Simplifying an Integral

Question

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.

Answer 1

Since $n \in \mathbb{N}$, you can expand $(1+t^b)^n$ using binomial formula, and use

$$ \int_0^\infty \frac{t^{b k}}{(1+t)^{b n+2}} \mathrm{d} t = B(b \, k+1, b \, (n-k)+1) $$ where $B(a,b)$ is a Euler Beta function. The answer than is

$$ \mathcal{I}_n = \sum_{k=0}^n \binom{n}{k} \frac{\Gamma(b\, k+1) \Gamma(b\, (n-k)+1)}{\Gamma(b\, n+2)} $$ I do not think this can be made much simpler.

Added: The sequence $\mathcal{I}_n$ can be efficiently evaluated numerically.

In[46]:= bn = 
 b /. FindRoot[Gamma[1 + 2 b] == (2 Gamma[b + 1])^2, {b, 3/2}, 
   WorkingPrecision -> 50]

Out[46]= 1.6467727665452770645330865956356454132544625403300

In[33]:= f[n_Integer] := 
 Sum[Binomial[n, k] Beta[bn k + 1, 1 + (n - k) bn], {k, 0, n}]