Paneer Definite Masala 5

Let $I = \int_0^\infty \dfrac{x\ln x}{(1+x^2)^2}\, dx$

Substituting $x=\dfrac 1t$ , $\mathrm dx = \dfrac{-1}{t^2}\mathrm dt$ $\begin{aligned}I &= \int_\infty^0\dfrac 1t \dfrac{\ln t}{(t^2+1)^2}\dfrac{t^4\mathrm dt}{t^2}\\&=\int_\infty^0\dfrac{t\ln t}{(1+t^2)^2}\, dt\\&=-I\end{aligned}$

Hence $I = 0$

$\text{Substitute x=tan}\theta$ , the integral turns $\displaystyle \mathfrak{I}=\int_{0}^{\frac{\pi}{2}} \sin\theta \cos\theta \ln(\tan\theta)d\theta$ .

By properties of definite integrals $\displaystyle \int_{0}^{a} f(x)dx = \int_{0}^{a}f(a-x)dx$ ,

$\displaystyle \mathfrak{I} = \int_{0}^{\frac{\pi}{2}} \sin\theta \cos\theta\ln(\tan\theta)d\theta = \int_{0}^{\frac{\pi}{2}} \sin\theta \cos\theta\ln(cot\theta)d\theta$

$\displaystyle 2\mathfrak{I}=\int_{0}^{\frac{\pi}{2}} \sin\theta \cos\theta(\ln(\tan\theta)+\ln(\cot\theta))d\theta = \boxed{0}$