I'm reading up on Fourier analysis and in my book it is stated, that, with the help of $ si(ax) = \frac{sin(ax)}{ax} $ you will have:
$$ a_n = 2A \cdot si(n\frac{\Pi}{2}) $$
$$ a_n = \frac{4A}{n\Pi} \cdot \sin(n\frac{\Pi}{2}) = 2A \cdot si(n\frac{\Pi}{2})$$
for $ n=1, 2, 3,..$. I don't see where the $ 4 $ on the left side goes to? Another understanding question: Would the $x$ be $n$ here? As you have $(ax)$ would this be $a = \Pi/2$ and $x=n$ ?
You can set $n=x$ and $a=\frac{\Pi}{2}$ if it is constant, for simplicity (and as $n=1,2,3,...)$. Then using $\text{si}(ax) = \frac{\sin(ax)}{ax} $ we have
$$\text{si}\left(n\frac{\Pi}{2}\right)=\frac{\sin\left(n\frac{\Pi}{2}\right)}{n\frac{\Pi}{2}}=\frac{2\sin\left(n\frac{\Pi}{2}\right)}{n\Pi}$$
so we have
$$ a_n = 2A \cdot \text{si}\left(n\frac{\Pi}{2}\right)=2A\cdot\left(\frac{2\sin\left(n\frac{\Pi}{2}\right)}{n\Pi}\right)=\frac{4A\sin\left(n\frac{\Pi}{2}\right)}{n\Pi}$$
Also check the sine integral.
You just have to write the equations one by one and pay attention. Read these carefully: $$ \text{Si}(x) = \frac{\sin x}{x} $$ therefore, $$ \tag{1} \text{Si}\left(n \frac{\pi}{2} \right) = \frac{\sin \left(n \frac{\pi}{2} \right)}{n \frac{\pi}{2}} = \frac{\sin \left(n \frac{\pi}{2} \right)}{n \pi \cdot \frac{1}{2}} = 2 \frac{ \sin( n\frac{\pi}{2} ) }{n \pi} $$ So what happens if you multiply Equation (1) by $2A$ ?
