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I'm trying to find a general formula for the $n$-th derivative of the following function $$ \frac{\sin(x)}{x} $$

I've computed the first five derivatives so far without finding any 'structure' in it.

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General Leibniz rule $$ \frac{d^n}{dx^n}f(x)g(x) = \sum \left(\matrix{n \\ k}\right)\frac{d^{n-k}}{dx^{n-k}}f(x)\frac{d^k}{dx^k}g(x) $$ let $$ f(x) = \frac{1}{x}\\ g(x) = \sin(x) =\mathcal{I}\left(\mathrm{e}^{ix}\right) $$ (using @RobertIsrael answer) therefore $$ \frac{d^{n-k}}{dx^{n-k}}\frac{1}{x} = (-1)^{n-k}\frac{(n-k)!}{x^{n-k+1}}\\ \frac{d^k}{dx^k}\mathcal{I}\left(\mathrm{e}^{ix}\right) = \mathcal{I}\left(i^k\mathrm{e}^{ix}\right) $$ thus $$ \frac{d^n}{dx^n}\frac{\sin x}{x} = \mathcal{I}\left[\sum \left(\matrix{n \\ k}\right)(-1)^{n-k}\frac{(n-k)!}{x^{n-k+1}}\left(i^k\mathrm{e}^{ix}\right)\right]\\ = \mathcal{I}\left[\sum_{k=0}^{n} \frac{n!}{k!}(-1)^n(-i)^k\frac{\mathrm{e}^{ix}}{x^{n-k+1}}\right] $$

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    $\begingroup$ Imaginary part, not real part. $\endgroup$ – Robert Israel Dec 4 '14 at 17:38
  • $\begingroup$ @robertisrael Oops! How silly of me. Thank you $\endgroup$ – Chinny84 Dec 4 '14 at 17:40
  • $\begingroup$ $\textit{I}(t)=-\frac{1}{2}i(t^2-1)t^{-1}$ $\endgroup$ – IV_ Mar 16 '19 at 19:00
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Consider $$F(x) = \dfrac{\exp(ix)}{x}$$ Then $$\dfrac{d^n}{dx^n} F(x) = P_n(1/x) \exp(ix)$$ where $$-t^2 P_n'(t) + i P_n(t) = P_{n+1}(t)$$

EDIT:

If we write $P_n(t) = Q_n(t) + i R_n(t)$ where $Q_n$ and $R_n$ have real coefficients, then

$$\dfrac{d^n}{dx^n} \text{sinc}(x) = Q_n(1/x) \sin(x) + R_n(1/x) \cos(x)$$

An exponential generating function comes from the Taylor series

$$F(x + t) = \sum_{n=0}^\infty \dfrac{t^n}{n!} \dfrac{d^n}{dx^n} F(x)$$

That is

$$\dfrac{s \exp(it)}{1+st} = \sum_{n=0}^\infty \dfrac{t^n}{n!} P_n(s) $$

We want the real and imaginary parts, so

$$ \eqalign{\dfrac{s \cos(t)}{1+st} &= \sum_{n=0}^\infty \dfrac{t^n}{n!} Q_n(s) \cr \dfrac{s \sin(t)}{1+st} &= \sum_{n=0}^\infty \dfrac{t^n}{n!} R_n(s) }$$

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  • $\begingroup$ $P_n$ is the Legendre polynomial, right? $\endgroup$ – user_of_math Dec 4 '14 at 16:33
  • $\begingroup$ Legendre? I don't think so. $\endgroup$ – Robert Israel Dec 4 '14 at 18:14
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If you're happy to accept integrals in your general formula, then a really simple expression (via T.A.E.) is \begin{align} \frac{d^{n}}{dx^{n}}\mathrm{sinc}(x) =\frac{d^{n}}{dx^{n}}\int_{-1}^{1}e^{ikx}\, \mathrm{d}k = \int_{-1}^{1}(ik)^{n}e^{ikx}\, \mathrm{d}k \end{align}

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$$\frac{d^n}{dx^n}\frac{\sin(x)}{x}=\frac{(-1)^{n}i}{2}(\Gamma(n+1,ix)-\Gamma(n+1,-ix))x^{-n-1}$$

$$=\frac{(-1)^{n}n!i}{2}\sum_{k=0}^{n}\frac{i^k}{k!}(e^{-ix}-(-1)^ke^{ix})x^{-n+k-1}$$

$$=\frac{(-1)^{n}n!i}{2}\sum_{k=0}^{n}\frac{i^k}{k!}((1-(-1)^k)\cos(x)-(1+(-1)^k)i\sin(x))x^{-n+k-1}$$

$$=(-1)^{n}n!2i\sum_{k=0}^{n}\frac{i^k}{k!}(\text{ frac}\left(\frac{k}{2}\right)\cos(x)-\text{ frac}\left(\frac{k+1}{2}\right)i\sin(x))x^{-n+k-1}$$

$$=(-1)^{n}n!2\sum_{k=0}^{n}\frac{i^{k+1}}{k!}\text{ frac}\left(\frac{k}{2}\right)\frac{\cos(x)}{x^{n-k+1}}+(-1)^{n}n!2\sum_{k=0}^{n}\frac{i^k}{k!}\text{ frac}\left(\frac{k+1}{2}\right)\frac{\sin(x)}{x^{n-k+1}}$$

$$=(-1)^{n}n!\sum_{k=0}^{\frac{n-1}{2}}\frac{(-1)i^{k+1}}{(2k+1)!}\frac{\cos(x)}{x^{n-2k}}+(-1)^{n}n!\sum_{k=0}^{\frac{n}{2}}\frac{(-1)^k}{(2k)!}\frac{\sin(x)}{x^{n-2k+1}}$$

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