calculus - Limit of a trigonometric integral: $limlimits_{x to infty} int_{0}^{pi} frac{sin (t)}{1+cos^{2}(xt)} , mathrm dt$

For all $x \in \mathbb{R}$, let
$$
{\rm f}\left(x\right)
=\int_{0}^{\pi}\frac{\sin\left(t\right)}{1 + \cos^{2}\left(xt\right)}\,{\rm d}t
$$

Compute the limit when $x\rightarrow +\infty$.

My attempt :

I tried the substitution $u=\sin(t)$ (and $u=\cos^2(xt)$) but it seems worse:
$$
\int _{0}^{1}\!{\frac {u}{ \left( 1+ \left( \cos \left( x\arcsin
\left( u \right) \right) \right) ^{2} \right) \sqrt {1-{u}^{2}}}}{du}
$$
I tried to use a subsequence $(x_n)$ which is tends to $+\infty$ and use the dominated convergence theorem but it didn't work either.

Sorry if my attempt doesn't make much sense.

Thank you in advance.

Answer

We can use a generalization of the Riemann-Lebesgue lemma that states that if $f(x)$ is a continuous function on $[a,b]$, where $0 \le a < b$, and $g(x)$ is a continuous $T$-periodic function on $[0, \infty)$, then $$\lim_{n \to \infty}\int_{a}^{b} f(x) g(nx) \, dx = \left(\frac{1}{T} \int_{0}^{T} g(x) \, dx \right) \left(\int_{a}^{b} f(x) \, dx \right).$$

A proof can be found in THIS PAPER.

The proof uses the fact that a primitive of $g(x)$ can be expressed as $$ \int_{0}^{x} g(t) \, dt = \frac{x}{T} \int_{0}^{T} g(t) \ dt + h(x),$$ where $h(x)$ is $T$-periodic function. (Lemma 2.2)

The regular Riemann-Lesbegue lemma is the case where the average value of $g(x)$ is $0$.

Applying the lemma to this particular integral, we get

$$ \lim_{x \to \infty} \int_{0}^{\pi} \frac{\sin t}{1+\cos^{2}(xt)} \ dt = \left( \frac{1}{\pi} \int_{0}^{\pi} \frac{1}{1+\cos^{2} t} \ dt \right) \left( \int_{0}^{\pi} \sin t \ dt \right). $$

The first integral can be evaluated by making the substitution $u = \tan t$.

This particular example actually appears in the paper, so I'll omit the details.

We end up with

$$ \lim_{x \to \infty} \int_{0}^{\pi} \frac{\sin t}{1+\cos^{2}(xt)} \ dt = \frac{1}{\pi} \left(\frac{\pi}{\sqrt{2}} \right) \left(2\right) = \sqrt{2} .$$