Evaluating $\lim\limits_{x\to\infty}{x\int_0^{\pi/4}{\ln(1+\tan^xt)dt}}$

Question

I'm trying to evaluate $$\lim_{x\to\infty}{x\int_0^{\pi/4}{\ln(1+\tan^xt)dt}}$$ I have no idea about the integral inside.
For the first time, I tried the substitute $x\to\frac1x$ and I got $$\lim_{x\to0}{\frac1x\int_0^{\pi/4}{\ln(1+\tan^{1/x}t)dt}}$$and then I used lhopital but it seems to be more complicated.
I think $\forall0\le\varepsilon<\frac\pi4 \ s.t.\ $$$\lim_{x\to\infty}{x\int_0^{\pi/4}{\ln(1+\tan^xt)dt}}=\lim_{x\to\infty}{x\int_\varepsilon^{\pi/4}{\ln(1+\tan^xt)dt}}$$but I cannot use this result.
The only thing I can do is proving this limit exists.

Answer 1

Applying the substitution $u = \tan^x t$, we find that

$$ x \int_{0}^{\frac{\pi}{4}} \log(1+\tan^x t) \, dt = \int_{0}^{1} \frac{\log (1+u)}{u} \cdot \frac{u^{1/x}}{1+u^{2/x}} \, du. $$

By the dominated convergence theorem, as $x\to\infty$ we have

$$ \lim_{x\to\infty} x \int_{0}^{\frac{\pi}{4}} \log(1+\tan^x t) \, dt = \frac{1}{2}\int_{0}^{1} \frac{\log (1+u)}{u} \, du = \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{24}. $$