Show $\lim\limits_{n\to\infty}\int_{0}^{\infty}e^{-x}\sin(\frac{n}{x})~\text{d}x=0$

I'm having trouble showing $$\lim_{n\to\infty}\int_{0}^{\infty}e^{-x}\sin\left(\frac{n}{x}\right)~\text{d}x=0$$ The integrand doesn't converge for any $x$ so I don't know how to use the standard Lebesgue convergence theorems. Thank you for any hints.

Solution 1:

Hint: change variables to ${\displaystyle {1 \over x}}$ and apply the Riemann-Lebesgue lemma.

Solution 2:

The change variables shouble be $u=\frac{n}{x}$. That way, $x=\frac{n}{u}$, and $dx = n\frac{du}{u^2}$. The integral then becomes: $\displaystyle\int_0^{+\infty} \frac{n}{u^2}e^{-\frac{n}{u}}\sin(u)du = \int_0^{+\infty} f_n(u)du$ with $f_n(u)=\frac{n}{u^2}e^{-\frac{n}{u}}\sin(u)$. $\forall n$, we have: $\displaystyle\lim_{u\rightarrow 0}f_n(u)=0$, $\displaystyle\lim_{u\rightarrow +\infty}f_n(u)=0$, $f_n$ is integrable on $\mathbb{R}^+$.

Let us now find $m_n$ such as $||f_n||_{\infty}\leq m_n$. For that we study $g_n(u)= \frac{n}{u^2}e^{-\frac{n}{u}}$. $g_n'(u) = (-\frac{2n}{u^3}+\frac{n^2}{u^4})e^{-\frac{n}{u}}= \frac{n}{u^3}(-2+\frac{n}{u})e^{-\frac{n}{u}}$, and so we see that $g_n$ reaches its maximun when $u=\frac{n}{2}$, and $g_n(\frac{n}{2}) = \frac{4}{n}e^{-2}$.

Finally, $\forall x\in \mathbb{R}^+$ we have $|f_n(x)|\leq\frac{4}{n}e^{-2}\rightarrow 0$, which enables us to use the theorem of the dominated convergence to say that $\displaystyle\lim_{n\rightarrow +\infty}\int_0^{+\infty}f_n = 0$

EDIT: Clarification on the theorem of the dominated convergence: It is easy to see that for all $u>1$, $g_n(u)$ is decreasing in $n$. Let's take an $\varepsilon>0$, and find $N$ and $A>1$ such as $\int_A^{+\infty}g_n(u)du\leq\frac{\varepsilon}{2}$ (a). Then we can apply without any problem the dominated convergence on $[0,A]$, so $\displaystyle\lim_{n\rightarrow +\infty}\int_0^A f_n(u)du=0$, so we can find $N'$ such as $\forall n\geq N'$, $|\int_0^Afn|\leq \frac{\varepsilon}{2}$ (b).

And when you add up inequalities (a) and (b), you get: $\forall n\geq\max(N,N')$, $|\int_0^{+\infty}f_n|\leq\varepsilon$, which achieves the demonstration.