modified gamma integral 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 $\Gamma(z)$.
 A: 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.)
A: 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,
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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 $\ds{\tt\mbox{Whittaker Function}}$.

