`IfI(m , n)=int_0^1x^(m-1)(1-x)^(n-1)dx ,(m , n in I ,m ,ngeq0),t h e n` (a) `I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m-n))dx` (b) `I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m+n))dx` (c) `I(m , n)=int_0^oo(x^(n-1))/((1+x)^(m+n))dx` (d) `I(m , n)=int_0^oo(x^n)/((1+x)^(m+n))dx`

Options

A. `I(m,n)=int_(0)^(oo) (x^(m-1))/((1+x)^(m-n))dx`
B. `I(m,n)=int_(0)^(oo) (x^(m))/((1+x)^(m+n))dx`
C. `I(m,n)=int_(0)^(oo) (x^(n-1))/((1+x)^(m+n))dx`
D. `I(m,n)=int_(0)^(oo)(x^(n))/((1+x)^(m+n))dx`

Answer

C

Explanation

Putting `x-1/(1+y), dx=-1/((1+y)^(2))dy`, we get
`I(m,n)=int_(0)^(1) x^(m-1)(1-x)^(n-1)dx`
`=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`
Since `I(m,n)=I(n,m)` we have
`I(m,n)=int_(0)^(oo) (x^(m-1))/((1+x)^(m+n))dx=int_(0)^(oo)(x^(n-1))/((1+x)^(m+n))dx`