# 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.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{t \equiv {1 \over 1 + x}\quad\iff\quad x = {1 \over t} - 1}$: \begin{align} &\color{#00f}{\large\int_{0}^{1}{x^{p}\,\dd x \over \pars{1 + x}^{q}}}= \int_{1}^{1/2}t^{q}\pars{1 - t \over t}^{p}\,\pars{-\,{\dd t \over t^{2}}} = \int_{1/2}^{1}t^{q - p -2}\pars{1 - t}^{p}\,\dd t \\[3mm]&=\int_{0}^{1}t^{q - p -2}\pars{1 - t}^{p}\,\dd t - \int_{0}^{1/2}t^{q - p -2}\pars{1 - t}^{p}\,\dd t \\[3mm]&={\rm B}\pars{q - p - 1,p + 1} - {\rm B}_{1/2}\pars{q - p - 1,p + 1} \\[3mm]&=\color{#00f}{\large{\rm B}\pars{q - p - 1,p + 1} - {2^{1 + p - q} \over q - p - 1}\ _{2}{\rm F}_{1}\pars{q - p -1,-p;q - p, \half}} \\[3mm]&\mbox{with}\ \Re\pars{p} > -1. \\[5mm]& \end{align}

${\rm B}\pars{a,b}$, ${\rm B}_{x}\pars{a,b}$ 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