trigonometry - How does a complex exponential turn into the sinc function?

Suppose I have a complex variable $j$ such that we have

$f(u) = \frac{1}{ju}[e^{\frac{ju}{2}} - e^{\frac{-ju}{2}}]$.

Could somebody please explain how this turns into a sinc function ?

I know I have to apply Euler's Formula $e^{jx} = \cos(x) + j\sin(x)$.

So applying the formula to $f(u) = \frac{1}{ju} [ (\cos(\frac{u}{2}) + j\sin(\frac{u}{2})) - (\cos(-\frac{u}{2}) + j\sin(-\frac{u}{2}))$

$= \frac{1}{ju} [ (\cos(\frac{u}{2}) + j\sin(\frac{u}{2})) - \cos(\frac{u}{2}) + j\sin(-\frac{u}{2}))$

$= \frac{1}{ju} [ j\sin(\frac{u}{2})) + j\sin(\frac{u}{2}))$

$= \frac{1}{ju} [ 2j\sin(\frac{u}{2}))$

$= \frac{1}{u} [ 2\sin(\frac{u}{2}))$

But how does this become the sinc function ?

Answer

Your last expression is equal to $\displaystyle \frac{\sin u/2}{u/2}$ which is equal to $sinc(u/2)$. Is this what you were looking for?