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