Simplifing integral to gamma function Can you please help me to write the following integral with gamma function using the proper change of variables:
$$ I = \int_0^\infty e^{-a \cdot r^{\alpha}-b \cdot r^{2}} \; \cdot r^{c} \; dr $$
With a, b, c and $ \alpha $ are real positive numbers, and $ 2<\alpha<6 $. I tried with $ t = a \cdot r^{\alpha}+b \cdot r^{2} $ , but I can't go so far with it, I can't write " r " as a function of " t ".
If it's not possible, at least I need it in the case when $ \alpha=4 $:
$$ I = \int_0^\infty e^{-a \cdot r^{4}-b \cdot r^{2}} \; \cdot r^{c} \; dr $$
Many thanks in advance.
 A: Let $r^{2}=x$, then $dx=2rdr$ then
$$I=2\int_{\mathbb{R}^{+}}x^{\frac{c-1}{2}}e^{-ax^{2}-bx}dx$$
Let $\xi=\sqrt{a}x+\frac{b}{2\sqrt{a}}$, then
$$I=\frac{2}{\sqrt{a}}e^{-\frac{b^{2}}{2a}}\int_{\mathbb{R}^{+}}{\Big(\frac{\xi}{\sqrt{a}}-\frac{b}{2a}\Big)^{\frac{c-1}{2}}}e^{-\xi^{2}}d\xi=$$
Now if $c=2l+1$ where $l\in\mathbb{N}$:
$$\Big(\frac{\xi}{\sqrt{a}}-\frac{b}{2a}\Big)^{l}=\sum_{k=0}^{l}\frac{l!(-1)^{k}}{k!(l-k)!}\Big(\frac{\xi}{\sqrt{a}}\Big)^{l-k}\Big(\frac{b}{2a}\Big)^{k}$$
And
$$I=\frac{2}{\sqrt{a}}e^{-\frac{b^{2}}{2a}}\sum_{k=0}^{l}\frac{l!(-1)^{k}}{k!(l-k)!}\Big(\frac{b}{2a}\Big)^{k}\Big(\frac{1}{\sqrt{a}}\Big)^{l-k}\int_{\mathbb{R}^{+}}\xi^{l-k}e^{-\xi^{2}}d\xi=$$
$$=\frac{2}{\sqrt{a}}e^{-\frac{b^{2}}{2a}}\sum_{k=0}^{l}\frac{l!(-1)^{k}}{k!(l-k)!}\Big(\frac{b}{2a}\Big)^{k}\Big(\frac{1}{\sqrt{a}}\Big)^{l-k}\frac{\Gamma\big(\frac{l-k+1}{2}\big)}{2}$$
Otherwise
$$\Big(\frac{\xi}{\sqrt{a}}-\frac{b}{2a}\Big)^{\frac{c-1}{2}}=\sum_{k=0}^{\infty}\frac{\big(\frac{c-1}{2}\big)_{k}(-1)^{k}}{k!}\Big(\frac{\xi}{\sqrt{a}}\Big)^{\frac{c-1}{2}-k}\Big(\frac{b}{2a}\Big)^{k}$$
and
$$I=\frac{2}{\sqrt{a}}e^{-\frac{b^{2}}{2a}}\sum_{k=0}^{\infty}\frac{\big(\frac{c-1}{2}\big)_{k}(-1)^{k}}{k!}\Big(\frac{1}{\sqrt{a}}\Big)^{\frac{c-1}{2}-k}\Big(\frac{b}{2a}\Big)^{k}\frac{\Gamma\Big(\frac{c+1-2k}{4}\Big)}{2}$$
Here you may have the convergence issues...
