i'm wondering how can i evaluate this integral using real methods:
\begin{equation*} \int_{0}^{\infty}\frac{\log x}{1+x^{2}}dx. \end{equation*}
I tried using mclaurin series of $\log x$ but really don't know how to proceed later and integrating by parts using trigonometrical substitution but i have a mess. Thanks for reading.
You may write $$ \begin{align} \int_{0}^{\infty}\frac{\log x}{1+x^{2}}dx&=\int_{0}^{1}\frac{\log x}{1+x^{2}}dx+\int_{1}^{\infty}\frac{\log x}{1+x^{2}}dx\\\\ &=\int_{0}^{1}\frac{\log x}{1+x^{2}}dx-\int_{1}^{\infty}\frac{\log (1/x)}{1+\frac1{x^2}}\frac{dx}{x^2}\\\\ &=\int_{0}^{1}\frac{\log x}{1+x^{2}}dx-\int_{1}^{\infty}\frac{\log (1/x)}{1+\frac1{x^2}}\frac{dx}{x^2}\\\\ &=\int_{0}^{1}\frac{\log x}{1+x^{2}}dx-\int_{0}^{1}\frac{\log u}{1+u^{2}}du\quad (u=1/x)\\\\ &=0. \end{align} $$
Setting $\displaystyle u=\frac1x$, that is $\displaystyle x=\frac1u$ we get $\displaystyle dx=-\frac{du}{u^2}$, $\displaystyle \log x=-\log u$, $1 \mapsto 1,\,\infty \mapsto 0$, then $$\color{blue}{\int_{1}^{\infty}\frac{\log x}{1+x^{2}}dx =\int_{1}^{0}\frac{-\log u}{1+\frac1{u^{2}}}\left(-\frac{du}{u^2}\right)=\int_{1}^{0}\frac{\log u}{1+u^{2}}du=-\int_{0}^{1}\frac{\log u}{1+u^{2}}du.}$$
No need to split the integral. Just effect the substitution $x=1/y$ so that $dx=-dy/y^2$ and the limits of integration flip. Thus, we have
$$\begin{align} I&=\int_0^{\infty}\frac{\log x}{1+x^2}dx\\\\ &=\int_{\infty}^0 \frac{\log (1/y)}{1+y^{-2}}\frac{-dy}{y^2}\\\\\ &=-\int_{\infty}^0 \frac{\log (1/y)}{y^2+1}dy\\\\ &=\int_{0}^{\infty} \frac{\log (1/y)}{y^2+1}dy=\\\\ &-\int_{0}^{\infty} \frac{\log y}{y^2+1}dy\\\\ &=-I \end{align}$$
Thus $I=-I$ which implies that $I=0$.
