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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
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2 Answers 2

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
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@bob: Actually it is free online: wolframalpha.com –  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
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@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
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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
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