# Closed form for $\int_0^1 \frac{x^a dx}{(1+x^b)^c}$.

By means of a substitution, the integral can be reduced to

$$\int_0^1 \frac{x^p dx}{(1+x)^q}$$ but no method that I've thusfar tried seems tenable.

This Beta-like integral cannot be reduced into the Beta function (i.e. $\int_0^\infty \frac{x^p dx}{(1+x)^q}$) by any immediately obvious method (the denominator ranges from $1$ to $2$, not $0$ to $1$, which no subsitution can resolve). Numerically the integral tends to be a fraction (suggesting that $\Gamma(x)$ may be involved), unless $c=1$, in which case the result is easy to work out.

• You can use power series approach and integrating term by term – Mhenni Benghorbal Dec 6 '13 at 20:24
• This is a particular case of hypergeometric function $_2F_1$ evaluated at $-1$ (compare with its integral representation). I am not sure that it reduces to simpler functions, although I wouldn't completely exclude that possibility. – Start wearing purple Dec 6 '13 at 20:48

As @OL points out the integral is a hypergeometric function:

$$\frac{\, _2F_1(p+1,q;p+2;-1)}{p+1}$$

You can get this by expanding the denominator in a series (by the binomial theorem), integrating term by term, and observing that you get a hypergeometric series. If there is a relationship of some sort between $p$ and $q,$ this might be simplifiable.


${\rm B}\pars{a,b}$, ${\rm B}_{x}\pars{a,b}$ and $_{2}{\rm F}_{1}\pars{a,b;c,d}$ are the Beta, Incomplete Beta and the Hypergeometric functions, respectively.

You can use the variable $t=1+x$ and use the binomial theorem (for extended binomials coefficient to real numbers)

• This is essentially what the other answers do, the equivalence between their forms and your idea basically comes from the definition of the hypergeometric function. – Meow Apr 17 '14 at 13:49