Simplifing integral to gamma function

Question

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.

Answer 1

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...

Answer 2

Since $$ \Gamma (z) = \int_0^\infty {e^{\, - x} x^{\,z - 1} dx} $$ then we can try and transform the exponential function into a series, as in Kyril's answer, but performing first some "massaging". So, standing that the given parameters (particularly $a$ and $b$) are positive

$$ \bbox[lightyellow] { \begin{gathered} I\left| {_{\alpha \, < \,2} } \right. = \int_0^\infty {\exp \left( { - \left( {ar^{\,\alpha } + br^{\,2} } \right)} \right)^\, r^{\,c} dr} = \int_0^\infty {\exp \left( { - ar^{\,\alpha } } \right)\exp \left( { - br^{\,2} } \right)r^{\,c - 1} r\,dr} = \hfill \\ = \int_0^\infty {\exp \left( { - \left( {ar^{\,\alpha } + br^{\,2} } \right)} \right)^\, r^{\,c} dr} = \int_0^\infty {\exp \left( { - ar^{\,\alpha } } \right)\exp \left( { - br^{\,2} } \right)r^{\,c - 1} r\,dr} = \hfill \\ = \frac{1} {{2bb^{\,\left( {c - 1} \right)/2} }}\int_0^\infty {\exp \left( { - br^{\,2} } \right)\exp \left( {\, - \frac{a} {{b^{\alpha /2} }}\left( {br^{\,2} } \right)^{\alpha /2} } \right)\left( {br^{\,2} } \right)^{\,\left( {c - 1} \right)/2} d\left( {br^{\,2} } \right)} = \hfill \\ = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\int_0^\infty {e^{\, - x} \exp \left( {\, - \frac{a} {{b^{\alpha /2} }}x^{\alpha /2} } \right)x^{\,\left( {c - 1} \right)/2} dx} = \hfill \\ \quad \left| {\;\beta = a/\,b^{\alpha /2} } \right. \hfill \\ = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\int_0^\infty {e^{\, - x} \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\beta ^{\,k} }} {{k!}}x^{\,\left( {\alpha \,k + c + 1} \right)/2 - 1} } dx} = \hfill \\ = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\beta ^{\,k} }} {{k!}}\Gamma \left( {\frac{{\alpha \,k + c + 1}} {2}} \right)} = \hfill \\ = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \beta ^{\,k} \frac{{\Gamma \left( {\frac{{\alpha \,}} {2}k + \frac{{c + 1}} {2}} \right)}} {{\Gamma \left( {k + 1} \right)}}} \hfill \\ \end{gathered} \tag {1} }$$ To better analyze the above result for convergence, we rewrite it as: $$ \begin{gathered} I\left| {_{\alpha \, < \,2} } \right. = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\beta ^{\,k} }} {{k!}}\Gamma \left( {\frac{{\alpha \,k + c + 1}} {2}} \right)} = \hfill \\ = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,2k} \frac{{\beta ^{\,2k} }} {{\left( {2k + 1} \right)!}}\left( {\left( {2k + 1} \right)\Gamma \left( {\frac{{\alpha \,}} {2}2k + \frac{{c + 1}} {2}} \right) - \beta \,\Gamma \left( {\frac{{\alpha \,}} {2}2k + \frac{{c + 1 + \alpha }} {2}} \right)} \right)} = \hfill \\ = \frac{1} {{2b^{\,\left( {c + 1} \right)/2} }}\sum\limits_{0\, \leqslant \,k} {\beta ^{\,2k} \frac{{\Gamma \left( {\alpha k + \frac{{c + 1}} {2}} \right)}} {{\Gamma \left( {2k + 1} \right)}}\left( {1 - \frac{\beta } {{\left( {2k + 1} \right)}}\,\frac{{\Gamma \left( {\alpha k + \frac{{c + 1 + \alpha }} {2}} \right)}} {{\Gamma \left( {\alpha k + \frac{{c + 1}} {2}} \right)}}} \right)} \hfill \\ \end{gathered} $$ which converges for $\alpha < 2$.

However the original integral clearly converges also for $2< \alpha$: in this case we have to invert the exponential splitting, obtaining $$ \bbox[lightyellow] { \begin{gathered} I\left| {_{2 \, < \,\alpha} } \right. = \int_0^\infty {\exp \left( { - \left( {ar^{\,\alpha } + br^{\,2} } \right)} \right)^\, r^{\,c} dr} = \int_0^\infty {\exp \left( { - ar^{\,\alpha } } \right)\exp \left( { - br^{\,2} } \right)r^{\,c - \alpha + 1} r^{\,\alpha - 1} \,dr} = \hfill \\ = \frac{1} {{a^{\,\left( {c + 1} \right)/\alpha } \,\alpha }}\int_0^\infty {\exp \left( { - ar^{\,\alpha } } \right)\exp \left( { - \frac{b} {{a^{\,2/\alpha } }}\left( {ar^{\,\alpha } } \right)^{\,2/\alpha } } \right)\left( {ar^{\,\alpha } } \right)^{\,\left( {c + 1} \right)/\alpha - 1} \,d\left( {ar^{\,\alpha } } \right)} = \hfill \\ = \frac{1} {{a^{\,\left( {c + 1} \right)/\alpha } \,\alpha }}\int_0^\infty {e^{\, - x} \exp \left( { - \frac{b} {{a^{\,2/\alpha } }}x^{\,2/\alpha } } \right)x^{\,\left( {c + 1} \right)/\alpha - 1} \,dx} = \hfill \\ \quad \left| {\;\gamma = \frac{b} {{a^{\,2/\alpha } }} = \frac{1} {{\beta ^{\,2/\alpha } }}} \right. \hfill \\ = \frac{1} {{a^{\,\left( {c + 1} \right)/\alpha } \,\alpha }}\int_0^\infty {e^{\, - x} \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\gamma ^{\,k} }} {{k!}}x^{\,\left( {2k + c + 1} \right)/\alpha - 1} } \,dx} = \hfill \\ = \frac{1} {{a^{\,\left( {c + 1} \right)/\alpha } \,\alpha }}\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\gamma ^{\,k} }} {{k!}}\Gamma \left( {\,\frac{{2k + c + 1}} {\alpha }} \right)} \hfill \\ \end{gathered} \tag{2} }$$ which in fact converges for $2< \alpha$.

