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I'm looking to simplify the integral

$$\int\nolimits_0^{\infty}\dfrac{(t^b+1)^n}{(1+t)^{nb+2}} dt$$.

(This arises out of the sum of a bunch of Beta functions, ie $\displaystyle\sum_{i=0}^{n} \binom{n}{i} B(1+ib,1+(n-i)b)$ with $b$ a probably irrational constant, $\approx 1.64677$. I already know about this version of the simplification.)

I'm not very experienced in simplifying integrals like this, and I would welcome any help or suggestions you would provide.

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    Anything special about $n$; e.g. it being an integer? And you expect $b$ to always take that value?2011-08-23
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    $n$ is an integer, $b$ is the solution to $\Gamma(1+2b)=4(\Gamma(1+b))^2$. I'd like to get a (simpler!) function for this integral in terms of $b$ and $n$ so I can see how incrementing $n$ by 1 changes the value and what happens with this expression in general.2011-08-23

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