Prove $3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx -\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx =\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3})$

Question

Prove that $$ I =3\int_{0}^{1}\frac{x\arctan x}{3x^2+1}\,\mathrm dx -\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx =\frac23 G-\frac {\pi}{12}\ln(2+\sqrt{3}). $$

where, $G$ is catalan's constant

Above two Integrals are a part of a integral which I was trying to solve.

Let $I=3I_{1}-I_{2}$

Attempt:-1

$$I_{1}=\int_{0}^{1}\frac{x\arctan x}{x^2+3}\,\mathrm dx$$

$$\implies I_{1}=\int_{0}^{1}\frac{x}{x^2+3}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1} x^{2n-1}\,\mathrm dx$$

$$\implies I_{1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\int_{0}^{1}\frac{x^{2n}}{x^2+3}\,\mathrm dx.$$

From my previous question 1 we have $$ \int_{0}^{1}\frac{x^{2n}}{x^{2}+3}\,\mathrm dx =(-3)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}\frac{(-3)^{n-1-k}}{2k+1}. $$

Therefore, $$ I_{1}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\bigg[(-3)^{n}\frac{\pi}{6\sqrt{3}}+\sum_{k=0}^{n-1}\frac{(-3)^{n-1-k}}{2k+1}\bigg], $$ $$ I_{1}=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{(-1)^{2n-k}\space 3^{n-1-k}}{(2k+1)(2n-1)}-\frac{\pi}{6\sqrt{3}}\sum_{n=1}^{\infty}\frac{3^{n}}{2n-1}. $$

Using Desmos Both of the series diverges so $I_{1}$ is of the form $\infty -\infty$ which have a finite answer. Same thing goes with $I_{2}$.

Attempt:- 2:

Try to convert one integral into another. Substitute $x=\frac{1}{x}$ in $I_{1}$, we get

$$I_{1}=-\frac{3\pi}{2}\int_{1}^{\infty}\frac{x}{x^2+3} \,\mathrm dx +3\int_{1}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx $$

$$\implies I=-\frac{3\pi}{2}\int_{1}^{\infty}\frac{x}{x^2+3} \,\mathrm dx -\int_{0}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx +4\int_{1}^{\infty}\frac{x \space tan^{-1}x}{x^2+3} \,\mathrm dx$$

Surprisingly all three integrals diverges and convergence of $I$ is maintained by negative and positive sign.

How can I prove the original result?

Thank you for your help!

Answer 1

Integrate by parts and then substitute $x=\tan \frac t2$ \begin{align} I=& \>\int_{0}^{1}\frac{x\arctan x}{x^2+\frac13}\,\mathrm dx -\int_{0}^{1}\frac{x\arctan x}{x^2+3}dx =\frac14 \int_0^{\frac\pi2} \ln\frac{1+\frac12 \cos t}{1-\frac12 \cos t}dt \end{align}

Let $J(a) = \int_0^{\frac\pi2}\ln(1+\cos a\cos t)dt$. Then $$J’(a) =-\tan a\left( \frac\pi2-\int_0^{\frac\pi2}\frac{1}{1+\cos a\cos t}dt\right) =a\sec a-\frac\pi2\tan a $$ and \begin{align} I&= \frac14\left(J(\frac\pi3)-J(\frac{2\pi}3)\right)=-\frac14 \int^{\frac{\pi}2}_{\frac\pi3} J’(a)da - \frac14\int^{\frac{2\pi}3}_{\frac\pi2} \overset{a\to \pi -a}{J’(a)da }\\ &=\frac12 \int_{\frac\pi3}^{\frac\pi2} (\frac\pi2-a)\sec a \>da \overset{a=\frac\pi2 - 2t}=2\int_0^{\frac\pi{12}} t\csc (2t) dt\\ &=- t\ln(\cot t)\bigg|_0^{\frac\pi{12}} + \int_{0} ^{\frac\pi{12}} {\ln(\cot t) dt}\\ &= -\frac\pi{12} \ln(2+\sqrt3)+\frac23 G \end{align} where $\int^{ \frac\pi{12}}_{0} \ln(\cot t)dt= \frac23G$

