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

Question

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.

Answer 1

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

Answer 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.

Answer 3

Let $$ I_n = \int_0^\infty \mathrm{e}^{-x} \sin\left(\frac{n}{x}\right) \mathrm{d} x = \Im\left( \underbrace{\int_0^\infty \exp\left(-x + i \frac{n}{x}\right)\mathrm{d}x}_{J_n} \right) $$ We can rotate the contour by angle $\phi$ clock-wise, so as to keep the real part of the argument of exponent negative, i.e. for $0 \leqslant \phi<\frac{\pi}{2}$: $$ J_n = \int_0^\infty \exp\left(-x + i \frac{n}{x}\right)\mathrm{d}x = \int_0^{\mathrm{e}^{-i \phi} \infty} \exp\left(-z + i \frac{n}{z}\right)\mathrm{d}z = \int_0^\infty \exp\left(-r \mathrm{e}^{-i \phi} + i \frac{n}{r}\mathrm{e}^{i \phi}\right) \mathrm{e}^{-i \phi} \mathrm{d}r $$ The above is a table integral (Gradshteyn&Ryzhik 3.471.9, also see question 128687): $$ J_n = 2 \sqrt{n} \mathrm{e}^{-i \frac{\pi }{4}} K_1\left(2 \sqrt{n} \mathrm{e}^{-i \frac{\pi }{4}}\right) = \sqrt{2n} (1-i) K_1\left(\sqrt{2n} (1-i)\right) $$ where $K_\nu(x)$ denoted the modified Bessel function of the second kind. The original integral is thus: $$ I_n = \sqrt{2n} \Im\left( (1-i) K_1\left(\sqrt{2n} (1-i)\right) \right) = \sqrt{2n} \left( \operatorname{kei}_1\left(2\sqrt{n}\right) - \operatorname{ker}_1\left(2\sqrt{n}\right) \right) $$ where $\operatorname{ker}_\nu(z)$ , $\operatorname{kei}_\nu(z)$ are Kelvin functions. The large argument expansion of $K_\alpha(x)$ is well-known: $$ J_n = \frac{\pi}{2} \sqrt{\sqrt{2n} (1-i)} \exp\left(-\sqrt{2n} (1-i)\right) \left( 1 + \mathcal{O}\left(\frac{1}{\sqrt{n}}\right)\right) = \sqrt{\pi} n^{1/4} \exp\left(-\sqrt{2n} + i \left(\sqrt{2n} - \frac{\pi}{8}\right)\right) $$ It now easily follows that $$ \lim_{n \uparrow \infty} I_n =\Im \lim_{n \uparrow \infty} J_n = 0 $$