1.

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)^(1)(x^(m-1)+x^(n-1))/((1+x)^(m+n))dx=

Answer»

g(N, m)
`g(m-1,n+1)`
`g(m-1,n-1)`
`g(m+1,n-1)`

Solution :`I=int_(0)^(1)(x^(m-1)+x^(n-1))/((1+x)^(m+n))DX`
`=int_(0)^(1)(x^(m-1))/((1+x)^(m+1))dx+int_(0)^(1)(x^(n-1))/((1+x)^(m+n))dx`
`=I_(1)+I_(2)`
In `I_(2)`, put `x=(1)/(t)`, then `I_(2)=int_(oo)^(1)((1)/(t^(n-1)))/((1+(1)/(t))^(m+n))dx`
`""=int_(1)^(oo)(x^(m-1))/((1+x)^(m+n))dx`
`therefore""I=int_(0)^(1)(x^(m-1))/((1+x)^(m+n))dx+int_(1)^(oo)(x^(m-1))/((1+x)^(m+n))dx`
`=int_(0)^(oo)(x^(m-1))/((1+x)^(m+n))dx=g(m,n)`


Discussion

No Comment Found

Related InterviewSolutions