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I have the following integral

$$\int_0^{+\infty} t^{z-1} e^{-t} \frac1{(kt + 1)^s}\mathrm dt$$

where $k>0, s > 0$. How would you suggest to solve it? Without $\frac1{(kt + 1)^s}$ it would be equal to $\Gamma(z)$.

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Is this related to one of your other recent questions? – cardinal Apr 23 '11 at 1:39
no, it's another problem, I actually fixed a typo – ACAC Apr 23 '11 at 1:54
up vote 3 down vote accepted

It depends what you want. I tried really hard for a few hours, and here are some things I derived.
Say $$I=\int_0^\infty t^{z-1}e^{-t}\frac{1}{(kt+1)^s}dt$$

The nicest other representations I found were $$I=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\text{B}(s-w,w)k^{-w}\Gamma(z-w)dw$$ Where $\text{B}(x,y)$ is the beta function. Also there is the more symmetric equation

$$I=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\int_{0}^{\infty}\mu^{s-1}\chi^{z-1}e^{-(\mu+\chi)}e^{-k\mu\chi}d\mu d\chi,$$ and

$$I=k^{-z}e^{\frac{1}{k}}\int_{0}^{1}\mu^{s-z-1}(1-\mu)^{z-1}e^{-\frac{1}{\mu k}}d\mu.$$

Now after the last one, and countless failed attempts I started to believe this had hypergeometric functions in it (which I have less experience with). However, I am aware that $$\int_0^1 e^{kt}t^{s-1}(1-t)^{z-1}dt$$ is a type of hypergeometric function, and the last integral was very close to this. Indeed, plugging everything into wolfram alpha yields $$I=k^{-s}\Gamma(z-s){}_1F_{1}\left(s;\ s-z+1;\ \frac{1}{k}\right)+\frac{k^{-z}\Gamma(z)\Gamma(s-z){}_1F_{1}\left(z;\ z-s+1;\ \frac{1}{k}\right)}{\Gamma(s)}$$ where ${}_1F_1$ is the Confluent Hypergeometric Series of the first kind.

Hope that helps,

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thanks a lot for your effort! I guess I will need to buy mathematica for future needs – ACAC Apr 23 '11 at 9:21
@bob: Actually it is free online: – Eric Naslund Apr 23 '11 at 19:14
In my case wolfram doesn't find any result. which equation did you plug in? – ACAC Apr 23 '11 at 22:55
@Bob: The same one. At first it didn't work, and timed out, but there is a button you can click to give it more time, and then it worked. – Eric Naslund Apr 23 '11 at 22:58
@Bob: that's why using Tricomi instead of Kummer here is profitable; the expression using Kummer has (removable) singularities. In the Tricomi solution, the only restriction is that $z$ not be a nonpositive integer. – J. M. Apr 27 '11 at 8:36

If you formally insert the binomial series

$$(1+kt)^{-s}=\sum_{j=0}^\infty \frac{(s)_j}{j!}(-kt)^j$$

into your integral, swap summation and integration, and integrate termwise (whose validity can be justified with Watson's lemma, but I shall skip the justification), we arrive at the formally divergent series

$$\sum_{j=0}^\infty \frac{(s)_j}{j!}(-k)^j \Gamma(z+j)=\Gamma(z){}_2 F_0(s,z;;-k)$$

The trick here is that the divergent hypergeometric series ${}_2 F_0(a,b;;z)$ can be formally shown to correspond to an asymptotic series for the integral mentioned in the OP, and can also be related to the (more familiar?) Tricomi confluent hypergeometric function $U(a,b,z)$, the other standard solution to the second-order differential equation satisfied by the Kummer confluent hypergeometric function mentioned in Eric's answer.

More directly, using a certain integral representation for the Tricomi confluent hypergeometric function, we have the "identity"

$${}_2 F_0(s,z;;-k)=k^{-s} U\left(s,1+s-z,\frac1{k}\right)=k^{-z} U\left(z,1+z-s,\frac1{k}\right)$$

The closed form $\Gamma(z)k^{-z} U\left(z,1+z-s,\frac1{k}\right)$ can then be shown to be equivalent to the expression in Eric's answer via this identity connecting the Kummer and Tricomi functions. (A justification is sketched in this reference.)

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Nice answer! – Eric Naslund Apr 24 '11 at 16:11

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_0^{\infty}t^{z - 1}\expo{-t}\,{1 \over \pars{kt + 1}^{s}}\,\dd t:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\int_0^{\infty}t^{z - 1}\expo{-t}\,% {1 \over \pars{kt + 1}^{s}}\,\dd t} =\int_0^{\infty}t^{z - 1}\pars{1 + {t \over 1/k}}^{-s}\expo{-t}\,\dd t \end{align}

By following the $\ds{\tt\mbox{MathWorld}}$ definition $\pars{5}$ of the Whittaker Function: $$ \mbox{}-q - \half + m = z - 1\,,\quad\mbox{}q - \half + m = -s\quad \mbox{we get}\quad \left\lbrace\begin{array}{rcl} q & = & {1 - z - s \over 2} \\[2mm] m & = & {z - s \over 2} \end{array}\right. $$ Then,

\begin{align}&\color{#c00000}{\int_0^{\infty}t^{z - 1}\expo{-t}\,% {1 \over \pars{kt + 1}^{s}}\,\dd t} \\[3mm]&=\Gamma\pars{\half - {1 - z - s \over 2} + {z - s \over 2}}\, {1 \over \expo{-\pars{1/k}/2}}\,{1 \over \pars{1/k}^{\pars{1 - z - s}/2}} {\rm W}_{\pars{1 - z -s}/2,\ \pars{z - s}/2}\pars{1 \over k} \end{align}

\begin{align}&\color{#66f}{\large\int_0^{\infty}t^{z - 1}\expo{-t}\,% {1 \over \pars{kt + 1}^{s}}\,\dd t} \\[3mm]&=\color{#66f}{\large\exp\pars{1 \over 2k}k^{\pars{1 - z - s}/2}\ \Gamma\pars{z}{\rm W}_{\pars{1 - z -s}/2,\ \pars{z - s}/2}\pars{1 \over k}} \end{align}

$\ds{\rm W}$ is the $\ds{\tt\mbox{Whittaker Function}}$.

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