Evaluating what looks like a Frullani integral Can the following integral be evaluated?
$$I=\int_{0}^{\infty} \frac{e^{-ax^\alpha}-e^{-bx^\beta}}{x}dx$$
It looks to me as though it could be a Frullani integral, but I can't seem to find a good way to solve it (if there is one). I've even tried differentiating under the integral sign, but I couldn't seem to get anywhere with that either. If someone could be nice and either point me in the correct direction or just answer the question entirely, that would be very much appreciated. Thanks!
 A: Let
$$I = \int_0^\infty \frac{e^{-ax^\alpha} - e^{-b x^\beta}}{x} \, dx.$$
I will assume $a,b, \alpha, \beta > 0$.
Integrating by parts we have
$$I = a \alpha \int_0^\infty \ln x \, x^{\alpha - 1} e^{-a x^\alpha} \, dx - b \beta \int_0^\infty \ln x\,  x^{\beta - 1} e^{-b x^\beta} \, dx = I_1 - I_2.$$
In $I_1$ setting $u = ax^\alpha$, gives $dx = (u/a)^{\frac{1}{\alpha} - 1} du/(a\alpha)$. The limits of integral are unchanged since $\alpha > 0$ (see below if $\alpha, \beta < 0$). Thus
\begin{align}
I_1 &= \frac{1}{\alpha} \int_0^\infty \ln \left (\frac{u}{a} \right ) e^{-u} \, du\\
&= \frac{1}{\alpha} \int_0^\infty \ln u e^{-u} \, du - \frac{\ln a}{\alpha} \int_0^\infty e^{-u} \, du\\
&= \frac{1}{\alpha} (-\gamma - \ln a).
\end{align}
Here the integral representation for the Euler–Mascheroni constant of 
$$\gamma = - \int_0^\infty \ln x e^{-x} \, dx,$$
has been used.
Similarly
$I_2 = \frac{1}{\beta} (-\gamma - \ln b).$
Thus
$$\int_0^\infty \frac{e^{-ax^\alpha} - e^{-b x^\beta}}{x} \, dx = \frac{1}{\alpha} (-\gamma - \ln a) - \frac{1}{\beta} (-\gamma - \ln b).$$
Note the result can be extended to all $\alpha, \beta \in \mathbb{R}$ provided the product $\alpha \beta$ is positive (needed if we are to have convergence in the integral). In this case the limits of integration become reversed in the substitution made in $I_1$. The final result is therefore
$$\int_0^\infty \frac{e^{-ax^\alpha} - e^{-b x^\beta}}{x} \, dx = \frac{1}{|\alpha|} (-\gamma - \ln a) - \frac{1}{|\beta|} (-\gamma - \ln b), \qquad a,b > 0, \alpha \beta > 0.$$
