I found the following integral which seems an extension of the gamma function $$ \int_{0}^{\infty}\left[1 - \mathrm{e}^{-\left(\large ux^{a} + vx^{b}\right)}\right] x^{-1 - c}\,\mathrm{d}x, $$ where $u,v,a,b,c$ are all positive constant such that $c \in \left(0,1\right)$ and $a > b > c$.

In the case of $v = 0$, I evaluated the integral thanks to the change of variable $y = x^{a}$ in terms of the Gamma function, but, when both $u$ and $v$ are strictly positive, I can not make the exponent linear in $x$ when $a \ne b$. I also try to use the series representation of $1 - \mathrm{e}^{-\left(ux^{a} + vx^{b}\right)}$ but it does not work. Do you have any suggestion or know some noteworthy extension of the Gamma function that can help me ?.

Thanks a lot !.

P.S. I'm new to "mathematics", so I also appreciate suggestions on how to ask questions effectively here :)

  • $\begingroup$ Can we know where did you found this integral? $\endgroup$ – user507623 Feb 2 '18 at 13:36
  • $\begingroup$ Sure @Pippo. I found the integral in probability theory. More precisely in the special case when $v=0$ and $a=1$ the integral, as function $u>0$, arises in the Laplace exponent of the marginal random variables of a stable process. This more general version arises as the two-dimensional extension (as function of $u >0$ and $v>0$) of the Laplace exponent of a modified stable process. $\endgroup$ – Gio Feb 2 '18 at 16:16

Integration by parts results in a linear combination of two integrals of the form \begin{align*} \int_0^\infty x^{d-1}e^{-ux^a-vx^b}\,dx&=\sum_{n=0}^\infty\frac{(-v)^n}{n!}\int_0^\infty x^{bn+d-1}e^{-ux^a}\,dx\\&=\frac1{u^{d/a}}\sum_{n=0}^\infty\frac1{n!}\left(\frac{-v}{u^{b/a}}\right)^n\Gamma\left(\frac{bn+d}{a}\right). \end{align*} This series converges, but there's no general closed form for it (other than just a name).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.