Is it $1^{\infty}$ ?

Note that the integral $\int_0^{\frac{1}{4}\pi}\left(\frac{1-\tan x}{1+\tan x} \right)^ndx\overbrace{=}^{ \text{}x=\frac{1}{4}\pi-y}\int_0^{\frac{1}{4}\pi}\tan^nxdx$ then substitute $u=\tan x\implies du=\sec ^2 x dx$ and hence $\int_0^{\frac{1}{4}\pi}\tan^nxdx=\int_0^1\frac{u^n}{1+u^2}du= \sum_{k=0}^{\infty}(-1)^k\left(\int_0^1 u^{2k+n}du\right)$ On Integration we yield $\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+n+1}=\sum_{k=0}^{\infty}\left(\frac{1}{n+4k+1}-\frac{1}{n+4k+3}\right)$ the latter series can be written as $\frac{1}{4}\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{4}{n+4k+3}\right)-\frac{1}{4}\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{4}{n+4k+1}\right)\\=\frac{1}{4}\sum_{k=0}^{\infty}\left(\frac{1}{k+1}-\frac{1}{\frac{n-1}{4}+k+1}-\left(\frac{1}{k+1}-\frac{1}{\frac{n-3}{4}+k+1}\right)\right)$ $\therefore \int_0^{\frac{1}{4}\pi}\left(\frac{1-\tan x}{1+\tan x}\right)^ndx=\frac{1}{4}\left(H_{\frac{n-1}{4}}-H_{\frac{n-3}{4}}\right)$ Using the fact ( the asymptotic expansion) of $H_n\approx \gamma +\ln n+\frac{1}{2n}-O(n^{-2})$ we yield $\frac{1}{4}\left(H_{\frac{n-1}{4}}-H_{\frac{n-3}{4}}\right)\approx \frac{1}{2n}\to 0 \;\text{as}\; n\to \infty$ and hence the required limit is $\sim\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^n=\sqrt{e}$ and hence $b\sqrt[b]{a}=2\sqrt{e}\approx 3.219744$ .