Evaluate the following integral by using beta and gamma functions $\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}.}$ How to evaluate this

Question

Evaluate the following integral by using beta and gamma functions $\int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}.}$

How to evaluate this?

I took $\frac{2x}{1+x}=y$ and trying to convert into beta function, but I could not get the answer.

Answer 1

Expanding Nemo's answer, setting $x = \frac1{y}$, so $dy = -\frac{dy}{y^2}$.

Then

$\begin{array}\\ \int_{0}^{1}\frac{x^{n-1}}{(1+x)^{m+n}}dx &=-\int_{\infty}^{1}\frac{1}{y^{n-1}(1+1/y)^{m+n}}\frac{dy}{y^2}\\ &=\int_1^{\infty}\frac{y^{m+n}dy}{y^{n+1}(y+1)^{m+n}}\\ &=\int_1^{\infty}\frac{y^{m-1}dy}{(y+1)^{m+n}}\\ \end{array} $

and his result is verified.

Note: Here is Nem's answer which, for some reason, was deleted:

\begin{align} \int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx&=\int_{0}^{1}\frac{x^{m-1}}{(1+x)^{m+n}}dx+\int_{0}^{1}\frac{x^{n-1}}{(1+x)^{m+n}}dx\\ &=\int_{0}^{1}\frac{x^{m-1}}{(1+x)^{m+n}}dx+\int_{1}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}dx\\ &=\int_{0}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}dx\\ &=B(m,n) \end{align}

Answer 2

$$\int\limits_{0}^{1} \frac{x^{m}+x^{n}}{(1+x)^{m+n}} \mathrm{d}x = \mathrm{B}(m,n)$$

Proof: We begin with the basic integral definition of the beta function

\begin{equation} \mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t \label{eq:bf1} \tag{1} \end{equation} for $\Re x \gt 0$ and $\Re y \gt 0$.

\begin{equation} \mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v \label{eq:bf2} \tag{2} \end{equation}

To derive this equation, we begin with \eqref{eq:bf1} and make the substitution

\begin{equation} t = \frac{v}{1+v} \end{equation}

to obtain

\begin{equation} \mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x-1}} \Big(1 - \frac{v}{1+v}\Big)^{y-1} \frac{1}{(1+v)^{2}} \mathrm{d} v \end{equation}

simplification yields \eqref{eq:bf2}.

\begin{equation} \mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{(v^{x-1}+v^{y-1})}{(1+v)^{x+y}} \mathrm{d} v \label{eq:bf3} \tag{3} \end{equation}

To obtain \eqref{eq:bf3} we begin with \eqref{eq:bf2} and break up the integral

\begin{equation} \mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v + \int\limits_{1}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v \end{equation}

and designate the last integral as I.

For I, we make the substitution $w = v^{-1}$

\begin{equation} \mathrm{I} = \int\limits_{0}^{1} \frac{1}{w^{x-1}} \frac{w^{x+y}}{(1+w)^{x+y}} \frac{1}{w^{2}} \mathrm{d} w \end{equation}

Simplifying and then making the substitution for I yields \eqref{eq:bf3}.

Note that equation \eqref{eq:bf3} for the beta function appears as Equation 3 on page 9 (pdf page 35) of Higher Transcendental Functions (Bateman Manuscript), Volume 1.