Originally Posted by

**Drexel28** You have a bit of a problem here.

Let us just consider this integral on the half-line $\displaystyle (-\infty,0)=\mathbb{R}^-$

Then a simple substitution leads us to $\displaystyle \int_{\mathbb{R}^-}\frac{x}{e^{-x}}dx=-\int_{\mathbb{R}^+}e^z\cdot z dz$ which obviously tends to negative infinity.

and since clearly $\displaystyle \int_{\mathbb{R}^+}\frac{x}{e^x}\le\int_{\mathbb{R }^+}\frac{dx}{x^2}$ converges we see that

$\displaystyle \int_{\mathbb{R}}\frac{x}{e^{-x}}dx=\int_{\mathbb{R}^-}\frac{x}{e^{-x}}dx+\int_{\mathbb{R}+}\frac{x}{e^x}dx$ diverges.

EDIT: Sorry, a little late.