1. ## Integral

Calculate : $\int\limits_0^\infty {\left( {\frac{{\ln \left( x \right)}}{{x^2 + 1}}} \right)} dx
$

2. With the substitution $\ln x=t \rightarrow x=e^{t} \rightarrow dx= t\cdot e^{t}$ the integral becomes...

$\int _{0}^{\infty} \frac{\ln x}{1+x^{2}}\cdot dx = \int_{-\infty}^{+\infty}\frac{t\cdot e^{t}}{1+e^{2t}}\cdot dt = \int_{-\infty}^{+\infty}\frac{t}{2\cdot \cosh t}\cdot dt$

But $t$ is an odd function and $\cosh t$ an even function, so that their ratio is an odd function, so that is...

$\int _{0}^{\infty} \frac{\ln x}{1+x^{2}}\cdot dx = 0$

Kind regards

$\chi$ $\sigma$

3. Originally Posted by dhiab
Calculate : $\int\limits_0^\infty {\left( {\frac{{\ln \left( x \right)}}{{x^2 + 1}}} \right)} dx
$
If you let $t=\frac{1}{x}$ then you get:
$\int_{\infty}^0 \frac{\log \left( \frac{1}{t}\right)}{t^2+1} dt = - \int_0^{\infty} \frac{\log t}{t^2+1}dt$

Thus, $\int_0^{\infty} \frac{\log t}{t^2+1}dt =- \int_0^{\infty} \frac{\log t}{t^2+1}dt \implies \int_0^{\infty} \frac{\log t}{t^2+1}dt = 0$