Hello, guys, I just want you to help me of finding the integral of $\displaystyle \int_{0}^{1}\dfrac{\ln(x)}{x^2 + 1}\mathrm{d}x$. I just know that is improper.
Is your difficulty with the anti-derivative $\displaystyle \displaystyle\int {\frac{{\log (x)}}{{{x^2} + 1}}dx}~? $
There is no elementary solution. The solution involves the Polylogarithm function: SEE HERE
integration by parts we get
$\displaystyle -\int_0^1 \frac{\tan ^{-1}x}{x} \, dx$
$\displaystyle -\int_0^1 \left(\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k}}{(2k+1)}\right) \, dx$
$\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)}\int_0^1 x^{2k} \, dx$
$\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)^2}$
$\displaystyle -Catalan$ constant
$-0.915966$