Integrate $\int_0^\infty \frac{e^{-x/\sqrt3}-e^{-x/\sqrt2}}{x}\,\mathrm{d}x$ I can't solve the integral
$$\int_0^\infty \frac{e^{-x/\sqrt3}-e^{-x/\sqrt2}}{x}\,\mathrm{d}x$$
I tried it by using Beta and Gamma function and integration by parts. Please help me to solve it.

An alternative approach to Marco Cantarini's perfectly sound answer.
If we set, for any $\alpha>1$,
$$ I(\alpha) = \int_{0}^{+\infty}\frac{e^{-x}-e^{-\alpha x}}{x}\,dx $$
differentation under the integral sign/Feynman's trick gives 
$$ I'(\alpha) = \int_{0}^{+\infty} e^{-\alpha x}\,dx = \frac{1}{\alpha}, $$
and since $\lim_{\alpha\to 1^+}I(\alpha) = 0$, it follows that $I(\alpha)=\log\alpha$. 
On the other hand, by setting $x=z\sqrt{6}$ in the original integral, we get:

$$ J = \int_{0}^{+\infty}\frac{e^{-z\sqrt{2}}-e^{-z\sqrt{3}}}{z}\,dz = I(\sqrt{3})-I(\sqrt{2}) = \color{red}{\frac{1}{2}\,\log\left(\frac{3}{2}\right)}.$$

