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This question already has an answer here:

I am trying to evaluate the following:

$$\int_{-\infty}^{\infty}\frac{\sin(x)}{x}\, dx$$

My first approach was to find the antiderivative but I can't seem to express it as I have not yet learnt about $\text{Si}(x)$. I then tried replacing the $\sin(x)$ with $(e^{ix}-e^{-ix})/(2i)$ but I just ended something even more complicated. Does making it go from $0$ to $\infty$ by multiplying by $2$ help?

Please help me in evaluating this integral.

By the way, I am familiar with substitution and integration by parts but not complex analysis or contour integration. However, if this question requires something I don't already know, I am willing to try and understand it.

Thanks.

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marked as duplicate by tired, mickep, Crostul, 3SAT, Kamil Jarosz Mar 3 '16 at 14:30

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.

  • $\begingroup$ This is well-known to be a tricky one (although classical). You most likely won't find the solution on your own. You can probably find solutions in undergrad courses, I don't know good references in english. $\endgroup$ – Captain Lama Mar 3 '16 at 12:03
  • $\begingroup$ Maybe this will help: math.stackexchange.com/questions/174072/… $\endgroup$ – Katie Imach Mar 3 '16 at 12:10
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Here's an approach I enjoy; maybe it's outwith the scope of your endeavour, so this may not pose as an appropriate answer notice $$\int \limits_0^\infty e^{-xy} \sin x dy = \frac{\sin x}{x}$$ Hence, \begin{align} \int \limits_{-\infty}^{+\infty}\frac{\sin x}{x}dx &=\int \limits_{-\infty}^{+\infty}\left(\int \limits_0^\infty e^{-xy} \sin x dy \right)dx \\ &= \int \limits_{-\infty}^{+\infty}\left(\int \limits_0^\infty e^{-xy} \sin x dx \right)dy \qquad \text{(Change order of integration)} \\ &=\int \limits_{-\infty}^{+\infty}\left(\frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2} \right)\Biggr \rvert_0^\infty dy \qquad \text{(Integrate inner brackets by parts)} \\ &=\int \limits_{-\infty}^{+\infty}\frac{1}{1+y^{2}}dy \\ &= \pi \end{align}

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  • $\begingroup$ Good answer!! But... Fix it! xD Lots of minus signs are missing! lol $\endgroup$ – Von Neumann Mar 3 '16 at 12:25
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    $\begingroup$ How do you justify changing the order of integration? Since $$\int_{-\infty}^{+\infty} \int_0^{+\infty} \lvert e^{-yx}\sin x\rvert\,dy\,dx = +\infty,$$ that's a delicate step. $\endgroup$ – Daniel Fischer Mar 3 '16 at 12:29
  • $\begingroup$ @1over137 Whoops! Many thanks for your check - All $-\infty$'s present and correct! $\endgroup$ – Kevin Mar 3 '16 at 12:29
  • $\begingroup$ @DanielFischer Great spot Daniel - I hope I'm correct in stating if you consider the strip $(0,a) \times (0,\infty)$ then observing that $\int_0^a \int_0^\infty \lvert e^{-xy} \sin x\rvert dx dy \leq a$ then apply Fubini's theorem. To get the result I quoted, take the limit $a \rightarrow \infty$ $\endgroup$ – Kevin Mar 3 '16 at 12:33
  • $\begingroup$ @Kevin You can change the order first and then let $a\to \infty$. But new problem comes. You need to justify the interchange of $\lim_{a\to \infty}$ and the intergral sign. $\endgroup$ – Sam Wong Oct 18 '18 at 10:48
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1. If you are familiar with Dirac Delta$$ \int_{-\infty}^{\infty}{\sin(x) \over x}\,{\rm d}x = \int_{-\infty}^{\infty}\left({1 \over 2}\,\int_{-1}^{1}{\rm e}^{{\rm i}kx}\,{\rm d}k\right) \,{\rm d}x = \pi\int_{-1}^{1}{\rm d}k \int_{-\infty}^{\infty}{{\rm d}x \over 2\pi}\,{\rm e}^{{\rm i}kx} = \pi\int_{-1}^{1}{\rm d}k\,\delta(k) = \pi $$

2. Trick Calculus way

$$\begin{align*} \int_{-\infty}^{\infty} \frac{\sin x}{x} \; dx &= 2 \int_{0}^{\infty} \frac{\sin x}{x} \; dx \\ &= 2 \int_{0}^{\infty} \sin x \left( \int_{0}^{\infty} e^{-xt} \; dt \right) \; dx \\ &= 2 \int_{0}^{\infty} \int_{0}^{\infty} \sin x \, e^{-tx} \; dx dt \\ &= 2 \int_{0}^{\infty} \frac{dt}{t^2 + 1} \\ &= \vphantom{\int}2 \cdot \frac{\pi}{2} = \pi. \end{align*}$$

You would love also Feynman differentiation under the integral sign

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    $\begingroup$ how do u justify the exchange of integrations in line one? i mean that is the only tricky part in this calculation $\endgroup$ – tired Mar 3 '16 at 12:20
  • $\begingroup$ @tired I don't :D There is not so much rigor in that calculation, indeed I'm thinking that it's quite wrong.. even if it works.. $\endgroup$ – Von Neumann Mar 3 '16 at 12:24
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    $\begingroup$ ok i will help u out: for example u could use an exponential regulator $\lim_{\epsilon \rightarrow 0}\int_R dxe^{-\epsilon |x|}...$ and take the limit in the end $\endgroup$ – tired Mar 3 '16 at 12:26
  • $\begingroup$ @tired Oh yes! This is what we do in QFT regularization processes, like to calculate self-energies! Haha here it is! Thanks! $\endgroup$ – Von Neumann Mar 3 '16 at 12:35
  • $\begingroup$ yeah i know, i am also part of this buisness $\endgroup$ – tired Mar 3 '16 at 12:38

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