I am reading a paper where the authors rewrite the following expression (for a continious function $\alpha$) into an integral:
$$\lim_{n\to \infty} \left(\min_{1\leq j \leq n}\left(\sum_{k=1}^{j-1} \frac{1-\alpha(k/n)}{n(1-\alpha(j/n))} + \frac{1}{n}\sum_{k=j}^n \prod_{\ell =1}^k \alpha(\ell/n)^{1/n}\right) \right) = \inf_{x\in [0,1]} \left( \int_{0}^x \frac{1-\alpha(y)}{1-\alpha(x)}dy + \int_{x}^1 e^{\int_0^y \log \alpha(w) dw}dy\right)$$
They quote the result as "standard rieman sum analysis".
I understand how they derived the first term (just a standard Rieman sum), but I don't see the later. Specifically, we have
$$\frac{1}{n}\sum_{k=j}^n \prod_{\ell=1}^k \alpha(\ell/n)^{1/n} = \frac{1}{n}\sum_{k=j}^n e^{\frac{1}{n} \sum_{\ell=1}^k \log \alpha(\ell/n) }$$
How do we conclude this is the same as the second term in the RHS?