Hallar la integral impropia

Question

Hallar la integral impropia $\int_{0}^{1}-\frac{\log(x)}{x^2}\cdot (xe)^\frac{1}{x}dx$ .

He utilizado la siguiente sustitución $t=\frac{1}{x}$ ,

$x=\frac{1}{t}\mspace{10mu},dx=-x^2dt,\mspace{10mu} \log(x)=-\log(t)$ .

$\int_{0}^{1}-\frac{\log(x)}{x^2}\cdot (xe)^\frac{1}{x}dx=\int_{1}^{\infty}\log(t)(\frac{e}{t})^t dt$

¿Cómo continúo?

Answer 1

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