Closed form for $\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$

Question

Please help me to find a closed form for this integral: $$\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$$

Answer 1

The integral can be trasformed into a "computable" form. Indeed, \begin{align} I&=\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx=\\ &=2\int_0^{\infty}\frac{\arctan y^2\ln(1+y^4)}{1+y^4}y^2dy=\\ &=\frac{1}{2i}\int_{-\infty}^{\infty}\frac{\ln^2(1+iy^2)-\ln^2(1-iy^2)}{1+y^4}y^2dy=\\ &=\frac{1}{4}\int_{-\infty}^{\infty}\left(\frac{1}{1+iy^2}-\frac{1}{1-iy^2}\right)\Bigl(\ln^2(1+iy^2)-\ln^2(1-iy^2)\Bigr)dy=\\ &=\frac12\mathrm{Re}\int_{-\infty}^{\infty}\frac{\ln^2(1+iy^2)-\ln^2(1-iy^2)}{1+iy^2}dy, \end{align} where the logarithms are defined on their main sheets. The remaining integrals can be evaluated by suitably deforming the contours (or using Mathematica). The final result is $$I=\frac{\pi}{6\sqrt{2}}\left(12G+9\ln^22+3\pi\ln2+\pi^2\right)\approx 11.7433,$$ where $G$ denotes the Catalan's constant.

Answer 2

OK, I worked out the second complex integral and here is my solution:

Step 1 - Contour Integration

Let $$\displaystyle I= \Re \int_{-\infty}^\infty \frac{\log^2 \left( 1-iz^2\right)}{1+i z^2}dz$$

and $\displaystyle f(z)=\frac{\log^2 \left( 1-iz^2\right)}{1+i z^2}$

Now integrate $f(z)$ around the following contour:

There is a branch point at $\displaystyle z= e^{\frac{3i\pi}{4}}$ and a pole at $\displaystyle z=e^{\frac{i\pi}{4}}$.

The residue at $\displaystyle z=e^{\frac{i\pi}{4}}$ is

$$\text{res}_{z=e^{\frac{i\pi}{4}}}f(z)=\frac{-e^{\frac{i\pi}{4}}}{2}\log^2(2)$$

Therefore \begin{align*} \int_{-\infty}^\infty \frac{\log^2 \left( 1-iz^2\right)}{1+i z^2}dz &= 2\pi i \left(\frac{-e^{\frac{i\pi}{4}}}{2}\log^2(2) \right) +\int_{e^{3i\pi /4}}^{\infty e^{3i\pi /4 }} \frac{\left( \pi i + \log|1-ix^2|\right)^2-\left( \log|1-ix^2|-i\pi \right)^2}{1+i x^2}dx \\ &= 2\pi i \left(\frac{-e^{\frac{i\pi}{4}}}{2}\log^2(2) \right) +4\pi i \int_{e^{3i\pi /4}}^{\infty e^{ 3i\pi /4 }} \frac{\log|1-ix^2|}{1+ix^2}dx \\ &= 2\pi i \left(\frac{-e^{\frac{i\pi}{4}}}{2}\log^2(2) \right) +4\pi i e^{\frac{3i\pi}{4}}\int_1^\infty \frac{\log(x^2-1)}{1+x^2}dx \end{align*}

Separating real parts, we have

\begin{align*} I &= \Re \int_{-\infty}^\infty \frac{\log^2 \left( 1-iz^2\right)}{1+i z^2}dz \\ &= \frac{\pi}{\sqrt{2}}\log^2(2)-\frac{4\pi}{\sqrt{2}}\int_1^\infty \frac{\log(x^2-1)}{1+x^2}dx \end{align*}

Step 2 - Evaluation using Fourier Series

The remaining integral can now be solved using fourier series of $\displaystyle \log(\cos \theta)$.

\begin{align*} \int_1^{\infty}\frac{\log \left( x^2-1\right)}{1+x^2}dx &= \int_0^1 \frac{\log \left( \frac{1-x^2}{x^2}\right)}{1+x^2}dx \\ &= \int_0^1 \frac{\log(1-x^2)-2\log(x)}{1+x^2} dx \\ &= \int_0^1 \frac{\log(1-x^2)}{1+x^2}dx+2G \\ &= 2G+\int_0^{\frac{\pi}{4}}\left( \log(\cos 2\theta)-2\log(\cos \theta) \right) d\theta \quad x= \tan \theta \\ &= 2G -\frac{\pi}{4}\log(2)-2\int_0^{\frac{\pi}{4}}\log(\cos \theta)d\theta \\ &= 2G -\frac{\pi}{4}\log(2)-2 \left( -\frac{\pi}{4}\log(2)-\sum_{j=1}^\infty \frac{(-1)^{j} }{j}\int_0^{\frac{\pi}{4}}\cos(2jx) dx\right) \\ &= 2G+\frac{\pi}{4}\log(2)+\sum_{j=1}^\infty \frac{(-1)^{j} \sin \left(\frac{\pi j}{2} \right)}{j^2} \\ &= 2G +\frac{\pi}{4}\log(2)-G \\ &= G+\frac{\pi}{4}\log(2) \end{align*}

Here, I used that

$\displaystyle \int_0^1 \frac{\log x}{1+x^2}dx =-G$

Step 3 - Final Answer

\begin{align*} I &= \frac{\pi}{\sqrt{2}}\log^2(2)-\frac{4\pi}{\sqrt{2}}\left(G+\frac{\pi}{4}\log(2) \right) \\ &= \boxed{\displaystyle \frac{\pi}{\sqrt{2}} \left( \log^2(2)-4G- \pi \log(2) \right)} \end{align*}