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$\mathrm{sinc}(x)$ is defined as $\frac{\sin(x)}{x}$ except continuous at $x=0$ (insert the removable singularity).

The derivative of $\mathrm{sinc}(x)$ is usually given as the derivative of $\frac{\sin(x)}{x}$, namely $$\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2},$$ but this has the same problem as $\frac{\sin(x)}{x}$. You have to re-insert the removable singularity at $x=0$.

Is there a more convenient form of the derivative of $\mathrm{sinc}(x)$, say perhaps using $\mathrm{sinc}(x)$ itself, that doesn't have this issue? (don't say piecewise; if piecewise was convenient we wouldn't have $\mathrm{sinc}$ in the first place.)

asmeurer
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    Yes, if you are willing to use alternative representations. Note that $f(z)=\frac{\sin(z)}{z}$ can be represented by the power series $$f(z)=\sum_{n=0}^\infty \frac{(-1)^{n}z^{2n}}{(2n+1)!}$$which is analytic. – Mark Viola Jun 24 '16 at 20:39
  • You could write it down as a power series? Not sure if that's more convenient. Edit: beat me to it @Dr.MV (+1) – parsiad Jun 24 '16 at 20:39
  • You need a function that fills in the singularity of $$ \frac{x\cos(x) - \sin(x)}{x} $$ at $x = 0$. Or, if you prefer, $$ \frac{x - \tan(x)}{x} $$ – Ben Grossmann Jun 24 '16 at 20:48
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    Here are five representation of the sinc function.

    $$\begin{align} \text{sinc}(x)&=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)1}\\ &=\prod_{n=1}^\infty \left(1-\frac{x^2}{(n\pi)^2}\right)\\ &=\frac{1}{\Gamma(1+x/\pi)\Gamma(1-x/\pi)}\\ &=\prod_{n=1}^\infty \cos(x/2^n)\\ &=\frac12 \int_{-1}^1 e^{ixt},dt \end{align}$$

     – Mark Viola Jun 24 '16 at 20:54
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    @par In the previous comment, I've posted five distinct representations. -Mark – Mark Viola Jun 24 '16 at 20:55
  • The gamma one is interesting. More complicated, but no singularities (well, at least if you assume $\frac{1}{\Gamma(-n)}$ is 0 for integer $n\geq 0$). – asmeurer Jun 24 '16 at 21:05
  • @asmeurer The most compelling and practical for ease of differentiation are the series and the integral representations. – Mark Viola Jun 24 '16 at 21:19

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