Let m,n be two positive real numbers and define `f(n)=int_(0)^(oo)x^(n-1)e^(-x)dx` and `g(m,n)=int_(0)^(1)x^(m-1)(1-m)^(n-1)dx`.
It is known that f(n) for n `gt` 0 is finite and g(m, n) = g(n, m) for m, n `gt` 0.
`int_(0)^(oo)(x^(m-1))/((1+x)^(m+n))dx=`

Options

A. g(m,n)
B. `g(m-1,n)`
C. `g(m-1,n-1)`
D. `g(m,n-1)`

Answer

A

Explanation

`g(m,n)=int_(0)^(1)x^(m-1)(1-x)^(n-1)dt`
Put `x=(1)/(1+y)`
`rArr" "g(m,n)=int_(oo)^(0)(1)/((1+y)^(m-1))(1-(1)/(1+y))^(n-1)(-(1)/((1+y)^(2)))dy`
`" "=int_(0)^(oo)(y^(n-1))/((1+y)^(m+n))dy`
`" "=int_(0)^(oo)(x^(n-1))/((1+x)^(m+n))dx`