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I have been working on this limit for days, but I am not getting it. The question is

Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$

Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since $$e^x + \sin tx > e^x-1 > x>0$$ And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.

WHAT I TRIED:

I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens: $$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$ Making the change of variables $u = 2n \pi x$ I get \begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge \sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&= \sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$ $$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$ I get the estimate \begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align} Summing all up, I got that $$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$ As $n \to \infty$ the series converges to the Riemann integral $$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$

Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.

Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.

Answers