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How can I show that $\int_0^\infty \frac{\sin(1/x)}{x}\ dx$ converges?

I have that $\sin(x)\leq x$ for $x\geq 0$ so then $\sin(1/x)\leq 1/x$ for $x\geq 0$. It follows then that $\int_1^\infty \frac{\sin(1/x)}{x}\ dx \leq \int_1^\infty \frac{1}{x^2}\ dx = 1$ so it doesn't diverge to $\infty$. I think it's also easy to show that it doesn't diverge in the oscillatory sense because of the dampening.

But how does one go about showing that $\int_0^1 \frac{\sin(1/x)}{x}\ dx$ converges?

I'm studying for an analysis exam so anything else that someone might want to mention in that sort of thing would be appreciated.

Olivier Oloa
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John Smith
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  • intuitively : note that $\int_0^1 \frac{1}{x} dx$ diverges, but not much is needed for it to converge. in particular, it diverges exactly in the same way that $\int_1^\infty \frac{1}{x} dx$, and you have to remember that $\int_1^\infty \frac{f(x)}{x} dx$ converges whenever $f(x)$ is periodic and zero-mean. in fact even $\int_1^\infty \frac{f(x)}{x^\epsilon} dx$ converges (it reduces to an alternated series !). finally, you'll have to see why $\int_0^1 \frac{f(1/x)}{x} dx$ is exactly the same, and hence $\int_0^1 \frac{\sin(1/x)}{x}dx$ converges. – reuns Apr 21 '16 at 22:40

5 Answers5

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But how does one go about showing that $\int_0^1 \frac{\sin(1/x)}{x}\ dx$ converges?

One may observe that, by the change of variable $\displaystyle u=\frac1x$, $$ \int_0^1 \frac{\sin(1/x)}{x}\ dx=\int_1^\infty\frac{\sin u}u\:du. $$ Then one may perform an integration by parts, $$ \int_1^M\frac{\sin u}u\:du=\left.\frac{-\cos u}u\right|_1^M-\int_1^M\frac{\cos u}{u^2}\:du $$ and the latter integral is absolutely convergent: $$ \left|\int_1^M\frac{\cos u}{u^2}\:du\right|\leq \int_1^M\frac1{u^2}\:du<\infty $$ giving the convergence of $\displaystyle \int_1^\infty\frac{\sin u}u\:du$ thus the convergence of $\displaystyle \int_0^1 \frac{\sin(1/x)}{x}\ dx$.

J.G.
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Olivier Oloa
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We can do a $u$-substitution with $u=\dfrac{1}{x} \quad du=-\dfrac{1}{x^2}dx=-u^2dx$

$\int_\infty^0 \sin(u)\cdot \dfrac{u}{-u^2}du=\int^\infty_0\dfrac{\sin u}{u}du$

While this is sufficient to show that the integral converges since $\sin(u)$ is bounded while $u$ is not, it is also interesting to show that the integral is precisely $\dfrac{\pi}{2}$.

You can read more about the sinc function.

Eliza
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For $\int_1^{\infty} \frac{\sin(1/x)}{x}dx,$ you really need absolute values. We actually have $|\sin u|\leq u$ for $u\geq 0$. Now, you may estimate as you have done to show that piece is finite.

Rick Sanchez
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Simple trick: If we set $ x=\frac{1}{2u}$ that is $\frac{1}{x} =2u$ then, $dx = -\frac{1}{2u^2}du$ and Given that $\lim_{u\to \infty}\frac{ \sin^2 u}{u} = \lim_{u\to 0}\frac{ \sin^2 u}{u} =0$ we get

\begin{split} \int_0^{+\infty} \frac{\sin(1/x)}{x}\ dx &=& \int_0^{+\infty} 2u \sin(2u) \left(\frac{1}{2u^2}du\right)= \int_0^{+\infty} \frac{\sin(2u)}{u} \ du\\&=& \int_0^{+\infty} \frac{2\cos u\sin u}{u} \ du= \int_0^{+\infty} \frac{(\sin^2 u)'}{u} \ du~~~~\text{since}~~~~(\sin^2 u)' =2\cos u\sin u\\&=& \left[\frac{ \sin^2 u}{u} \right]_0^\infty+\int_0^{+\infty} \frac{ \sin^2 u}{u^2} \ du~~~\text{by integration by part}~\\&=&\color{blue}{\int_0^{+\infty} \frac{ \sin^2 u}{u^2}du=\int_0^{+\infty} \frac{ \sin u}{u}du =\frac{π}{2 }} \end{split}

you Can get the last line from the first line or see this post here and use the following:Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

Proof of the convergence However, For $u>1$ $$\frac{\sin^2u}{u^2}\le \frac{1}{u^2} \implies \int_1^{+\infty} \frac{ \sin^2 u}{u^2}du \le \int_0^{+\infty} \frac{ 1}{u^2}du$$ and For $u<1$ $$ |\sin u|\le |u|\implies \frac{\sin^2 u}{u^2}\le 1 \implies \int_0^{1} \frac{ \sin^2 u}{u^2}du\le 1.$$

Thus, \begin{split} \color{red}{\int_0^{+\infty} \frac{\sin(1/x)}{x}\ dx = \int_0^{+\infty} \frac{ \sin^2 u}{u^2}du<\infty} \end{split}

Guy Fsone
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$$ \begin{align} \int_0^\infty\frac{\sin(1/x)}{x}\,\mathrm{d}x &=\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\int_0^1\left(\frac{\sin(\pi(x+2k))}{\pi(x+2k)}+\frac{\sin(\pi(x+2k+1))}{\pi(x+2k+1)}\right)\mathrm{d}\pi x\\ &=\sum_{k=0}^\infty\int_0^1\left(\frac{\sin(\pi x)}{x+2k}-\frac{\sin(\pi x)}{x+2k+1}\right)\mathrm{d}x\\ &=\sum_{k=0}^\infty\int_0^1\frac{\sin(\pi x)}{(x+2k)(x+2k+1)}\,\mathrm{d}x\\ &\le\pi\log(2)+\frac2\pi\sum_{k=1}^\infty\frac1{2k(2k+1)}\\ &=\pi\log(2)+\frac2\pi(1-\log(2)) \end{align} $$

robjohn
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