Equivalent $ f (t) = \int \limits_t^{2t} \ln (2 + \sin x) \, dx $, where $ t \to \infty $

Question

Find the equivalent to the function $ f (t) = \int\limits_t^{2t} \ln (2 + \sin x) \, dx $, where $ t \to +\infty $

I tried to find some exact estimates, but something simple is not observed. I will be glad to any help.

Answer 1

Let $h$ be a periodic function with period $\xi$, and let $\mu = \frac{1}{\xi} \int_0^\xi h(t)\, dt$ be its mean value. Now consider functions of the form $$f(t) = \int_{a(t)}^{b(t)} h(x)\, dx.$$

Claim: If $\lim_{t \to \infty} f(t) = \infty$, then $f$ is asymptotically equivalent to $g(t) = \mu b(t) - \mu a(t)$, i.e., $\lim_{t \to \infty} \frac{g(t)}{f(t)} = 1$.

Proof: Define \begin{align*} \eta(t) &= f(t) - g(t) \\ &= \int_{a(t)}^{b(t)} (h(x) - \mu)\, dx. \end{align*} The integrand $h(x) - \mu$ is a periodic function of period $\xi$ and mean value zero; that is, $\int_y^{y + \xi} (h(x) - \mu)\, dx = 0$ for any real number $y$. The value of an integral over $h(x) - \mu$ is therefore determined by the limits of integration modulo $\xi$. So for all $t$, $$|\eta(t)| \leq \max_{\alpha, \beta \in [0, \xi]} \left| \int_\alpha^\beta (h(x) - \mu)\,dx \right| = K$$ and $K$ is clearly finite. We have thus shown $f(t) - g(t)$ is bounded, which implies asymptotic equivalence, as $$\lim_{t \to \infty} \frac{f(t) - K}{f(t)} \leq \lim_{t \to \infty} \frac{g(t)}{f(t)} \leq \lim_{t \to \infty} \frac{f(t) + K}{f(t)}$$ and if $\lim_{t \to \infty} f(t) = \infty$, then the limits on the outside must equal $1$.

Application of this claim to your specific problem should be straightforward, because $\ln (2 + \sin x)$ is periodic with period $2\pi$. WolframAlpha gives a complicated antiderivative of $\ln (2 + \sin x)$ in terms of the dilogarithm $\operatorname{Li}_2$, and I believe that the $\operatorname{Li}_2$ terms should cancel in an integral over a complete period.