Alternative methods of solving $\int_{0}^{1}x^{\ln^n(x)}\ dx$

Question

I know how to solve this problem as follows: $$\int_{0}^{1}x^{\ln^n(x)}\ dx\\ =\int_{0}^{1}e^{\ln^{n+1}(x)}\ dx\\ =\int_{0}^{1}\sum_{k=0}^\infty\frac{\left[\ln^{n+1}(x)\right]^k}{k!}dx\\ =\sum_{k=0}^\infty\frac{1}{k!}\int_{0}^{1}\ln^{k(n+1)}(x)\ dx\\ =\sum_{k=0}^\infty\frac{\left[\ k(n+1)\ \right]!}{k!}$$ Are there any other methods to getting to this result? Possibly even a closed form of the equation?

Answer 1

With a simple substitution $x=e^u$ the integral takes the form

$$I=\int_{-\infty}^0du\exp(u+u^{n+1})$$

This integral clearly diverges for positive odd integers, since by changing $u\to -u$ we see that:

$$I=\int^{\infty}_0du ~\exp(-u+u^{2m})$$

The integral converges for non-negative even integers, since by the same change of variables one can see that the integrand decays super-exponentially at infinity. In fact, the integral also converges for negative even integers, since the integrand decays exponentially fast at $0$ and $-\infty$, and is otherwise finite. All these are easy to show, but it takes more work to find out if the integral converges, for other, potentially complex values of $n$.

I will restrict myself to even positive integers from now on, setting $\alpha=n+1$ to reduce clutter. The series obtained in the question above diverges for most values on $n$. To obtain a convergent series expand the $e^{-u}$ term and integrate term by term using the definition of the Gamma function:

$$I=\int^{\infty}_0du ~\exp(-u-u^{\alpha})=\sum_{m=0}^{\infty}\frac{1}{m!}\int_{0}^{\infty}u^m\exp(-u^\alpha)=\frac{1}{\alpha}\sum_{m=0}^{\infty}\frac{\Gamma(\frac{m+1}{\alpha})}{m!}$$

This series expansion converges actually for any $\alpha>1$ (whether it actually reflects the actual value of the function defined in the question is not clear to me).

You can obtain the previous divergent result by expanding $\exp(u^{n+1})$ in a power series instead and integrating term by term again.