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.
 A: \begin{align}
\int_0^{\infty}e^{-yx}\ \mathrm dx
&=\frac{1}{y}\\[9pt]
\int_b^a\int_0^{\infty}e^{-yx}\ \mathrm dx\ \mathrm dy
&=\int_b^a\frac{\mathrm dy}{y}\\[9pt]
\int_0^{\infty}\int_b^ae^{-xy}\ \mathrm dy\ \mathrm dx&=\ln a-\ln b\\[9pt]
\int_0^{\infty}\left[-\frac{e^{-xy}}{x}\right]_b^a\ \mathrm dx
&=\ln\left(\frac{a}{b}\right)\\[9pt]
\int_0^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\ \mathrm dx
&=\ln\left(\frac{b}{a}\right)\\[9pt]
\end{align}

$$
\int_0^{\infty}\frac{e^{-x/\sqrt{3}}-e^{-x/\sqrt{2}}}{x}\ \mathrm dx
=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{1}{2}\ln\left(\frac{3}{2}\right)}}
$$
A: 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:

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

A: By Frullani's theorem we have $$\int_{0}^{\infty}\frac{e^{-x/\sqrt{3}}-e^{-x/\sqrt{2}}}{x}dx=\frac{1}{2}\log\left(\frac{3}{2}\right).$$
