Doing some exercises i found this function expressed by power series, someone recognize a friend? $$F_{n,m}(x)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{x^{2(k+n+1)+1}}{2(k+n+1)+1}m^{2k+1}$$ It's possible to get out a different rappresentation for $m=0,1,2,3$ at least?
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Note that
$$\frac{d}{dx} F_{n,m}(x) = x^{2 n+2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)!} (m x)^{2 k+1} = x^{2 n+2} \sin{(m x)} $$
We may then integrate to find $F_{n,m}(x)$; when we do, we find that it satisfies a recurrence:
$$ F_{n,m}(x) = -\frac{1}{m} x^{2 n+2} \cos{m x} + \frac{2 n+2}{m^2} x^{2 n+1} \sin{m x} - \frac{(2 n+2) (2 n+1)}{m^2} F_{n-1,m}(x)$$
$$F_{0,m}(x) = \frac{\left(2-m^2 x^2\right) \cos (m x)+2 m x \sin (m x)-2}{m^3}$$
Ron Gordon
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