I'm trying to see if the following statement is true:
For $x>1$, as $t\to\infty$ $$\int \frac{1}{x+\sin(tx)}dx
\rightarrow \int \frac{1}{\sqrt{x^{2}-1}}dx \tag{1}$$
This means the convergence of the family of antiderivatives
I was inspired by the answer in this question. This is the result from that answer:
$$ \lim_{t \to \infty}\int_a^b f(x,\sin(tx)) dx = \frac{1}{2\pi}\int_a^b \int_0^{2\pi} f(x,\sin(s)) ds dx. $$
Using that result, I let $f(x,y)=\frac{1}{x+y}$ and computed $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{x+\sin(s)}ds=\frac{1}{\sqrt{x^{2}-1}},(x>1)$, then I was left with: $$\lim_{t\to\infty}\int_{a}^{b}\frac{1}{x+\sin(tx)}dx=\int_{a}^{b}\frac{1}{\sqrt{x^{2}-1}}dx \ ,(1<a\leq b) \tag{2}$$ Let's say $F_{t}(x)$ is an antiderivative of $\frac{1}{x+\sin(tx)}$ and $G(x)=\ln(x+\sqrt{x^{2}-1})$ is an antiderivative of $\frac{1}{\sqrt{x^{2}-1}}$, from $(2)$, we get $$\lim_{t\to\infty}\left((F_t(b)-G(b))-(F_t(a)-G(a)\right)=0 \tag{3}$$ Question: Does this mean $\exists C\in \mathbb{R},\forall x>1:\lim_{t\to\infty}(F_t(x)-G(x))=C$ ? And thus $(1)$ is true?
I took a look at the graph and I don't see how the shape of $F_{t}(x)$ tends to be the shape of $G(x)$. Although they kind of resemble each other, I see $F_{t}(x)$ still behaves differently with its alternating steep and gradual increases.
