# Loop Limit of improper integral

I need a hint on starting to find the value of the limit of this integral: $$\int_0^1\left(\ln(1+x^{-n})\right)^n\,dx.$$ This is part of a problem of finding the limit. So either a recursive technique to start or finding a lower and upper bound is very helpful.

• I would start by formally messing with the series $\ln(1+a) = a - \frac{1}{2}a^2 + \frac{1}{3}a^3 - \cdots$ to get an idea of what the integral should be doing. Of course, that's not a proof, but it's something to start with. Nov 15 '13 at 8:07
• We are dealing with a sub-unitary number raised to a negative power, which is the same as a super-unitary number raised to a positive power, meaning that the quantity inside the logarithm grows with n, rising towards infinity as n increases. The fact that it's afterwards risen to a positive power, which also rises towards infinity, does not make things any better, i.e., your function, as well as its integral, rise towards infinity as n increases. Nov 15 '13 at 9:39
• As for its minimum, derive under the integral sign with regards to n. There's no other way, I'm afraid. Then, whether the zeroes of the new expression can be found analytically, or only using something like Newton's method is beyond me. Nov 15 '13 at 9:45

Hint. By letting $t=1/x$, we get $$\int_0^1\left(\ln(1+x^{-n})\right)^n\,dx=\int_1^\infty\frac{\left(\ln(1+t^{n})\right)^n}{t^2}\,dt\geq \int_e^\infty\frac{\left(\ln(t^{n})\right)^n}{t^2}\,dt\geq n^n\int_e^\infty\frac{1}{t^2}\,dt=\frac{n^n}{e}.$$
Hint: Note that the integrand is nonnegative, and $\ln(1+x^{-n})>2$ if $\displaystyle 0<x<\frac{1}{\sqrt[n]{e^2-1}}$.