Showing that $\int_0^{+\infty} \frac{\sin(1/x)}{x}\ dx$ converges 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.
 A: 
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 $\displaystyle \int_0^1 \frac{\sin(1/x)}{x}\ dx$.
A: 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.
A: 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.
A: 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}

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