While messing around in Desmos just now, I noticed a strange property of the sine integral.

First recall $$\operatorname{sinc}(x)=\begin{cases}1 & x=0 \\ \frac{\sin x}{x}&\text{otherwise}\end{cases}$$ The cardinal sine. Now let $$\operatorname{Si}(x)=\int_0^x \operatorname{sinc}(\xi)\mathrm d\xi$$ be the sine integral.

Now define $$A(x)=\int_{\pi(x-1/2)}^{\pi(x+1/2)}\operatorname{sinc}(\xi)\mathrm d\xi=\operatorname{Si}\big(\pi(x+1/2)\big)-\operatorname{Si}\big(\pi(x-1/2)\big)$$

I noticed two rather strange things.

  • The positive roots of $A$ are very well approximated by $n+\frac{1}{\pi^2 n}$, where $n\in\mathbb N$.
  • $\operatorname{sinc}(x)$ is reasonably well approximated by $\frac{1}{A(0)}A(x/\pi)$.

To see this for yourself, please go to my Desmos graph.

Question: Can anyone explain these weird and unexpected properties?

  • $\begingroup$ Instead of using $10$, use $\pi^2$. It looks better at matching the roots of $A$. $\endgroup$
    – PC1
    Commented Jun 11, 2023 at 17:36
  • $\begingroup$ @PC1 Oh, nice find! I'll edit it. $\endgroup$
    – K.defaoite
    Commented Jun 11, 2023 at 18:42

2 Answers 2


For more terms and just for your curiosity.

If you perform a series expansion around integer values of $x=n$ and follow by a power series reversion, the estimate of the solution is $$x_{(n)}=n+\frac{1}{\pi ^2 n}+\frac{3 \pi ^2-32}{6 \pi ^4 n^3}+\frac{13384-1500 \pi ^2+15 \pi ^4}{120 \pi ^6 n^5}+O\left(\frac{1}{n^7}\right)$$ which is good even for small values of $n$ $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 1.09758291620 & 1.09754802090 \\ 2 & 2.05016021032 & 2.05015991299 \\ 3 & 3.03362365061 & 3.03362363291 \\ 4 & 4.02526671391 & 4.02526671153 \\ 5 & 5.02023161919 & 5.02023161869 \\ 6 & 6.01686796808 & 6.01686796794 \\ 7 & 7.01446254778 & 7.01446254774 \\ 8 & 8.01265716788 & 8.01265716788 \\ \end{array} \right)$$

This also shows how excellent is your approximation since $$\frac{3 \pi ^2-32}{6 \pi ^4 n^3} \sim -\frac 1 {245n^3}$$

Now, concerning $$f(x)=\text{sinc}(x)-\frac{A\left(\frac{x}{\pi }\right)}{A(0)}$$ a series expansion around $x=0$ gives $$f(x)=\left(\frac{2}{\pi ^2 \text{Si}\left(\frac{\pi }{2}\right)}-\frac{1}{6}\right) x^2+\left(\frac{1}{120}-\frac{\pi ^2-8}{2 \pi ^4 \text{Si}\left(\frac{\pi }{2}\right)}\right) x^4+O\left(x^6\right)$$ which, converted to rational is $$f(x)=-\frac {x^2}{53}+\frac {x^4}{750}+O\left(x^6\right)$$ which is another good approximation.

  • $\begingroup$ I'm not familiar with the process of series reversion that you used to estimate the roots - could you please direct me to some more information on it? $\endgroup$
    – K.defaoite
    Commented Jun 12, 2023 at 19:00
  • $\begingroup$ Sorry for the delay. Have a look at mathworld.wolfram.com/SeriesReversion.html $\endgroup$ Commented Jun 16, 2023 at 5:13

I define $$B(x)=\frac{A(x/\pi)}{A(0)}$$ Using Mathematica (Simplify[Series[B[1/u],u->0],u>0]//Normal), I can see that the Taylor series at $u\to0$ of $B(1/u)$ is: $$B(1/u)\sim\frac1{\mathrm{Si}(\pi/2)}\left(u\sin(1/u)-u^2\cos(1/u)\right)$$

Taking the limit $u\to0$, I see that the term in $\cos$ vanishes and I retrieve that: $$B(x)\sim\frac{\mathrm{sinc}(x)}{\mathrm{Si}(\pi/2)}$$ so the approximation is not scaled properly for $x\to\infty$, by a factor of $\mathrm{Si}(\pi/2)\sim 1.37076$.

You can indeed see on your graph that the approximation undershoots $\mathrm{sinc}(x)$ by that factor. A better approximation (for $x\to\infty$) is $B(x)=\frac12A(x/\pi)$. However it doesn't agree well for $x\to0$. If you are interested to have a good approximation for both $x\to0$ and $x\to\infty$ then maybe you could investigate doing a Padé approximant which agrees with these two points. That's a bit more involved and it really depends what's your objective here.

To show that the roots are approximated by $n+\frac1{\pi^2n}$, you might take the development of $B(1/u)$ and consider that $\cos(u)\sim1$ when $\sin(u)\sim0$ (near a zero) then expand. I think you will find that it matches your approximation.


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