# How to show that $\lim_{\varepsilon \rightarrow 0^+} \int_0^\infty \frac{\varepsilon}{\varepsilon^2+x}\sin(1/x) \, dx=0$?

So I tried integration by parts and substitution but I couldn't bound the integral on the inside so that the limit would tend towards zero. Any hints on how to do this?

Here is the work I have done so far

\begin{align} \int_0^t \frac{\varepsilon}{\varepsilon^2+x} \sin(1/x) \, dx & = \int_0^t \frac{\varepsilon x}{\varepsilon^2+x^2}\sin(1/x^2) \, dx \\ &= \left.\arctan \left(\frac x {\varepsilon}\right) x\sin(1/x^2)\right|_0^t-\int_0^t\arctan \left(\frac x \varepsilon \right) \frac d {dx}(x\sin(1/x^2)) \, dx\\ &=-\int_0^t\arctan\left(\frac x \varepsilon \right)\frac{d}{dx}(x\sin(1/x^2)) \, dx \end{align}

• >> $\int_0^t \frac{\epsilon}{\epsilon^2+x}\sin(1/x)dx=\int_0^\infty\frac{\epsilon x}{\epsilon^2+x^2}\sin(1/x^2)dx$<< where did your $t$ go? Commented Sep 13, 2017 at 18:26
• For small $\epsilon$, only the contribution to the integral from a small neighborhood of $0$, something like $(0,M\epsilon^2)$ for some $M$ not depending on $\epsilon$, can be significant (except perhaps in the case of an unbounded domain where you must be more careful). The integral on a small neighborhood of zero should be nearly zero because of the oscillations of $\sin(1/x)$.
– Ian
Commented Sep 13, 2017 at 18:26
• the first row equality seems incrrect Commented Sep 13, 2017 at 18:30
• I have corrected it. It was a typo Commented Sep 13, 2017 at 18:39
• As $\varepsilon\,\downarrow\,0$ you have $\arctan\left( \dfrac x \varepsilon \right) \to \dfrac \pi 2. \qquad$ Commented Sep 13, 2017 at 18:51

Using the change of variable $$u= \frac{\varepsilon}{x}$$ i.e $$dx = -\frac{\varepsilon}{u^2}du$$ leads to

$$\int_0^\infty \frac{\varepsilon}{\varepsilon^2+x}\sin(1/x) \, dx=\int_0^\infty \frac{\sin(x/\varepsilon)}{x^2+x/\varepsilon}\, dx$$

Use the convergence dominated theorem as follow

Fix $$0<\varepsilon \le 1$$

Let $$f_\varepsilon(x) = \frac{\sin(x/\varepsilon)}{\varepsilon^2+x/\varepsilon}$$

then, for every $$x>0$$ $$\lim_{\varepsilon \rightarrow 0}f_\varepsilon(x) = \frac{\sin(x/\varepsilon)}{x^2+x/\varepsilon} = 0$$ Since $$|\sin(x/\varepsilon)|\le 1$$

1. For $$x>1$$ we have that

$$|f_\varepsilon(x)| = | \frac{\sin(x/\varepsilon)}{x^2+x/\varepsilon}| \le \frac{1}{x^2}$$ Since $$|\sin a|\le 1$$.

1. For $$0 using the fact that $$|\sin a|\le |a|$$ we have

$$|f_\varepsilon(x)| = |\frac{\sin(x/\varepsilon)}{x^2+x/\varepsilon}| \le\frac{x/\varepsilon}{x^2+x/\varepsilon} \le 1$$

1. +2) imply that $$|f_\varepsilon(x)|\le \frac{1}{x^2}\chi_{(1,\infty)}(x)+\chi_{(0,1)}(x)$$

Setting $$\phi(x)=\frac{1}{x^2}\chi_{(1,\infty)}(x)+\chi_{(0,1)}(x)$$

Where $$\chi_B$$ is the characteristic function of the set B i.e $$\chi_B(x) = 1$$ if $$x\in B$$ and $$\chi_B(x) = 0$$ elsewhere.

Then, it is easy to check that, $$\phi \in L^1(0,\infty)$$.

Conclusion $$f_\varepsilon(x)\to 0$$ as $$\varepsilon \to0$$ and $$|f_\varepsilon(x)|\le \phi(x)$$ with $$\phi \in L^1(0,\infty)$$. hence by convergence dominated theorem,

$$\lim_{\varepsilon \to 0} \int_{0}^{\infty} f_\varepsilon(x) dx =0$$

## Here is another and more direct way to come across this issue. Still we use the convergence dominated theorem as follows

Set $h(x,\varepsilon) = \frac{\varepsilon}{\varepsilon^2+x}\sin(1/x) \,$

Obviously for every $x>0$

$$\lim_{\varepsilon \rightarrow 0^+} \frac{\varepsilon}{\varepsilon^2+x}\sin(1/x) =0$$

Let set $$g(x) = \sup_{\varepsilon >0} \{|h(x,\varepsilon)|\} = \sup_{\varepsilon >0} \{\frac{\varepsilon}{\varepsilon^2+x}|\sin(1/x)|\} = \frac{|\sin(1/x)|}{2\sqrt{x}}$$

with $g\in L^1(0,\infty)$

Thus for very $\varepsilon>0$, $$|h(x,\varepsilon)|\le g(x) =\frac{|\sin(1/x)|}{2\sqrt{x}}$$ with $g\in L^1(0,\infty)$ hence by convergence dominated theorem,

$$\lim_{\varepsilon \rightarrow 0^+} \int_0^\infty \frac{\varepsilon}{\varepsilon^2+x}\sin(1/x) \, dx=0$$

Indeed,Let prove that $g(x)= \frac{|\sin(1/x)|}{2\sqrt{x}}$ and $g\in L^1(0,\infty)$

$$\frac{\partial }{\partial \varepsilon}\left(\frac{\varepsilon}{\varepsilon^2+x}\right) = \frac{x-\varepsilon^2}{(\varepsilon^2+x)^2} = 0\Longleftrightarrow \varepsilon = \pm\sqrt{x}$$ Therefore for $\varepsilon\in (0,\infty)$ the only critical value of the function $\varepsilon \mapsto \frac{\varepsilon}{\varepsilon^2+x}$ is given by, $\varepsilon = \sqrt{x}$. Moreover, we have that

$$\lim_{\varepsilon \rightarrow {\infty,}{0^+} } \frac{\varepsilon}{\varepsilon^2+x} \, =0$$ we then, conclude that, $\varepsilon = \sqrt{x}$ is the global maximum of the function $\varepsilon \mapsto \frac{\varepsilon}{\varepsilon^2+x}$ i.e

$$\sup_{\varepsilon>0}\{\frac{\varepsilon}{\varepsilon^2+x}\} =\frac{\sqrt{x}}{\sqrt{x}^2+x}=\ = \frac{1}{2\sqrt{x}}\,$$

Now let show that $g\in L^1(0,\infty)$. $$\int_{0}^{\infty}g(x) dx =\int_{0}^{\infty}\frac{|\sin(1/x)|}{2\sqrt{x}} dx =\int_{1}^{\infty}\frac{|\sin(1/x)|}{2\sqrt{x}} dx +\int_{0}^{1}\frac{|\sin(1/x)|}{2\sqrt{x}} dx\le \int_{1}^{\infty}\frac{1}{2x^{\frac{3}{2}}} dx +\int_{0}^{1}\frac{1}{2\sqrt{x}} dx <\infty.$$

Where in the first integral we used $|\sin a|\le |a|$ , in the second we used $|\sin a|\le 1$