The question asks:
Find the finite Fourier sine transform of: $F(x) = x^2$ where $0<x<l$.
My tutors solution is simply:
$f_s(n) = \int_{0}^{l}x^2 sin(\frac{\pi n x}{l})dx = \frac{2\pi l^3}{n^3 \pi^3}(cos(n\pi-1)-\frac{l^3}{n\pi}cos(n\pi)$.
Is there some trick that I'm missing? How the hell did he go from that integral to that so quickly – there's so many rules with Fourier transforms, so thats why I ask. So far, I think he just skipped a bunch of steps, which is annoying given I get stuck:
My workings:
We have:
$f_s(n) = \int_{0}^{l}x^2 sin(\frac{\pi n x}{l})dx$
let $u = x^2, v = sin(\frac{\pi n x}{l})$ and use integration by parts:
$x^2sin(\frac{\pi n x}{l}) – \int v dv$, where $\frac{dv}{du} = cos(\frac{\pi n – \sqrt{u}}{l}) + \frac{\sqrt{u}}{2}$, and then I get stuck. Any help would be appreciated.
Best Answer
For this question, we use Integration by parts twice, \begin{matrix}x^2&\sin\dfrac{\pi nx}{l}\\2x&-\dfrac l{\pi n}\cos\dfrac{\pi nx}{l}\\2&-\dfrac{l^2}{\pi^2n^2}\sin\dfrac{\pi nx}{l}\end{matrix} Therefore the integral is then $$-\dfrac{x^2l}{\pi n}\cos\dfrac{\pi nx}{l}+\dfrac{2xl^2}{\pi^2n^2}\sin\dfrac{\pi nx}{l}\bigg|_0^l-\dfrac{2l^2}{\pi^2n^2}\int_0^l \sin\dfrac{\pi nx}{l}dx$$ The rest should be easy for you.