$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\frac{x}{\sqrt{t^2+2}}}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $

Question

I'm trying to solve $$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $$

The value should be $\displaystyle \frac{\pi}{2}\ln(2)$. I tried different substitutions, but it doesn't work out. Changing the integration order seems not really nice as below: $$\int_1^{\infty} \frac{2\sqrt{z+1}\arctan\left(\sqrt{\frac{z-1}{z+1}}\right)+\sqrt{z-1}(\ln(2z)-\ln(z-1))}{4z(z-1)(z+1)} dz$$

Has anyone a good start for me? Maybe Feynman? Thanks!

Answer 1

Suppose $\displaystyle\int\limits_0^x\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt$. It's well known that $\arctan z=\dfrac{\pi}{2}-\text{arccotan} z$, so our integral is $$\dfrac{\pi}{2}\int\limits_0^x\frac{dt}{(t^2+1)\sqrt{t^2+2}}\, dt-\int\limits_0^x\frac{\text{arccotan}\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt.$$

It's obvious that $\dfrac{1}{z}\text{arccotan}\dfrac{x}{z}=\dfrac{\pi}{2z}-\displaystyle\int_0^x\dfrac{dy}{y^2+z^2}$. Given that $z=\sqrt{2+t^2}$ and substituting the last equality in second integral, we get that our two integrals turns into $$\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+t^2+2)}=I.$$ Further, it's easy to see that $I=\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}-\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}$.

Since in the second integral everything is symmetric, $\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}=I$ (we can also make a replacement $t\to y$ and $y\to t$). Thus $$I=\dfrac{1}{2}\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}=\dfrac{1}{2}\left(\displaystyle\int\limits_0^x\dfrac{dt}{t^2+1}\right)^2=\dfrac{\arctan^2x}{2}.$$

It remains to calculate the integral $\dfrac{1}{2}\displaystyle\int\limits_0^\infty\dfrac{\arctan^2x}{x^2}dx=\displaystyle\int\limits_0^\infty\dfrac{\arctan x}{x(1+x^2)}dx$ (here we applying integrate by parts). Can you finish it yourself? Hint: you can consider the integral $\displaystyle\int\limits_0^\infty\dfrac{\arctan ax}{x(1+x^2)}dx$ and differentiate it by the parameter.

Answer 2

$$ \Omega=\int_0^{\infty} \frac{1}{x^2} \int_0^x \frac{\arctan \left(\frac{x}{\sqrt{2+t^2}}\right)}{\left(1+t^2\right) \sqrt{2+t^2}} d t d x $$

Rewrite $: \arctan \left(\frac{x}{\sqrt{2+t^2}}\right)=\int_0^1 \frac{x \sqrt{2+t^2}}{2+t^2+x^2 y^2} d y$

$$ \begin{gathered} \Omega=\int_0^{\infty} \int_0^1 \int_0^x \frac{1}{x\left(1+t^2\right)\left(2+t^2+x^2 y^2\right)} d t d y d x \quad \text { Make Sub : } t=x z \\ \Omega=\int_0^{\infty} \int_0^1 \int_0^1 \frac{1}{\left(1+x^2 z^2\right)\left(2+\left(y^2+z^2\right) x^2\right)} d z d y d x \end{gathered} $$

Exploiting Symmetry of $y, z$

$$ \begin{gathered} \Omega=\int_0^{\infty} \int_0^1 \int_0^1 \frac{1}{\left(1+x^2 y^2\right)\left(2+\left(y^2+z^2\right) x^2\right)} d z d y d x \\ \Omega=\frac{1}{2} \int_0^{\infty} \int_0^1 \int_0^1 \frac{1}{2+\left(y^2+z^2\right) x^2}\left(\frac{1}{1+x^2 y^2}+\frac{1}{1+x^2 z^2}\right) d z d y d x \\ \Omega=\frac{1}{2} \int_0^{\infty}\left(\int_0^1 \frac{d z}{1+x^2 z^2} \int_0^1 \frac{d y}{1+x^2 y^2}\right) d x=\frac{1}{2} \int_0^{\infty} \frac{\arctan ^2(x)}{x^2} d x \\ \Omega=\frac{1}{2}\left[-\frac{\arctan ^2(x)}{x}\right]_0^{\infty}+\int_0^{\infty} \frac{\arctan (x)}{x\left(1+x^2\right)} d x=\int_0^{\infty} \frac{\arctan (x)}{x\left(1+x^2\right)} d x \\ \Omega=\int_0^{\frac{\pi}{2}} \theta \cot \theta d \theta=[\theta \ln \sin \theta]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \ln \sin \theta d \theta=\frac{\pi}{2} \ln 2 \\ \int_0^{\infty} \frac{1}{x^2} \int_0^x \frac{\arctan \left(\frac{x}{\sqrt{2+t^2}}\right)}{\left(1+t^2\right) \sqrt{2+t^2}} d t d x=\frac{\pi}{2} \ln 2 \end{gathered} $$

