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)(log_(e).(1)/(x))dx=`

Options

A. `(f(n+1))/((m+1)^(n))`
B. `(f(n))/((m+1)^(n+1))`
C. `(f(n+1))/((m+1)^(n+1))`
D. `g(m+1),n+1)`

Answer

C

Explanation

Putting `log_(e)..(1)/(x)=t`
`rArr" "x=e^(-t)`
`rArr" "int_(0)^(1)x^(m)(log_(e)(1)/(x))^(n)dx`
`" "=int_(oo)^(0)e^(-mt)t^(n)(-e^(t))dt`
`" "=int_(0)^(oo)t^(n)e^(-(m+1))dt`
`" "=(1)/((m+1)^(n+1))int_(0)^(oo)t^(n)e^(-y)dy" (putting (m + 1) t = y)"`
`" "=(f(n+1))/((m+1)^(n+1))`