integration - Are there any ways to evaluate $int^infty_0frac{sin x}{x}dx$ 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.

Answer

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

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

$$2\int_0^\infty\frac{\sin x}{x}=\pi\Longrightarrow \int_0^\infty\frac{\sin x}{x} dx=\frac{\pi}{2}$$