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

Question

I am trying to prove that

$$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 2 +G\log 2$$

where $G$ is the Catalan's Constant. Numerically, it's value is $-0.199739$.

Using the substitution $x=\tan \theta$, it can be written as

$$ \begin{align*} I &= \int_0^{\frac{\pi}{4}}\theta \tan \theta \log(\cos 2\theta) d\theta-2\int_0^{\frac{\pi}{4}}\theta \tan \theta \log(\cos \theta)d\theta \end{align*} $$

Can anyone suggest a good approach to evaluate it?

Answer 1

Expanding the inverse tangent in logarithms, writing $\frac{x}{1+x^2}=\Re\frac1{x-i}$, and expanding $\log(1-x^2)=\log(1-x)+\log(1+x)$, each of the resulting four indefinite integrals has a closed form. Each term is amenable to automatic integration, (an example), which means that after taking limits, slogging through simplifications and special values, such as those found here, the closed form can be computed.

For example, for the term above, $$ \int_0^1\frac{\log(1-ix)\log(1-x)}{x-i}\,dx = -\frac{K\pi }{4}-\frac{17 i \pi ^3}{384}-\frac{1}{2} i K \log2+\frac{13}{192} \pi ^2 \log2+\frac{3}{32} i \pi (\log2)^2-\frac{(\log2)^3}{48}+3 \,\text{Li}_3({\textstyle\frac{1+i}{2}})-\frac{45 \zeta(3)}{32}. $$

Now, the integrand of the integral in question is the real part of the sum $$ \frac i2 \frac{\log(1-ix)\log(1-x)}{x-i} - \frac i2\frac{\log(1+i x)\log(1-x)}{x-i}+\frac i2\frac{\log(1-ix)\log(1+x)}{x-i}-\frac i2\frac{\log(1+ix)\log(1+x)}{x-i}, $$ where each term has a closed form for its integral, as above, in terms of $\pi$, $K$, $\log 2$ and $\text{Li}_3$.

After sufficient simplification, the integral of that sum is $$\begin{aligned} &\int_0^1 \frac{\arctan x\log(1-x^2)}{x-i}\,dx = \\ &-\frac{1}{4} i K\pi -\frac{\pi ^3}{48}+\frac{1}{32} i \pi ^2 \log2-\frac{1}{8} \pi (\log2)^2+K \log2+\frac{7}{32} i \zeta(3), \end{aligned}$$ of which the real part gives the answer $$ -\frac{\pi ^3}{48}-\frac{1}{8} \pi (\log2)^2+ K \log2$$

Answer 2

\begin{align} \displaystyle I&=\int_0^1 \dfrac{x\ln(1-x^2)\arctan x}{1+x^2}dx\\ \displaystyle &=\int_0^1 \dfrac{x\ln(1+x)\arctan x}{1+x^2}dx+\int_0^1 \dfrac{x\ln(1-x)\arctan x}{1+x^2}dx \end{align}

Let, \begin{equation} \displaystyle F=\int_0^1 \dfrac{x\ln(1-x)\arctan x}{1+x^2}dx \end{equation}

Perform the change of variable $y=\dfrac{1-x}{1+x}$ in the latter integral,

$\displaystyle F=\int_0^1 \dfrac{\Big(\ln 2-y\ln 2+(1-y)\ln y+(y-1)\ln (1+y)\Big)\arctan\left(\dfrac{1-y}{1+y}\right)}{y^3+y^2+y+1}dy$

For $y\neq -1$, define the function $H$, \begin{equation} \displaystyle H(y)=\dfrac{\Big(\ln 2-y\ln 2+(1-y)\ln y+(y-1)\ln (1+y)\Big)\arctan\left(\dfrac{1-y}{1+y}\right)}{y^3+y^2+y+1} \end{equation}

Since for $0<y<1$,

\begin{equation} \arctan\left(\dfrac{1-y}{1+y}\right)=\dfrac{\pi}{4}-\arctan y \end{equation}

and, for $y\neq 1$,

$\dfrac{1}{y^3+y^2+y+1}=\dfrac{1}{(1+y)(1+y^2)}=\dfrac{1}{2(1+y)}+\dfrac{1-y}{2(1+y^2)}$

