln(1+y)}{1+y^2}+\frac{\ln(2)}{2(1+y^2)}\right)dy$$

$$=-\frac{\pi^3}{192}-\color{blue}{\int_0^1\frac{\ln(y)\ln(1+y)}{1+y^2}dy}-\frac12\ln(2)\ G\tag2$$

By plugging $(2)$ in $(1)$, the blue integral magically cancels out and we get $I=\frac12G\ln2-\frac{\pi^3}{64}$.