# 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!

• Is it meant to be $\alpha x^\alpha$ or is it intentionally two different constants? – Henry Lee Jan 21 at 0:17
• Two different constants, sorry my code was a little bad so the integral looks tiny for some reason. – DerpyMcDerp Jan 21 at 0:31
• I don’t believe is it of this form as we cannot define a function f(x) that fits both parts of the equation – Henry Lee Jan 21 at 0:48

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 $$\int_0^\infty \frac{e^{-ax^\alpha} - e^{-b x^\beta}}{x} \, dx = \frac{1}{|\alpha|} (-\gamma - \ln a) - \frac{1}{|\beta|} (-\gamma - \ln b), \qquad a,b > 0, \alpha \beta > 0.$$