Evaluate $\int \limits _{0}^{\infty}\ln\left({x+\frac{1}{x}}\right)\cdot\frac{\mathrm dx}{1+x^2}$

$$\int \limits _{0}^{\infty}\ln\left({x+\frac{1}{x}}\right)\cdot\frac{dx}{1+x^2}$$ we are asked to solve this definite integral so here's what i did $$\int \limits_{0}^{\infty}\ln \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2} = \int\limits_{0}^{\infty}\ln \left(x^2+1\right)\cdot\frac{dx}{1+x^2} - \int \limits_{0}^{\infty}\ln (x).\frac{dx}{1+x^2}$$ now how to proceed after that ? should i intregrate both seperate functions by substituting $(x^2+1)$ and what should i susbtitute in other integral , by-parts integration is making troubles when substituting $\infty$ , now what to do ?

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Are you sure the integral converge? –  Danis Fischer Aug 24 '13 at 10:57
@CoarguAliquis what do you mean ? –  Tesla Aug 24 '13 at 11:00
Maple produces $\pi\ln(2).$ –  user64494 Aug 24 '13 at 12:18

Make substitution $x=\tan t$ with $t\in(0,\pi/2)$ and then take a look at this answer

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well then let me try it out –  Tesla Aug 24 '13 at 11:03
$\int_{0}^{\pi/2}ln(2cosec(2\psi)d\psi$ again a bump –  Tesla Aug 24 '13 at 11:12
Note that $\ln(2\csc(2\psi))=\ln 2-\ln\sin 2\psi$ –  no identity Aug 24 '13 at 11:14

As you have done the integral can be fairly easy evaluated by splitting it into two easier integrals. $$\int_{0}^{\infty}\log \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2}\mathrm{d}x = \int_{0}^{\infty}\frac{\log\left(x^2+1\right)}{1+x^2}\,\mathrm{d}x - \int_{0}^{\infty}\frac{\log x}{1+x^2}\,\mathrm{d}x$$ For the last integral we have that $$\int_0^\infty R(x) \log x = 0$$ given that $R(x)$ is a rational function satisfying $R(x)=R(1/x)/x^2$. I will leave it to you to checki that $R(x) = 1/(1+x^2)$ satisfies this functional relation. Alternatively one can see that the last integral is zero by splitting the integral by splitting it into $[0,1]\cap[1,\infty)$ $$\int_0^1 \frac{\log x}{1+x^2}\mathrm{d}x + \int_1^\infty \frac{\log x}{1+x^2}\mathrm{d}x$$ and using $x = 1/u$ in the last integral. Now we have that $$\int_{0}^{\infty}\log \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2}\mathrm{d}x = \int_{0}^{\infty}\frac{\log\left(x^2+1\right)}{1+x^2}\,\mathrm{d}x$$ which can be evaluated by introducing a variable $\alpha$, and differentiating under the integral sign. By looking at the following integral $$I(\alpha) := \int_{0}^{\infty}\frac{\log\left(1 + \alpha^2x^2\right)}{1+x^2}\,\mathrm{d}x$$ we see that our integral equals $I(1)$. Now by differentiating by $\alpha$ we obtain \begin{align} \frac{\mathrm{d}}{\mathrm{d}\alpha}I(\alpha) & = \frac{\mathrm{d}}{\mathrm{d}\alpha} \int_{0}^{\infty}\frac{\log\left(1 + \alpha^2x^2\right)}{1+x^2}\,\mathrm{d}x \\ & = \int_{0}^{\infty} \frac{\partial}{\partial \alpha}\frac{\log\left(1 + \alpha^2x^2\right)}{1+x^2}\,\mathrm{d}x \\ & = \int_0^\infty \frac{2a x^2}{(1+\alpha^2x^2)(1+x^2)}\,\mathrm{d}x \\ & = \frac{2\alpha}{\alpha^2-1} \int_0^\infty \frac{1}{1+x^2} - \frac{1}{1 +\alpha^2x^2} \,\mathrm{d}x \\ & = \frac{\pi}{a+1} \end{align} Integrating both sides now gives $$I(\alpha) = \pi \log(1+\alpha) + \mathcal{C}$$ But since $I(\alpha)=0$, then $\mathcal{C}=0$. We can now "finaly" conclude that $$\int_0^\infty \left( \alpha^2 x + \frac{1}{x}\right) \frac{\mathrm{d}x}{1+x^2} = \pi \log(1+\alpha)$$ Your integral is now evaluated by simply plugging in $\alpha = 1$. The switching of the derivation is legal since we have that $$\frac{\log(1+\alpha x^2)}{1+x^2} < \frac{\log(\alpha x^2+\alpha x^2)}{1+x^2} \, \forall \, x>1$$ and the last integral converges (why?). By similar means we have that $\pi/(1+\alpha)$ converges as well.

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By symmetry the integral is $$2\int_1^{\infty}\frac{\ln(x+1/x)}{x+1/x}\frac{dx}{x}$$ Make the substitution $$x=e^{\cosh^{-1}\csc t}$$ to get $$\frac{\pi}{2}\ln 2-\int_0^{\pi/2}\ln(\sin t)dt=\pi\ln 2$$

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