$$\int_0^\infty\left(\dfrac{\pi}{1 + \pi^2x^2}- \dfrac{1}{1+x^2}\right)\ln x dx$$
Attempt:
$x \to \dfrac{1}{x}$
gives:
$\displaystyle\int_\infty^0 \left(\dfrac{\pi}{\pi^2+x^2}- \dfrac{1}{1+x^2}\right)\ln x ~dx$
So its basically reduced to evaluating: $I (k)= \displaystyle\int_\infty^0\dfrac{k}{x^2 +k^2}\ln x ~dx$
For this I tried:
-
Differentiation wrt k using Lebnitz Rule
-
Integration by parts
Both these methods don't seem to work because of the presence of messy inverse functions. Please let me know how to continue.
Best Answer
You can let $x= k t$ to find $$ I(k) \equiv\int \limits_0^\infty \frac{-k \ln(x)}{k^2+x^2} \, \mathrm{d} x = \int \limits_0^\infty \frac{-\ln(t)}{1+t^2} \, \mathrm{d} t + \int \limits_0^\infty \frac{-\ln(k)}{1+t^2} \, \mathrm{d} t \, .$$ Your substitution $t \to \frac{1}{t}$ shows that the first integral vanishes and the second integral can be solved by a trigonometric substitution. In the end you should get $$ I(k) = -\frac{\pi}{2} \ln(k) \, . $$