$\displaystyle \displaystyle \[I = \int\limits_0^{ + \infty } {\frac{{\ln x}}{x}} dx = \mathop {\lim }\limits_{\gamma \to + \infty } \left( {\int\limits_{1/\gamma }^\gamma {\frac{{\ln x}}{x}dx} } \right)\]$

Because $\displaystyle \displaystyle \[\int {\frac{{\ln x}}{x}dx = \int {\ln x \cdot d\left( {\ln x} \right)} } = \frac{{{{\left( {\ln x} \right)}^2}}}{2}\]$

$\displaystyle \displaystyle \[I = \frac{1}{2}\mathop {\lim }\limits_{\gamma \to + \infty } \left( {{{\left( {\ln \gamma } \right)}^2} - {{\left( {\ln \frac{1}{\gamma }} \right)}^2}} \right) = \frac{1}{2}\mathop {\lim }\limits_{\gamma \to + \infty } \left( {{{\left( {\ln \gamma } \right)}^2} - {{\left( { - \ln \gamma } \right)}^2}} \right)\]$

So it seems like zero, bus this feels kind of wrong to me. Besides, computer software assumes that integral diverges.

What should I do?