Compute $\lim_{n \rightarrow \infty} \lim_{R \rightarrow \infty} \int_0^R \sin{(x/n)} \sin{(e^x)}dx$.

Question

Another practice preliminary question for you all. This time, a double limit of an integral.

Problem Compute $\lim_{n \rightarrow \infty} \lim_{R \rightarrow \infty} \int_0^R \sin{(x/n)} \sin{(e^x)}dx$. Hint: Integrate by parts.

My issue is the order of the limits. I can't get a nice closed form solution that does not blow up to infinity in the first limit. I've performed integration by parts in order to try to find something that is more easily approximated or to see if the integral "repeats itself", so to say. What I've tried doesn't seem to go anywhere.

My Attempt Define $f_n(x) = \sin(x/n) \sin (e^x)$. For any fixed $x \in \mathbb{R}$ we have that $f_n(x) \rightarrow 0$. Additionally, $|f_n(x)| \leq 1$ for all $n$ and $x$. So we have that $f_n$ is bounded, measurable, and converges pointwise to $0$ on $\mathbb{R}$. At this point, I'd love to conclude that the integral is zero from the Bounded Convergence Theorem and that $\mathbb{R} = \cup_{m \in \mathbb{N}}[m-1,m]$. On every interval as such we have $\lim_{n \rightarrow \infty} \int_{[m-1,m]}f_n = 0$ by the BCT. However, the conclusion seems like taking the limits in reverse order. Is it the case that $\sum_{m \in \mathbb{N}} \lim_{n \rightarrow \infty} \int_{[m-1,m]}f_n = \lim_{n \rightarrow \infty} \sum_{m \in \mathbb{N}} \int_{[m-1,m]}f_n $?

Otherwise I think a solution might be found from the integral over the ascending union $\cup_{n \rightarrow \infty} [0,n]$. I'm certain this problem will require use of the Lebesgue Dominated Convergence Theorem, but I am missing the integrable function that bounds $f_n$.

Thanks in advance for any hints or nudges in the right direction.

Answer 1

$$ \lim_{R\to\infty}\int_0^R \sin{(x/n)} \sin{(e^x)}dx = \lim_{R\to\infty}\int_0^R \underbrace{\sin{(x/n)} e^{-x}}_{u} \cdot \underbrace{e^x \sin{(e^x)}dx}_{dv} $$ $$ =\lim_{R\to\infty}\left. - e^{-x}\sin(x/n) \cos(e^x)\right|_{0}^{R} + \int _0^R \cos(e^x)\cdot(-e^{-x}\sin(x/n)+\frac{1}{n}e^{-x}\cos(x/n))\,dx $$ $$ =\lim_{R\to\infty}\int _0^R \cos(e^x)\cdot(-e^{-x}\sin(x/n)+\frac{1}{n}e^{-x}\cos(x/n))\,dx $$Now use $|\sin(\theta)|<|\theta|$ and take absolute values: $$ \lim_{R\to\infty} \int _0^R \left|\cos(e^x)\cdot\left(-e^{-x}\sin(x/n)+\frac{1}{n}e^{-x}\cos(x/n)\right)\right|\,dx $$ $$ \leq \lim_{R\to\infty} \int _0^R e^{-x} \left|\sin(x/n)+\frac{1}{n}\cos(x/n)\right|\,dx $$ $$ \leq \lim_{R\to\infty} \int _0^R e^{-x} \left(\left|\sin(x/n)\right|+\left|\frac{1}{n}\cos(x/n)\right|\right)\,dx$$ $$ \leq \lim_{R\to\infty} \frac{1}{n}\int _0^R e^{-x}(x+\left|\cos(x/n)\right|)\,dx $$ $$ \leq \lim_{R\to\infty} \frac{1}{n}\int _0^R e^{-x}(x+1)\,dx = \frac{2}{n} $$

Answer 2

Using the same integration by part scheme as @integrand used, namely $u=e^{-x}\sin(x/n)$ and $v=-\cos(e^x)$, we have

$$\begin{align} \lim_{R\to\infty}\int_0^R \sin(x/n)\sin(e^x)\,dx&=\underbrace{\lim_{R\to\infty}\left.\left(-e^{-x}\cos(e^x \right)\sin(x/n)\right)|_{0}^{R}}_{=0}\\\\ &+\lim_{R\to\infty}\int_0^R e^{-x}\left(\frac1n \cos(x/n)-\sin(x/n)\right)\cos(e^x)\,dx\\\\ &=\lim_{R\to\infty}\int_0^R e^{-x}\left(\frac1n \cos(x/n)-\sin(x/n)\right)\cos(e^x)\,dx \end{align}$$

Noting that $|\sin(x/n)\cos(e^x)e^{-x}|\le e^{-x}$, the Dominated Convergence Theorem guarantees that

$$\lim_{n\to\infty}\int_0^\infty e^{-x}\left(\sin(x/n)\right)\cos(e^x)\,dx=\int_0^\infty \lim_{n\to\infty}e^{-x}\left(\sin(x/n)\right)\cos(e^x)\,dx=0$$

Similarly, we find that

$$\left|\int_0^\infty e^{-x}\left(\frac1n \cos(x/n)\right)\cos(e^x)\,dx\right|\le \frac1n \to 0 \,\,\text{as}\,\,n \to \infty$$

We conclude therefore, that

$$\lim_{n\to\infty}\lim_{R\to\infty}\int_0^R \sin(x/n)\sin(e^x)\,dx=0$$

And we are done.