For $\alpha = 2$ we instead have $$ \bbox[lightyellow] { \begin{gathered} I\left| {_{\alpha \, = \,2} } \right. = \int_0^\infty {\exp \left( { - \left( {a + b} \right)r^{\,2} } \right)^\, r^{\,c - 1} rdr} = \hfill \\ = \frac{1} {{2\left( {a + b} \right)^{\,\left( {c + 1} \right)/2} }}\int_0^\infty {\exp \left( { - \left( {a + b} \right)r^{\,2} } \right)\left( {\left( {a + b} \right)r^{\,2} } \right)^{\,\left( {c + 1} \right)/2 - 1} d\left( {\left( {a + b} \right)r^{\,2} } \right)} = \hfill \\ = \frac{1} {{2\left( {a + b} \right)^{\,\left( {c + 1} \right)/2} }}\Gamma \left( {\frac{{c + 1}} {2}} \right) \hfill \\ \end{gathered} \tag{3} }$$

---- addendum -----

For the case $\alpha = 4$ you can, among other ways, "complete the square": $$ \bbox[lightyellow] { \begin{gathered} I_{\,\alpha \, = \,4} = \int_{r = 0}^\infty {\exp \left( { - \left( {ar^{\,4} + br^{\,2} } \right)} \right)^\, r^{\,c} dr} = \hfill \\ = \int_{r = 0}^\infty {\exp \left( { - a\left( {r^{\,2} + \left( {\frac{b} {{2a}}} \right)} \right)^{\,2} + \frac{{b^{\,2} }} {{4a}}} \right)^\, r^{\,c} dr} = \hfill \\ = \frac{1} {{2\sqrt a ^{\,\frac{{c + 1}} {2}} }}\exp \left( {\frac{{b^{\,2} }} {{4a}}} \right)\int_{r = 0}^\infty {\exp \left( { - \left( {\sqrt a \,r^{\,2} + \left( {\frac{b} {{2\sqrt a }}} \right)} \right)^{\,2} } \right)^\, \left( {\sqrt a r^{\,2} } \right)^{\,\frac{{c - 1}} {2}} d\left( {\sqrt a r^{\,2} } \right)} = \hfill \\ = \frac{1} {{2\sqrt a ^{\,\frac{{c + 1}} {2}} }}\exp \left( {\frac{{b^{\,2} }} {{4a}}} \right)\int_{x = 0}^\infty {\exp \left( { - \left( {x + \left( {\frac{b} {{2\sqrt a }}} \right)} \right)^{\,2} } \right)^\, x^{\,\frac{{c - 1}} {2}} dx} = \hfill \\ = \frac{1} {{4a^{\,\frac{{c + 1}} {4}} }}\left( {\Gamma \left( {\frac{{c + 1}} {4}} \right){}_1F_1 \left( {\frac{{c + 1}} {4}\;;\;\frac{1} {2}\;;\;\frac{{b^{\,2} }} {{4a}}} \right) - \frac{b} {{\sqrt a }}^\, \;\Gamma \left( {\,\frac{{c + 3}} {4}} \right)\;{}_1F_1 \left( {\frac{{c + 3}} {4}\;;\;\frac{3} {2}\;;\;\frac{{b^{\,2} }} {{4a}}} \right)} \right) \hfill \\ \end{gathered} \tag{4} }$$

You obtain the same result if you rewrite the (2) as $$ \begin{gathered} I_{\,2\, < \,\alpha \;} = \frac{1} {{a^{\,\left( {c + 1} \right)/\alpha } \,\alpha }}\;\sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\gamma ^{\,k} }} {{k!}}\Gamma \left( {\,\frac{{2k + c + 1}} {\alpha }} \right)} = \hfill \\ = \frac{1} {{a^{\,\left( {c + 1} \right)/\alpha } \,\alpha }}\;\left( {\sum\limits_{0\, \leqslant \,k} {\frac{{\gamma ^{\,2k} }} {{\left( {2k} \right)!}}\Gamma \left( {\,\frac{{4k + c + 1}} {\alpha }} \right)} - \sum\limits_{0\, \leqslant \,k} {\frac{{\gamma ^{\,2k + 1} }} {{\left( {2k + 1} \right)!}}\Gamma \left( {\,\frac{{4k + c + 3}} {\alpha }} \right)} } \right) \hfill \\ \end{gathered} $$ and note that for $\alpha = 4$, the ratio $t_{k+1}/t_{k}$ among the terms in each sum is a rational function in $k$, which allows to tranform the sum into a hypergeometric function.