Find the generating function (in closed form) of (2, 0, $\frac{-2}{3!}$, 0, $\frac{2}{5!}$, $\dots$)

Question

I am looking for the closed form for the sequence, (2, 0, $-\frac{2}{3!}$, 0, $\frac{2}{5!}$, $\dots$).

For now I have $$\frac{e^{-x} \left(e^{2 x}-1\right)}{x} \longleftrightarrow (2, 0, \frac{2}{3!}, 0, \frac{2}{5!}, \dots).$$

What can I do to make the 2nd, 6th, 10th... elements be negative?

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Mathematica gives me

$$-\frac{i \left(-1+e^{i x}\right)}{x}-\frac{i e^{-i x} \left(-1+e^{i x}\right)}{x}$$

How can I possibly get this by hand? Or is there a solution that does not involve complex number?

Answer 1

Hint: $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\ .$$