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I have seen a question asking to find the value of $\int_{-100}^{100} \frac{\sin{x}}{x} dx$.

I have to confess that I didn't think this was possible.

If I expand the $\sin$ using Taylor series, then unless the endpoints of the domain lie inside $(-1,1)$, the result will diverge. And in fact, I think it might even diverge if they lie inside this domain - is that correct?

On the other hand, we can use complex contour integrals to show the are bounded by the graph and the entire $x$ axis is equal to $\pi$. Therefore, we should expect the area bounded by the graph and this finite part of the $x$ axis to be both finite and less than $\pi$.

I know the antiderivative is $\operatorname{Si}(x)$ but I thought this was only defined for $x>0$. Perhaps we could try evaluating $\operatorname{Si}(400)-\operatorname{Si}(0)$ and doubling it or something.

In short, I have no idea what to do here.

Thanks for your help.

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  • $\begingroup$ Observe that $\frac{\sin(x)}{x}$ is a bounded function (as $x$ approaches $0$, the value approaches $1$). Therefore, the integral certainly exists. The Taylor series for $\sin(x)$ also converges everywhere, so why do you need the $(-1,1)$ condition? $\endgroup$ Feb 13, 2016 at 22:00
  • $\begingroup$ The Taylor series for $\sin(x)$ converges for every value of $x$, thanks to the factorial at the denominator. It even converges if you take all the terms with the positive sign. It even converges if you add the even terms. Indeed $e^x= 1+x+x^2/2! +x^3/3!+\cdots$ always converges. $\endgroup$ Feb 13, 2016 at 22:07
  • $\begingroup$ @GiovanniResta Ok. So I get $2.100-\frac{2}{3} \frac{100^3}{3!} + \frac{2}{5} \frac{100^5}{5!} - \dots$. I guess that's the best we can do since it doesn't simplify beyond that? Secondly, to see convergence are you just comparing adjacent terms e.g. $\frac{a_{n+1}}{a_n}=\frac{n!}{(n+1)!} x$ and saying that for fixed $x$ and large enough $n$, this ratio drops below 1? $\endgroup$
    – user11128
    Feb 13, 2016 at 22:21
  • $\begingroup$ @user11128 In general if you have an alternating series whose terms tend to zero and are decreasing in absolute value (as here it happens at a certain point) then the Leibniz criterion tells us that the series converges. The fact that this series converges absolutely can also be achieved by comparing it with the expansion of $2e^{100}$, if one already knows that the series for $e^x$ converges everywhere. $\endgroup$ Feb 13, 2016 at 22:30
  • $\begingroup$ @GiovanniResta, you realize that you can place your comment as an answer. $\endgroup$ Feb 13, 2016 at 22:34

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Instead of using Taylor series which is incredibly inefficent for values as large as $x=100$ (test for yourself how many terms you need to get an precision of say $10^{-3}$) you should derive an asymptotic expansion for this integral. An asymptotic expansion is something like the evil and not well behaved but powerful brother of an usual series expansion. It doesn't converge in a usual sense but gives incredible accurate results for large arguments as long as one takes the right number of terms (which means terminate it at a meaningful order). I will not go into any formal details here, but just show you how to apply this procedure for the example at hand. Write

$$ I(z)=\int_{-z}^z \frac{\sin(x)}{x}=\int_{-\infty}^{\infty} \frac{\sin(x)}{x}-2\int_{z}^{\infty} \frac{\sin(x)}{x} $$

The first integral is a classic and yields $\pi$, for the second let's integrate by parts with $\sin(x)=u'$ and $1/x=v$

$$ \int_{-z}^z \frac{\sin(x)}{x}=\pi-2\frac{\cos(z)}{z}+2\int_{z}^{\infty} \frac{\sin(x)}{x^2} $$

we see immediatly that the remaining integral is much smaller then the rest of the rhs. so just we are lazy we just stop here and neglect the integral. Now doing the numbers (putting $z=100$) we get

$$ I(100)\approx 3.124346276144 $$

which is incredibly close to the real value $3.124450933778$ keeping in mind that we did just one integration by parts (the error is $\approx 10^{-5}$).

Integrating by parts $N-$ times we could derive the coefficent of the resulting series go as $N!$ which means that it is badly divergent! This is what i meaned by 'bad brother': An divergent series (for any fixed $x$) which gives incredibly exact result if we truncate it at a senseful point!

Fascinating,right?

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  • $\begingroup$ Thanks for the great reply. I understand why further integration by parts lead to terms with coefficient $n!$ but I'm confused why this is happening. We know the answer should be convergent (see earlier posts) and yet now it seems to be diverging? $\endgroup$
    – user11128
    Feb 14, 2016 at 12:44
  • $\begingroup$ There is an answer in terms of a convergent series and an answer in terms of an asymptotic series which gives accurate value as long as the number of terms is not too large (note also that it stays finite if $N$ is fixed) . This are just two different ways to approach this problem whereas the second one is much more effective for the value of $z$ at hand. $\endgroup$
    – tired
    Feb 14, 2016 at 12:54
  • $\begingroup$ For formal in depth discussion of a very related integral (the so called exponential integral), see here: macs.hw.ac.uk/~simonm/ae.pdf $\endgroup$
    – tired
    Feb 14, 2016 at 12:57
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If you know the function $Si(x)$, then the solution should be fairly simple.

In fact, the function is bounded from above and continuous, so there should be no Problem evaluating the integral, and also the Taylor Series converge everywhere. Your guess that the integral should be approximately $\pi$ is also correct.

The function $Si(x)$ is used for positive x because you can "mirror" the function to only use with positive values.

Observe, that for all x: $\frac{Sin(x)}{x} = \frac{Sin(-x)}{-x}$, so your integral evaluates to:

$\int \limits_{-100}^{100} \frac{Sin(x)}{x} = 2 Si(100)$

This is already not far from $\pi$, and I don't know if there is a very nice term for it.

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