Why is the following integral written with Fourier series development like this?
$\displaystyle h_i\sum_{k,l=1}^{\infty}a_k a_l \int_0^1 \sin(k\pi t)\sin(l\pi t) \,dt = \frac{h_i}{2}\sum_{k,l=1}^{\infty} a_k a_l \delta_{kl}$
What is used to move from left limb to right limb?
Thanks! :)
It is simply because, when $k\neq l$, we have $$ \int_0^1 \sin (\pi k t) \sin (\pi l t) \, dt = \frac{l \sin (\pi k) \cos (\pi l)-k \cos (\pi k) \sin (\pi l)}{\pi k^2-\pi l^2} = 0 $$ And when $k=l$, we have $$ \int_0^1 \sin (\pi k t) \sin (\pi k t) \, dt = \frac{1}{2}-\frac{\sin (2 \pi k)}{4 \pi k} = \dfrac{1}{2} $$ Hence $$ \int_0^1{\sin \left( k\pi t \right)}\sin \left( l\pi t \right) \,dt = \dfrac{1}{2} \delta_{kl} $$ Thus your relation holds.