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I was having trouble with the following integral: $\int_{0}^\infty \frac{\sin(x)}{x}dx$. My question is, how does one go about evaluating this, since its existence seems fairly intuitive, while its solution, at least to me, does not seem particularly obvious.

BelowAverageIntelligence
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1 Answers1

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Let $I(s)$ be given by

$$I(s)=\int_0^\infty \frac{e^{-sx}\sin(x)}{x}\,dx \tag1$$

for $s\ge 0$. Note that $\lim_{s\to \infty}I(s)=0$ and that $I(0)=\int_0^\infty \frac{\sin(x)}{x}\,dx$ is the integral of interest.

Differentiating $I(s)$ as given by $(1)$ (this is justified by uniform convergence of the "differentiated" integral for $s\ge \delta>0$) reveals

$$\begin{align} I'(s)&=-\int_0^\infty e^{-sx}\sin(x)\,dx\\\\ &=-\frac{1}{1+s^2} \tag 2 \end{align}$$

Integrating $(2)$, we find that $I(s)=\pi/2-\arctan(s)$ whence setting $s=0$ yields the coveted result

$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\frac\pi2$$

Mark Viola
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  • How do you compute the integral of $e^{-sx}\sin(x)$? – user39082 Jan 03 '21 at 12:33
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    @user39082 There are a variety of ways. One way forward, we write $\sin(x)=\text{Im}(e^{ix})$. – Mark Viola Jan 03 '21 at 15:18
  • integrating $-\dfrac{1}{1+s^2}$ just equals $-\arctan(s)$. How should this equal $\pi/2-\arctan(s)$ ? – Leon Jun 20 '21 at 15:15
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    @leon Clearly we have $$I(s)=\int_s^\infty \frac1{s'^2+1},ds'=\frac\pi2 -\arctan(s)$$ – Mark Viola Jun 20 '21 at 15:36
  • But don't we got $\displaystyle{\int_s^{\infty} -\dfrac{1}{{s'}^2+1},\mathrm{ds'} = -\frac{\pi}{2}+\arctan{s}}$ ? – Leon Jun 20 '21 at 15:46
  • @leon Leon, it's evident that we have $$\int_s^\infty -\frac1{s'^2+1},ds'=-I(s)$$ – Mark Viola Jun 20 '21 at 20:06
  • @MarkViola could you explain why it becomes $I(s)=\int_s^\infty \frac{1}{s'^2+1}ds^\prime$ instead of $\int_0^s \frac{1}{s' ^2+1}ds^\prime$? Thanks in advance. – Messi Lio Nov 21 '22 at 12:33
  • @MessiLio $I(0)\ne 0$, which is the implication of the statement $I(s)=\int_0^s \frac1{s'^2+1},ds'$. $I(s)=\lim_{s\to\infty}\int_s^\infty \frac1{s'^2+1},ds'$ implies that $I(s)\to 0$ as $s\to \infty$ is consistent. Does that make sense? – Mark Viola Nov 21 '22 at 14:26
  • I think this argumentation is not completely finished: As you write, $I'(s)$ can be only computed for $s > 0$, but after integrating you plug in $s = 0$, however the equality $I(s) = \pi / 2 - \arctan(s)$ has only been shown for $s > 0$. You need some additional machinery to make this rigorous, i.e. show that the integral $$\int_0^b e^{-sx} \frac{\sin(x)}{x}$$ converges uniformly in $s> 0$ to $$\int_0^{\infty} e^{-sx} \frac{\sin(x)}{x}$$ for $b \to \infty$. This requires a bit of work. Correct me if I'm wrong. – Richard Mar 15 '23 at 11:30
  • @Richard Actually, the improper integral $\int_0^\infty e^{-sx}\frac{\sin(x)}{x},dx$ converges uniformly for $s\ge 0$. Hence, $I(s)$ is continuous for $s\in [0,\infty)$ and we have $\lim_{s\to 0^+}I(s)=I(0)$. So, there is no issue with the analysis herein. – Mark Viola Mar 20 '23 at 03:41
  • @MarkViola ok, I have somehow overread that you wrote this in your answer, sorry ... – Richard Mar 21 '23 at 12:09
  • @user39082 Integration by parts should do nicely. – homosapien Jan 02 '24 at 17:46