$\dfrac{y}{y^3+y^2+y+1}=\dfrac{y}{(1+y)(1+y^2)}=\dfrac{1+y}{2(1+y^2)}-\dfrac{1}{2(1+y)}$

then, for $y\neq -1$,

\begin{align*} H(y)&=\dfrac{\Big(\big(-\ln 2-\ln y+\ln(1+y)\big)y+\big(\ln 2+\ln y-\ln(1+y)\big)\Big)\Big(\dfrac{\pi}{4}-\arctan y\Big)}{y^3+y^2+y+1}\\ &=-\dfrac{y\arctan y\ln(1+y)}{1+y^2}+\dfrac{\arctan y\ln(1+y)}{1+y}+ \dfrac{\pi y\ln(1+y)}{4(1+y^2)}-\dfrac{\pi \ln(1+y)}{4(1+y)}+\\ &\dfrac{y\arctan y\ln y}{1+y^2}-\dfrac{\arctan y\ln y}{1+y}-\dfrac{\pi y\ln y}{4(1+y^2)}+\dfrac{\pi \ln y}{4(1+y)}+\dfrac{y\ln 2\arctan y}{1+y^2}-\\ &\dfrac{\ln 2\arctan y}{1+y}-\dfrac{\pi y\ln 2}{4(1+y^2)}+\dfrac{\pi \ln 2}{4(1+y)} \end{align*}

Let, \begin{align*} \displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\ \displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\ \displaystyle C&=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx\\ \displaystyle J&=\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx \end{align*}

Thus,

\begin{equation} (1)\boxed{\displaystyle I=A-C+J-\dfrac{5}{384}\pi^3-\dfrac{7}{32}\pi\left(\ln 2\right)^2+\dfrac{1}{2}G\ln 2} \end{equation}

$G$, being the Catalan constant.

$$\int_0^1 \dfrac{\arctan x\ln(1-x)}{1+x}dx-\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx=\int_0^1 \dfrac{\arctan x\ln\left(\dfrac{1-x}{1+x}\right)}{1+x}dx$$

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

Therefore,

\begin{align} \int_0^1 \dfrac{\arctan x\ln(1-x)}{1+x}dx-\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx&=\dfrac{\pi}{4}\int_0^1\dfrac{\ln x}{1+x}dx-\int_0^1\dfrac{\ln x\arctan x}{1+x}dx\\ &=-\dfrac{\pi^3}{48}-C \end{align}

Therefore,

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

Perform the change of variable $y=1-x$,

$$\int_0^1 \dfrac{\arctan x\ln(1-x)}{1+x}dx=\int_0^1 \dfrac{\arctan(1-x) \ln(x)}{2-x}dx$$

Define the function $R$ on $[0;1]$,

$$R(x)=\int_0^x \dfrac{\ln t}{2-t}dt=\int_0^1 \dfrac{x\ln(tx)}{2-tx}dt$$

\begin{align} \int_0^1 \dfrac{\arctan(1-x) \ln(x)}{2-x}dx&=\Big[R(x)\arctan(1-x)\Big]_0^1+\int_0^1\int_0^1\dfrac{x\ln(tx)}{(2-tx)(1+(1-x)^2)}dtdx-\int_0^1\left[\dfrac{\ln x\ln(2-tx)}{1+(1-x)^2}\right]_{t=0}^{t=1}dx+\\ &\displaystyle\int_0^1\left[\dfrac{\ln t\ln(x^2-2x+2)}{2(1+(1-t)^2)}-\dfrac{\ln t\ln(2-tx)}{1+(1-t)^2}-\dfrac{t\ln t\arctan(x-1)}{1+(1-t)^2}+\dfrac{\ln t\arctan(x-1)}{1+(1-t)^2}\right]_{x=0}^{x=1}dt\\ &=\ln 2\int_0^1\dfrac{\ln x}{1+(1-x)^2}dx-\int_0^1 \dfrac{\ln x\ln(2-x)}{1+(1-x)^2}dx-\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln t}{1+(1-t)^2}dt+\\ &\ln 2\int_0^1\dfrac{\ln t}{1+(1-t)^2}dt-\int_0^1\dfrac{\ln t\ln(2-t)}{1+(1-t)^2}dt-\dfrac{\pi}{4}\int_0^1 \dfrac{t\ln t}{1+(1-t)^2}dt+\dfrac{\pi}{4}\int_0^1 \dfrac{\ln t}{1+(1-t)^2}dt \end{align}

