Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral? 
Are there any ways to evaluate $\int^\infty_0\frac{\sin x}{x}dx$ without using double integral?

I can't find any this kind of solution. Can anyone please help me? Thank you.
 A: By continuity of the function $\frac{1}{\sin(\theta/2)} - \frac{2}{\theta}$ on $[-\pi,\pi]$ we know that it is integrable on there. Furthermore the identity $\sin ((N+\frac{1}{2})\theta) = \sin N \theta \cos \frac{\theta}{2} + \cos N \theta \sin \frac{\theta}{2}$ shows we can apply Riemann - Lebesgue below to get that
$${\int}_{-\pi}^\pi \sin\left(\left(N + \frac{1}{2}\right)\theta\right)\left(\frac{1}{\sin(\theta/2)} - \frac{2}{\theta} \right)d\theta \rightarrow 0.$$
This also means to say that $$ \int_{-\pi}^\pi 2\frac{\sin \left((N + \frac{1}{2})\theta\right)   }{\sin \theta} d \theta \rightarrow \pi$$
since we already know what the integral of the Dirichlet Kernel is on $[-\pi,\pi]$. Making a change of variables, we get that 
$$\int_{- (N+1/2)\pi}^{(N+1/2)\pi} \frac{\sin \theta}{\theta} d \theta \rightarrow \pi.$$
However the term in the integrand is symmetric about the line $x = 0$ and so we get that 
$$\lim_{N \rightarrow \infty} \int_{0}^{(N+1/2)\pi} \frac{\sin \theta}{\theta} \,  d\theta = \pi/2.$$
$\hspace{6in}  \square$
A: Here is a complex integration without the S.W. theorem. Define
$$f(z):=\frac{e^{iz}}{z}\Longrightarrow\,\, Res_{z=0}(f)=\lim_{z\to 0}\,zf(z)=e^{i\cdot 0}=1$$
Now we choose the following contour (path to line-integrate the above complex function):
$$\Gamma:=[-R,-\epsilon]\cup\gamma_\epsilon\cup[\epsilon,R]\cup\gamma_R\,\,,\,\,0<e<<R\in\Bbb R^+$$
With $\,\displaystyle{\gamma_M:=\{z=Me^{it}\;:\;M>0\,\,,\,\,0\leq t\leq \pi\}}\,$
Since our function $\,f\,$ has no poles within the region enclosed by $\,\Gamma\,$, the integral theorem of Cauchy gives us
$$\int_\Gamma f(z)\,dz=0$$
OTOH, using the lemma and its corollary here and the residue we got above , we have
$$\int_{\gamma_\epsilon}f(z)\,dz\xrightarrow[\epsilon\to 0]{}\pi i$$
And by Jordan's lemma we also get
$$\int_{\gamma_R}f(z)\,dz\xrightarrow [R\to\infty]{}0$$
Thus passing to the limits $\,\epsilon\to 0\,\,,\,\,R\to\infty\,$
$$0=\lim_{\epsilon\to 0}\lim_{R\to\infty}\int_\Gamma f(z)\,dz=\int_{-\infty}^\infty\frac{e^{ix}}{x}dx-\pi i$$
and comparing imaginary parts in both sides of this equation (and since $\,\frac{\sin x}{x}\,$ is an even function) , we finally get
$$ 2\int_0^\infty\frac{\sin x}{x}=\pi\Longrightarrow \int_0^\infty\frac{\sin x}{x} dx=\frac{\pi}{2} $$
A: You should see this link about Feynman way, a very elegant way. Laplace transforms is also a very fast and nice way. Enter here.
A: Using the following version of the Fourier transform,
$$F(\omega)=(\mathcal{F}f)(\omega)=\int_{-\infty}^\infty e^{-i\omega t}f(t)\,dt,$$
then Parseval's theorem looks like
$$\int_{-\infty}^\infty \overline{f(t)}\,g(t)\,dt=\frac{1}{2\pi}\int_{-\infty}^\infty \overline{F(\omega)}\,G(\omega)\,d\omega.$$
Let $H(t)$ and $\delta(t)$ denote the Heaviside step function and the Dirac delta, respectively. Then
$$H(t+1)-H(t-1)\stackrel{\mathcal{F}}{\longmapsto}2\frac{\sin{\omega}}{\omega}$$
and
$$\delta(t)\stackrel{\mathcal{F}}{\longmapsto}1.$$
Now we use Parseval's theorem and get
$$\int_0^\infty \frac{\sin{x}}{x}\,dx=\frac{1}{2}\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx=\frac{1}{4}\int_{-\infty}^\infty 2\frac{\sin{\omega}}{\omega}\cdot 1\,d\omega =$$
$$=\frac{\pi}{2}\cdot\frac{1}{2\pi}\int_{-\infty}^\infty 2\frac{\sin{\omega}}{\omega}\cdot 1\,d\omega = [\text{Parseval}] = $$
$$=\frac{\pi}{2}\int_{-\infty}^\infty (H(t+1)-H(t-1))\delta(t)\,dt=\frac{\pi}{2}.$$
A: $$\int_{0}^{\infty}\frac{\sin x}xdx=L[\sin x]=\int_{0}^{\infty}\frac 1{s^{2}+1}ds=\lim_{s\to\infty}\arctan s=\pi/2$$
