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

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

## marked as duplicate by Guy Fsone, Arnaud D., Raskolnikov, J. M. is a poor mathematician, Rolf HoyerNov 10 '17 at 16:23

• You can find a Complex proof and a Differentiation-under-the-integral-sign proof here en.wikipedia.org/wiki/Dirichlet_integral – Alexander Vlasev Sep 8 '12 at 7:33
• There is the one using Riemann - Lebesgue.... – user38268 Sep 8 '12 at 7:37

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

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

• In your answer can you expand out what "the S.W. theorem" means without using abbreviations? All that comes to mind when I see "S.W. theorem" is Stone--Weierstrass theorem and that doesn't seem relevant to this task at all. – KCd Aug 9 '15 at 8:01

You should see this link about Feynman way, a very elegant way. Laplace transforms is also a very fast and nice way. Enter here.

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

$$\lim_{N \rightarrow \infty} \int_{0}^{(N+1/2)\pi} \frac{\sin \theta}{\theta} \, d\theta = \pi/2.$$

$\hspace{6in} \square$

$$\int_{0}^{\infty}\frac{\sin x}xdx=L[\sin x]=\int_{0}^{\infty}\frac 1{s^{2}+1}ds=\lim_{s\to\infty}\arctan s=\pi/2$$

• Could you please elaborate on what is meant by $L[\sin x]$? – Mårten W Jan 19 '13 at 22:32
• that's Laplace transformation of f(x)=sinx @Mårten W – Harshiv Feb 2 at 7:09
• @Harshiv: But the (one-sided) Laplace transform of $\sin{x}$ is given by $\int_0^\infty e^{-sx}\sin{x}\,dx$, and certainly not by $\int_0^\infty \frac{\sin{x}}{x}\,dx$. Something is missing here. – Mårten W Feb 2 at 11:51
• well , let $f(x)=sin\mathit x$ ,now division by x rule: $L[\frac{f(x)}{x}] = \int_s^\infty F(s) ds$ thus LHS of question can be equated to $\int_s^\infty \frac{1}{s^2 + 1} ds$ , solve that with $s=0$ , this is me first time using math formulas here so it may not be as readable as I wish. – Harshiv Feb 3 at 4:57
• not the entire LHS of question, $\frac{sin x}{x}$ is replaced by $\frac{1}{s^2 +1} ds$ I suppose. – Harshiv Feb 3 at 5:16

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 $$\int_0^\infty \frac{\sin{x}}{x}\,dx=\frac{1}{2}\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx=\frac{1}{4}\int_{-\infty}^\infty 2\frac{\sin{\omega}}{\omega}\cdot 1\,d\omega =$$ $$=\frac{\pi}{2}\cdot\frac{1}{2\pi}\int_{-\infty}^\infty 2\frac{\sin{\omega}}{\omega}\cdot 1\,d\omega = [\text{Parseval}] =$$ $$=\frac{\pi}{2}\int_{-\infty}^\infty (H(t+1)-H(t-1))\delta(t)\,dt=\frac{\pi}{2}.$$

• Very nice! I haven't seen this before. +1 – Potato Jan 20 '13 at 0:45