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I've been searching the web for a way to prove that $\int^{\infty}_{-\infty}{\sin(x)/x} = \pi$ with complex analysis, because I have a problem of consistency.

I found two, carried in the following link : Computing $\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x$ with residue calculus. But then I wanted to find it by using the fact that :

$\sin(x)/x = \text{Im}(e^{ix}/x)$

To do so, I shifted the integration by $-i$ in the complex plane, showing that the integral on $[-\infty , \infty]$ is equal to the one on $[-\infty -i, \infty -i]$, since the integrand vanish on the vertical borders of the rectangle when we tend to infinity. I did this to get rid of the pole on the contour of integration.

Then, by using the residues theorem on the contour $C_1 : z = y-i, y \in [-R,R]$ and $C_2 : z = Re^{it}-i, t \in [0,\pi]$. Given the integral on $C_2$ vanishes, we have then that :

$\int^{\infty}_{-\infty}{\sin(x)/x} = \text{Im}(\int^{\infty}_{-\infty}{e^{ix}/x}) = \text{Im}(2\pi i\,\text{Res}(e^{iz}/z, 0)) = 2 \pi$

Which gives me the answer with a factor of $2$. I don't understand were did I go wrong ? I think it has something to do with the fact that I take the imaginary part of the integral, but I don't really know... Can someone spot my mistake ?

If needed, I can provide further detail on my calculations.

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marked as duplicate by Guy Fsone, Sahiba Arora, Ethan Bolker, Robert Soupe, Namaste Feb 5 '18 at 23:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Note also that $$\frac{\sin(\pi x)}{\pi x} = \int_{-1/2}^{1/2} \exp(2 \pi i x f) \; df = \int_{-\infty}^{\infty}Rect(f) \exp(2 \pi i x f) \; df \; ,$$ so the Fourier transform gives $$\int_{-\infty}^{\infty} \frac{\sin(\pi x)}{\pi x} \exp(-2\pi i f x) \; dx = Rect(f).$$ Scaling $x$ and evaluating at $f=0$ gives the expected result. $\endgroup$ – Tom Copeland May 12 '15 at 20:16
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The problem with your computation is that $$ \int_{-\infty-i}^{\infty-i}\frac{\sin(x)}{x}\,\mathrm{d}x \ne\mathrm{Im}\left(\int_{-\infty-i}^{\infty-i}\frac{e^{ix}}{x}\,\mathrm{d}x\right) $$ since the integrand on the left has no singularities and the integrand on the right has one at $0$. One way to get around this is to use $\mathrm{Im}\left(\frac{e^{ix}-1}{x}\right)$ which also has no singularities. Another is to use the whole of the sine function as I did in this answer to your cited question.


Using the contours $$ \gamma_1=[-R,R]\tag{1} $$ and $$ \gamma_2=Re^{i[0,\pi]}\tag{2} $$ Consider the integral $$ \begin{align} \int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x &=\mathrm{Im}\left(\int_{-\infty}^\infty\frac{e^{ix}-1}{x}\,\mathrm{d}x\right)\\ &=\mathrm{Im}\left(\int_{\gamma_1\cup\gamma_2}\frac{e^{iz}-1}{z}\,\mathrm{d}x-\int_{\gamma_2}\frac{e^{iz}-1}{z}\,\mathrm{d}z\right)\tag{3} \end{align} $$ Since there are no singularities of $\frac{e^{ix}-1}{x}$, the integral over $\gamma_1\cup\gamma_2$ is $0$. However, the integral over $\gamma_2$ is a bit more interesting.

Note that $$ \begin{align} \left|\,\int_{\gamma_2}\frac{e^{iz}}{z}\,\mathrm{d}z\,\right| &\le\int_0^\pi e^{-R\sin(\theta)}\,\mathrm{d}\theta\\ &=2\int_0^{\pi/2} e^{-R\sin(\theta)}\,\mathrm{d}\theta\\ &\le2\int_0^{\pi/2} e^{-2R\theta/\pi}\,\mathrm{d}\theta\\[3pt] &=\frac\pi{R}(1-e^{-R})\\[3pt] &\le\frac\pi{R}\\[6pt] &\to0\tag{4} \end{align} $$ and $$ \begin{align} \int_{\gamma_2}\frac1z\,\mathrm{d}z &=\int_0^\pi\frac1{Re^{i\theta}}iRe^{i\theta}\,\mathrm{d}\theta\\ &=\int_0^\pi i\,\mathrm{d}\theta\\[6pt] &=i\pi\tag{5} \end{align} $$ Putting together $(3)$, $(4)$, and $(5)$, we get $$ \begin{align} \int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x &=\mathrm{Im}(i\pi)\\[6pt] &=\pi\tag{6} \end{align} $$

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