Perform the change of variable $y=1-x$,

\begin{align} \displaystyle \int_0^1 \dfrac{\arctan(x) \ln(1-x)}{1+x}dx&=\ln 2\int_0^1\dfrac{\ln(1-x)}{1+x^2}dx-\int_0^1 \dfrac{\ln (1-x)\ln(1+x)}{1+x^2}dx-\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln (1-t)}{1+t^2}dt+\\ &\ln 2\int_0^1\dfrac{\ln(1-t)}{1+t^2}dx-\int_0^1 \dfrac{\ln (1-t)\ln(1+t)}{1+t^2}dt-\dfrac{\pi}{4}\int_0^1\dfrac{(1-t)\ln(1-t)}{1+t^2}dt+\\ &\dfrac{\pi}{4}\int_0^1 \dfrac{\ln (1-t)}{1+t^2}dt\\ \displaystyle &=\dfrac{3}{2}\ln 2\int_0^1 \dfrac{\ln (1-t)}{1+t^2}dt+\dfrac{\pi}{4}\int_0^1 \dfrac{t\ln (1-t)}{1+t^2}dt-2\int_0^1 \dfrac{\ln (1-x)\ln(1+x)}{1+x^2}dx\\ \displaystyle&=\dfrac{3}{2}\ln 2\left(\dfrac{\pi\ln 2}{8}-G\right)+\dfrac{\pi}{4}\left(\dfrac{(\ln 2)^2}{8}-\dfrac{5\pi^2}{96}\right)-2\int_0^1 \dfrac{\ln (1-x)\ln(1+x)}{1+x^2}dx \end{align}

Thus,

\begin{equation*} (3)\boxed{\displaystyle \int_0^1 \dfrac{\arctan(x) \ln(1-x)}{1+x}dx=\dfrac{7}{32}\pi(\ln 2)^2-\dfrac{3}{2}G\ln 2-\dfrac{5\pi^3}{384}-2\int_0^1 \dfrac{\ln (1-x)\ln(1+x)}{1+x^2}dx\\} \end{equation*}

Apply the integration by parts formula in the following integral,

\begin{equation*} \displaystyle \int_0^1 \dfrac{(\ln(1+x))^2}{1+x^2}dx=\Big[\arctan x\left(\ln\left(1+x\right)\right)^2\Big]_0^1-2\int_0^1 \dfrac{\arctan x\ln(1+x)}{1+x}dx \end{equation*}

\begin{equation*} (4)\boxed{\displaystyle \int_0^1 \dfrac{(\ln(1+x))^2}{1+x^2}dx=\dfrac{\pi}{4}(\ln 2)^2-2J} \end{equation*}

Perform the change of variable $y=\dfrac{1-x}{1+x}$ in the following integral,

\begin{align} \displaystyle \int_0^1 \dfrac{\ln(1+x)\ln x}{1+x^2}dx&=\int_0^1\dfrac{(\ln(1+x))^2}{1+x^2}dx-\int_0^1 \dfrac{\ln(1-x)\ln(1+x)}{1+x^2}dx+\ln 2\int_0^1 \dfrac{\ln(1-x)}{1+x^2}dx-\\ &\ln 2\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx \end{align}

Therefore,

\begin{align*} \displaystyle \int_0^1 \dfrac{\ln(1-x)\ln(1+x)}{1+x^2}dx&=\int_0^1\dfrac{(\ln(1+x))^2}{1+x^2}dx-\int_0^1 \dfrac{\ln(1+x)\ln x}{1+x^2}dx+\ln 2\int_0^1 \dfrac{\ln\left(\tfrac{1-x}{1+x}\right)}{1+x^2}dx\\ &=\int_0^1\dfrac{(\ln(1+x))^2}{1+x^2}dx-\int_0^1 \dfrac{\ln(1+x)\ln x}{1+x^2}dx-G\ln 2\\ \end{align*}

Thus, $$(5)\boxed{\displaystyle \int_0^1 \dfrac{\ln(1-x)\ln(1+x)}{1+x^2}dx=\dfrac{\pi}{4}(\ln 2)^2-G\ln 2-\int_0^1 \dfrac{\ln(1+x)\ln x}{1+x^2}dx-2J}$$

Plug (5) into (3), it follows,

\begin{align*} \displaystyle \int_0^1 \dfrac{\arctan(x) \ln(1-x)}{1+x}dx=-\dfrac{9}{32}\pi(\ln 2)^2+\dfrac{1}{2}G\ln 2-\dfrac{5\pi^3}{384}+2\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx+4J \end{align*}

Using (2), it follows that,

\begin{equation*} (6)\boxed{\displaystyle J=\dfrac{\pi^3}{384}-\dfrac{G\ln 2}{3}+\dfrac{3}{32}\pi(\ln 2)^2-\dfrac{2}{3}\int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx} \end{equation*}

From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,

$$\displaystyle \int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx=A-\dfrac{1}{2}B-C-2G\ln 2+\beta(3)$$

and, $\displaystyle \beta(3)=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)^3}$

It follows that,

$$(7)\boxed{J=\dfrac{\pi^3}{384}+G\ln 2+\dfrac{3\pi\left(\ln 2\right)^2}{32}-\dfrac{2}{3}A+\dfrac{1}{3}B+\dfrac{2}{3}C-\dfrac{2}{3}\beta(3)}$$

