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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 suggestions or know some noteworthy extensions 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 :)

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  • $\begingroup$ Can we know where did you found this integral? $\endgroup$
    – user507623
    Commented Feb 2, 2018 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. $\endgroup$
    – Jim
    Commented Feb 2, 2018 at 16:16

1 Answer 1

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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).

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