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=

Let m,n be two positive real numbers and define f(n)=int_(0)^(oo)x^(n-

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)(log_(e).(1)/(x))dx=
Answer» `(f(n+1))/((m+1)^(n))` `(f(n))/((m+1)^(n+1))` `(f(n+1))/((m+1)^(n+1))` `g(m+1),n+1)` Solution :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))`

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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=

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