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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?

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  • 1
    $\begingroup$ The tag definite-inte should be definite-integral rather than definite-inte! $\endgroup$ – Harry Peter Oct 28 '13 at 13:09
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    $\begingroup$ Great question! I tried using series expansions but ran into some nasty functions in the sum. I find it very interesting that $$ I = -\frac{\pi}{8}\log^22 - \frac{J}{3}, $$ where $J$ is the integral encountered here. We know roughly how to evaluate $J$ now, so if someone can show $I+J/3$ is equal to the remainder somehow then they'll have done it. $\endgroup$ – Bennett Gardiner Nov 4 '13 at 1:10
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    $\begingroup$ I don't know if it's "good" or not, but what I find odd is that on one hand $\frac{\arctan x}{1+x^2}$ can be viewed as $\frac12[\arctan^2(x)]'$ , and on the other hand, $\frac{x}{1+x^2}$ can be written as $\frac12[\ln(1+x^2)]'$, which might be connected to the $\ln(1-x^2)$ also present there. $\endgroup$ – Lucian Nov 6 '13 at 3:33
  • $\begingroup$ I think this lemma may be useful: $$\frac{x\arctan x}{1+x^2}=\sum_{k=1}^{+\infty}\left(H_{2k}-\frac{H_k}{2}\right)(-1)^k x^{2k}$$ Now we only need to find expressions for $\int_{0}^{1}x^{2k}\log(1-x^2)dx$, that can be derived from $\int_{0}^{1}x^{k}\log(1-x)dx=-\frac{H_{k+1}}{k+1}.$ $\endgroup$ – Jack D'Aurizio Nov 7 '13 at 23:43
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\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,

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

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  • $\begingroup$ finally a clear demonstration (+1) $\endgroup$ – user178256 Sep 12 '16 at 9:51
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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 $$ -\frac{\pi ^3}{48}-\frac{1}{8} \pi (\log2)^2+ K \log2$$

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