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.

  • $\begingroup$ You can use power series approach and integrating term by term $\endgroup$ – Mhenni Benghorbal Dec 6 '13 at 20:24
  • $\begingroup$ 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. $\endgroup$ – 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)

  • $\begingroup$ 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. $\endgroup$ – Meow Apr 17 '14 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.