TITLE: Evaluating $\int _0^{\infty }\frac{e^{-ax^m}-e^{-bx^n}}{x^p}\:dx$
QUESTION [6 upvotes]: I was able to evaluate this using Feynman's trick and managed to find a closed form though it has strict conditions that make this integral converge but the main thing is how can one evaluate this using other techniques? i find it very hard to come up with anything else, this time ill checkmark the best approach in my opinion.
My attempt.
To find the closed form of the integral i heavily relied on the following identity,
$$\int _0^{\infty }x^ne^{-ax^b}\:dx=\frac{\Gamma \left(\frac{n+1}{b}\right)}{b\:a^{\frac{n+1}{b}}}$$
Now resuming on the integral.
$$I\left(a\right)=\int _0^{\infty }\frac{e^{-ax^m}-e^{-bx^n}}{x^p}\:dx$$
$$I'\left(a\right)=-\int _0^{\infty }x^{m-p}\:e^{-ax^m}\:dx$$
$$I'\left(a\right)=-\frac{\Gamma \left(\frac{1-p}{m}+1\right)}{m\:a^{\frac{1-p}{m}+1}}$$
$$\int _{\infty }^aI'\left(a\right)\:da=-\frac{\Gamma \left(\frac{1-p}{m}+1\right)}{m}\int _{\infty }^aa^{\frac{p-1}{m}-1}\:da$$
We can also use the same identity we used earlier to calculate $I\left(\infty \right)$ so,
$$I\left(\infty \right)=-\int _0^{\infty }x^{-p}e^{-bx^n}dx=-\frac{\Gamma \left(\frac{1-p}{n}\right)}{n\:b^{\frac{1-p}{n}}}$$
Resuming on the original expression we now have:
$$I\left(a\right)+\frac{\Gamma \left(\frac{1-p}{n}\right)}{n\:b^{\frac{1-p}{n}}}=-\left(\frac{1-p}{m}\right)\frac{\Gamma \left(\frac{1-p}{m}\right)}{m}\left(\frac{m}{p-1}\:a^{\frac{p-1}{m}}\right)$$
$$I\left(a\right)=\frac{\Gamma \left(\frac{1-p}{m}\right)}{m}\:a^{\frac{p-1}{m}}-\frac{\Gamma \left(\frac{1-p}{n}\right)}{n}\:b^{\frac{p-1}{n}}$$
Meaning that:
$$\boxed{I\left(a\right)=\int _0^{\infty }\frac{e^{-ax^m}-e^{-bx^n}}{x^p}\:dx=\frac{\Gamma \left(\frac{1-p}{m}\right)}{m}\:a^{\frac{p-1}{m}}-\frac{\Gamma \left(\frac{1-p}{n}\right)}{n}\:b^{\frac{p-1}{n}}}$$
I tried using this to calculate for some values and in all cases it agrees with mathematica even when the integral diverges.
Noticed immediately after posting that i couldve bring the $x^p$ up and use the same identity not having to go through all of Feynman's trick, -.- at least its more fancy.

REPLY [3 votes]: Using Fubini's theorem seems to work too:
\begin{aligned}
\int^\infty_0 \frac{e^{-ax^m}-e^{-bx^n}}{x^p}dx&=-\int^\infty_0 x^{-p}\Big(\int^\infty_0 \,e^{-t}\mathbb{1}_{(ax^m,bx^n)}(t)\,dt\Big)dx\\
&=-\int^\infty_0 e^{-t}\Big(\int^{(t/a)^{1/m}}_{(t/b)^{1/n}}x^{-p}\,dx\Big)dt\\
&=\frac{a^{-\tfrac{1}{m}(1-p)}}{p-1}\int^\infty_0 e^{-t} t^{\tfrac{1}{m}(1-p)}dt-
\frac{b^{-\tfrac{1}{n}(1-p)}}{p-1}\int^\infty_0 e^{-t} t^{\tfrac{1}{n}(1-p)}dt\\
&=\frac{a^{\tfrac{1}{m}(p-1)}}{p-1}\Gamma\Big(\frac{1}{m}(p-1)+1\Big) - \frac{b^{\tfrac{1}{n}(p-1)}}{p-1}\Gamma\Big(\frac{1}{n}(p-1)+1\Big)\\
&=\frac{a^{\tfrac{1}{m}(p-1)}}{m}\Gamma\Big(\frac{1}{m}(p-1)\Big) - \frac{b^{\tfrac{1}{n}(p-1)}}{n}\Gamma\Big(\frac{1}{n}(p-1)\Big)
\end{aligned}