Answer 2

First $(*)$ substitute $x=\sqrt{\dfrac{1-y}{1+y}}$, recalling that $\arctan\sqrt{\dfrac{1-y}{1+y}} = \dfrac12\arccos y$ for $y>-1$, then $(**)$ integrate by parts:

$$\begin{align*} I &= 3\int_{0}^{1}\frac{x\arctan x}{3x^2+1} \, dx - \int_{0}^{1}\frac{x\arctan x}{x^2+3} \, dx \\ &= \int_0^1 \frac{8x}{3x^4+10x^2+3} \arctan x \, dx \\ &= \int_0^1 \frac{8\sqrt{\frac{1-y}{1+y}}}{3\left(\frac{1-y}{1+y}\right)^2 + 10\frac{1-y}{1+y} + 3} \arctan\sqrt{\frac{1-y}{1+y}} \frac{dy}{\sqrt{\frac{1-y}{1+y}} (1+y)^2} \tag{$*$} \\ &= \int_0^1 \frac{\arccos y}{4-y^2} \, dy \\ &= \frac12 \int_0^1 \frac{\operatorname{artanh}\frac y2}{\sqrt{1-y^2}} \, dy \tag{$**$} \end{align*}$$

Now let

$$I(a) = \int_0^1 \frac{\operatorname{artanh}(ay)}{\sqrt{1-y^2}} \, dy,$$

observing that $I(0)=0$. Differentiate w.r.t. $a$, integrate w.r.t. $y$ (e.g. via Euler substitution), and integrate a second time w.r.t. $a$:

$$I'(a) = \int_0^1 \frac{y}{\left(1-a^2y^2\right) \sqrt{1-y^2}} \, dy = \frac{\arcsin a}{a\sqrt{1-a^2}} \\ I(a) = \int_0^a \frac{\arcsin b}{b \sqrt{1-b^2}} \, db \stackrel{(**)}= \frac{\arcsin^2a}{2a} + \frac12 \int_0^a \frac{\arcsin^2b}{b^2} \, db$$

To evaluate the remaining integral, expand the numerator with

$$\arcsin^2x = \sum_{n\ge1} \frac{2^{2n-1}}{n^2\binom{2n}n} x^{2n}$$

so that

$$\int_0^a \frac{\arcsin^2b}{b^2} \, db = \sum_{n\ge1} \frac{2^{2n-1}}{n^2 \binom{2n}n} \int_0^a b^{2n-2} \, db = \sum_{n\ge1} \frac{(2a)^{2n-1}}{n^2 (2n-1) \binom{2n}n}.$$

Now,

$$I = \frac12 I\left(\frac12\right) = \frac{\pi^2}{72} + \frac14 \sum_{n\ge1} \frac{1}{n^2 (2n-1) \binom{2n}n}$$

Partial fraction expansion of the summand yields two series, one which can be evaluated using the expansion of $\arcsin^2$, while the other has an equivalent form per S.Ramanujan.

$$\begin{align*} & \sum_{n\ge1} \frac{1}{n^2 (2n-1) \binom{2n}n} \\ &= \sum_{n\ge1} \frac2{n (2n-1) \binom{2n}n} - \sum_{n\ge1} \frac1{n^2\binom{2n}n} \\ &= \sum_{n\ge0} \frac1{(2n+1)^2 \binom{2n}n} - 2 \arcsin^2 \frac x2\bigg|_{x=1} \\ &= \frac83 G - \frac\pi3 \log\left(2+\sqrt3\right) - \frac{\pi^2}{18} \end{align*}$$

It follows that

$$I = \boxed{\frac23 G - \frac\pi{12} \log\left(2+\sqrt3\right)}$$