Answer 3

\begin{align} &\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\frac{x}{\sqrt{t^2+2}}}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx \\ = &\int_{0}^{\infty} d\left(-\frac1x\right)\int_{0}^{x}\frac{\arctan\frac{x}{\sqrt{t^2+2}}}{(t^2+1)\sqrt{t^2+2}}\, dt \\ \overset{ibp}= &\int_{0}^{\infty} \bigg( \frac{\arctan\frac{x}{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}}+\int_{0}^{x}\frac{1}{(t^2+1)(x^2+t^2+2)}\, dt \bigg) \frac{dx}x\\ =&\int_0^\infty \frac{\arctan x}{(x^2+1)x}dx =\int_0^\infty \int_0^1 \frac1{(x^2+1)(1+x^2y^2)}dy \ dx\\ =&\ \frac\pi2\int_0^1\frac1{1+y}dy =\frac\pi2\ln2 \end{align} where $$\int_{0}^{x}\frac{1}{(t^2+1)(x^2+t^2+2)}dt = \frac{\arctan x}{x^2+1}-\frac{\arctan\frac{x}{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}} $$ is used above.

Answer 4

\begin{align}J&=\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx\\ &=\int_0^\infty \frac1{x^2}\int_0^x \left(\underbrace{\int_0^x\frac{1}{(1+t^2)(2+y^2+t^2)}dy}_{u(y)=\frac yx}\right)dtdx\\ &\int_0^\infty \frac 1x\underbrace{\left(\int_0^x\left(\int_0^1\frac1{(1+t^2)(2+x^2u^2+t^2)}du\right)dt\right)}_{v(t)=\frac tx}dx\\ &=\int_0^\infty\left(\int_0^1 \left(\int_0^1 \frac1{(1+x^2v^2)(2+x^2u^2+v^2x^2 )}du\right)dv\right)dx\\ &=\int_0^1\int_0^1 \left[\frac{v\arctan(vx)}{v^2-u^2}-\frac{\sqrt{u^2+v^2}\arctan\left(\frac{x\sqrt{u^2+v^2}}{\sqrt{2}}\right)}{(v^2-u^2)\sqrt{2}}\right]dudv\\ &=\frac{\pi}2\int_0^1\int_0^1\left(\frac{v}{v^2-u^2}-\frac{\sqrt{u^2+v^2}}{(v^2-u^2)\sqrt{2}}\right)dudv\\ &=\frac{\pi}4\int_0^1\int_0^1\left(\frac{v}{v^2-u^2}-\frac{\sqrt{u^2+v^2}}{(v^2-u^2)\sqrt{2}}\right)dudv+\\&\frac{\pi}4\int_0^1\int_0^1\left(\frac{u}{u^2-v^2}-\frac{\sqrt{u^2+v^2}}{(u^2-v^2)\sqrt{2}}\right)dvdu\\ &=\frac{\pi}4\int_0^1\int_0^1\frac{1}{u+v}dudv=-\frac{\pi}4\int_0^1\ln\left(\frac{v}{1+v}\right)dv\overset{w=\frac{v}{1+v}}=-\frac{\pi}4\int_0^{\frac12}\frac{\ln w}{(1-w)^2}dw\\ &\overset{\text{IBP}}=-\frac{\pi}4\left[\left(\frac1{1-w}-1\right)\ln w\right]_0^{\frac12}+\frac{\pi}4\int_0^{\frac12}\frac{1}{1-w}dw\\&=\frac{\pi\ln 2}4-\frac{\pi}4\Big[\ln(1-w)\Big]_0^{\frac12}=\boxed{\frac{\pi\ln 2}2}\\ \end{align}