$$\int_0^1 (x\ln x)^n dx$$
How should I proceed with the above integral? I'm actually supposed to solve it for n = 50, but I don't think reduction formulae would be a good idea (or would they?)
I couldn't establish a closed form for the above integral in terms of $n$. How do I do that? Could some please guide me in the right direction, or provide a solution for better understanding? Thanks!
On
Hint: Use substitution $\ln x=-u$ and then apply the Gamma function definition. The result will simplify to $\dfrac{(-1)^nn!}{(n+1)^{n+1}}$.
It can be done by reduction. In fact, integrate by parts to get $$ \int_0^1 x^m (\log x)^n\;dx = -\frac{n}{m+1}\int_0^1 x^m (\log x)^{n-1}\;dx $$ Using this $n$ times, we get $$ \int_0^1 x^m (\log x)^n\;dx = \frac{(-1)^n n!}{(m+1)^n}\int_0^1 x^m\;dx =\frac{(-1)^n n!}{(m+1)^{n+1}} $$ The case $m=n$ is your integral.
side note
This also works when $m$ is not an integer... $$ \int_0^1 x^{1/2} (\log x)^n\;dx = \frac{(-1)^n n!}{(3/2)^{n+1}} $$