How to simplify the following $$\int_0^b\sum_{i=1}^N Y_i(x)Y_j(x)A(t)\delta(x-vt)dx=?$$ where $$ \{Y_i(x)\}_{i=1}^\infty=\Bigg\{\sqrt{\dfrac{2}{b}}\sin(c_ix)\Bigg\}_{i=1}^\infty$$ and $c_i=i\pi/b.$
The attemps: I know that
$\int_0^bY_i(x)Y_i(x)dx=\sqrt{\dfrac{2}{b}}\sqrt{\dfrac{2}{b}}\int_0^b\sin(i\pi x/b)\sin(i\pi x/b)dx=\dfrac{2}{b}\dfrac{b}{2}=1$
if $0<vt<b$,
$\int_{0}^{b}A(t)\delta(x-vt)dx=A(vt)$
else it is zero.
But how can I use them in this question? Also, is there any possibility for getting rid of summations like $\sum_{i=1}^N$ bla bla...?
Thank you.
The Dirac Delta function "picks out" the value of the integrand where the argument of the delta function is zero. So I would say that \begin{align*}\int_0^b\sum_{i=1}^N Y_i(x)\,Y_j(x)\,A(t)\,\delta(x-vt)dx&=\begin{cases} \displaystyle A(t)\,Y_j(vt)\sum_{i=1}^N Y_i(vt), \quad &0<vt<b \\ 0,\quad \text{otherwise} \end{cases}\\ &=\begin{cases} \displaystyle A(t)\,\sqrt{\dfrac{2}{b}}\sin(c_jvt)\sum_{i=1}^N \sqrt{\dfrac{2}{b}}\sin(c_ivt), \quad &0<vt<b \\ 0,\quad \text{otherwise} \end{cases}\\ &=\begin{cases} \displaystyle \dfrac{2A(t)}{b}\,\sin(c_jvt)\sum_{i=1}^N \sin(c_ivt), \quad &0<vt<b \\ 0,\quad \text{otherwise} \end{cases}\\ &=\begin{cases} \displaystyle \dfrac{2A(t)}{b}\,\sin\left(\frac{j\pi vt}{b}\right)\sum_{i=1}^N \sin\left(\frac{i\pi vt}{b}\right), \quad &0<vt<b \\ 0,\quad \text{otherwise} \end{cases} \\ &=\begin{cases} \displaystyle \dfrac{A(t)}{b}\,\sin\left(\frac{j\pi vt}{b}\right)\csc\left(\frac{\pi v t}{2b}\right)\left[\cos\left(\frac{\pi v t}{2b}\right)-\cos\left(\frac{(1+2N)\pi v t}{2b}\right)\right], \quad &0<vt<b \\ 0,\quad \text{otherwise} \end{cases}. \end{align*}