Limit of integral $\lim_{n \to \infty} n\int_0^{\frac{\pi}{4}} \tan^n x\,dx$

Question

Compute

$$\lim_{n \to \infty}n \int_0^{\frac{\pi}{4}}\tan^n x\,dx$$

I tried to define a recurrence with $I_n=\int_0^{\frac{\pi}{4}} \tan^n x$ :

$I_0 = \frac{\pi}{4}, I_1=\ln\sqrt{2}$

and $I_{n}=\frac{1}{n-1}-I_{n-2}$, but I can't complete it.

Answer 1

Define $f : \mathbb{R} \to \mathbb{R}, f(x)=\dfrac{1}{1+x^2}$. Now, integrating by parts:

$$ \begin{aligned} \int_0^1 (n+1)x^nf(x)\,dx &= \left[x^{n+1}f(x)\right]_0^1-\int_0^1x^{n+1}f'(x)\,dx \\ &= f(1)+2\int_0^1 \frac{x^{n+2}}{(1+x^2)^2}dx \end{aligned} $$

Using $1\leq 1+x^2\leq 2$ over $[0,1]$ we get

$$\frac{1}{2(n+3)}=\frac{1}{2}\int_0^1 x^{n+2}dx\leq 2\int_0^1\frac{x^{n+2}}{(1+x^2)^2}dx \leq 2\int_0^{1} x^{n+2}dx=\frac{2}{n+3}$$

and squeezing, we can easily see that:

$$2\int_0^1\frac{x^{n+2}}{(1+x^2)^2}dx \to 0$$

and therefore:

$$\lim_{n\to \infty}n\int_0^1 x^nf(x)\,dx=\lim_{n\to \infty}\frac{n}{n+1}\int_0^1 (n+1)x^nf(x)\,dx=f(1)=\frac{1}{2}$$

Now substitute $x \to \tan x$ to get that the limit equals $\dfrac{1}{2}$.