Find improper integral Find improper integral  $\int_{0}^{1}-\frac{\log(x)}{x^2}\cdot (xe)^\frac{1}{x}dx$.
I used the following substitution  $t=\frac{1}{x}$,
$x=\frac{1}{t}\mspace{10mu},dx=-x^2dt,\mspace{10mu} \log(x)=-\log(t)$.
$\int_{0}^{1}-\frac{\log(x)}{x^2}\cdot (xe)^\frac{1}{x}dx=\int_{1}^{\infty}\log(t)(\frac{e}{t})^t dt$
How do I continue? 

Put $x=e^z$ and $dx=e^zdz$. Then
$$
-\int _{0}^{1} \frac{\log(x)}{x^2} (xe)^{x^{-1}}\,dx =
$$
$$
=- \int _{-\infty}^{0} e^{e^{-z} (z+1)-z} z\,dz 
$$
$$
= \left.e^{e^{-z} (z+1)}\right|_{-\infty}^{0} = e-0=e
$$