From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,

$$(8)\boxed{A=\dfrac{1}{64}\pi^3-B-G\ln 2}$$

It follows that,

$$(9)\boxed{J=\dfrac{5}{3}G\ln 2-\dfrac{\pi^3}{128}+\dfrac{3\pi\left(\ln 2\right)^2}{32}+B+\dfrac{2}{3}C-\dfrac{2}{3}\beta(3)}$$

Plug (8) and (9) into (1) it follows that,

$$(10)\boxed{I=-\dfrac{1}{192}\pi^3+\dfrac{7}{6}G\ln 2-\dfrac{1}{8}\pi\left(\ln 2\right)^2-\dfrac{1}{3}C-\dfrac{2}{3}\beta(3)}$$

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

$$C=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$$

and, knowing that $\beta(3)=\dfrac{\pi^3}{32}$,

it follows that,

$$\boxed{I=G\ln 2-\dfrac{1}{48}\pi^3-\dfrac{1}{8}\pi\left(\ln 2\right)^2}$$

Answer 3

A solution proposed by Cornel Ioan Valean

As also shown in this post, an immediate connection can be made with the integral in the current OP. So, integrating by parts, rearranging, and employing the needed results, we arrive at once at the desired closed form

$$\int_0^1 \frac{x \arctan(x) \log(1-x^2)}{1+x^2}\textrm{d}x=\frac{1}{2}\int_0^1 \left(\log\left(\frac{1+x^2}{2}\right)\right)' \arctan(x) \log(1-x^2)\textrm{d}x$$ $$=\frac{1}{2}\log(2)\underbrace{\int_0^1 \frac{\log(1-x^2)}{1+x^2}\textrm{d}x}_{\displaystyle \log(2)\pi/4-G}-\underbrace{\int_0^1 \frac{x\arctan(x) \log(2/(1+x^2))}{1-x^2}\textrm{d}x}_{\displaystyle \pi^3/192}-\frac{1}{2}\underbrace{\int_0^1\frac{\log(1-x^2)\log(1+x^2)}{1+x^2}\textrm{d}x}_{\displaystyle(\pi ^3+16 \log ^2(2)\pi -96 \log (2) G)/32}=\log(2)G-\log^2(2)\frac{\pi}{8}-\frac{\pi^3}{48}.\tag1$$

The first remaining integral in $(1)$ may be viewed as a sums of two well-known classical (and trivial) integrals, one of them being a Putnam question, that is $\displaystyle\int_0^1 \frac{\log(1+x)}{1+x^2}\textrm{d}x=\frac{\pi}{8}\log(2)$.

The second resulting integral in $(1)$ is calculated magically here.

As regards the last integral in $(1)$, we have the simple (and magical) fact that $$\int_0^1 \frac{\log(1-x^2)\log(1+x^2)}{1+x^2}\textrm{d}x$$ $$ =\lim_{b\to 0} \, \left(\frac{1}{16} \frac{\partial ^2}{\partial b^2}B\left(\frac{1}{4},b\right)-\frac{1}{16} \frac{\partial ^2}{\partial b^2}B\left(\frac{3}{4},b\right)\right)-\frac{1}{64} \left(\lim_{a\to \frac{1}{2}} \, \frac{\partial ^2}{\partial a^2}B\left(a,\frac{1}{2}\right)\right)$$ $$=\frac{1}{32} \left(\pi ^3+16 \pi \log ^2(2)-96 \log (2) G\right).$$

A note: Observe that for the last integral it is enough to use the algebraic identity $ab=1/4 ((a + b)^2 - (a - b)^2)$, where $a=\log(1-x^2)$ and $b=\log(1+x^2)$. Thus, we have $$\int_0^1 \frac{\log(1-x^2)\log(1+x^2)}{1+x^2}\textrm{d}x$$ $$=\frac{1}{4} \int_0^1 \left(\frac{\log ^2\left(1-x^4\right)}{1-x^4}-\frac{x^2 \log ^2\left(1-x^4\right)}{1-x^4}\right) \textrm{d}x-\frac{1}{4} \int_0^1 \frac{\displaystyle \log^2((1-x^2)/(1+x^2))}{1+x^2} \textrm{d}x,$$ where in the first integral we may let the variable change $x^4=y$, and in the second integral we may act in various ways, like letting $(1-x^2)/(1+x^2)=t^{1/2}$, which together lead us to the Beta function form above. So, a straightforward story essentially.

End of story

More details will be given in the sequel of (Almost) Impossible Integrals, Sums, and Series.