# Closed form or expansion of integral - $\int_{e^{-1/e}}^{N^N} \frac{\arcsin(\cos u)}{\mathrm{W}(\ln u) + 1} \frac{\mathrm{d}u}{u}$

I asked this question a couple days ago, where the solution was brought down to

$$\lim_{N \to \infty} \left( \pi \left(N-\frac{1}{e}\right) - 2\int_{e^{-1/e}}^{N^N} \frac{\arcsin(\cos u)}{\mathrm{W}(\ln u) + 1} \frac{\mathrm{d}u}{u} \right)$$

by @user3002473. I'm stuck on the integral part,

$$\int_{e^{-1/e}}^{N^N} \frac{\arcsin(\cos u)}{\mathrm{W}(\ln u) + 1} \frac{\mathrm{d}u}{u}$$

I'd like to find a closed form or its asymptotic behavior at $\infty$ to evaluate the limit. I was incapable of finding accurate numerical approximations via mathematica or scipy due to the rapid oscillations within the integrand. I also tried IBP, but got stuck there as well.

## 1 Answer

We have that $\arcsin(\cos u)$ is a triangle wave and the term $W(\log u)$ is approximately constant on short intervals, since it behaves like $\log\log u$. The similar integral $$I= \int_{0}^{+\infty}\frac{\arcsin\sin u}{u}\,du$$ has a simple closed form due to a well-known Fourier series: $$I = \frac{4}{\pi}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}\int_{0}^{+\infty}\frac{\sin((2n+1)u)}{u}\,du =2\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}=\color{red}{2G}$$ where $G$ is Catalan's constant. By using integration by parts and this lemma you can achieve good numerical approximation of your integral.