Consider this example,
Two infinite grounded plates lie parallel to the $xz$ plane, one at $y = 0$, the other ar $y = a$ plane. The left end at $x = 0$ is closed off with an infinite strip insulated from two plates and maintained at a specific potential $V_0(y)$. Find potential inside this slot.
To solve for potential I need to solve
$$\dfrac{\partial^2 V }{\partial x^2} +\dfrac{\partial^2 V }{\partial y^2} = 0$$
For boundary conditions,
$$\begin {cases}\text{V = 0 when y = 0}\\\text{ V = 0 when y = a}\\\text{V = $V_0$(y) when x = 0}\\\text{$V \to 0$ as $x \to \infty$}\end{cases}$$
Using separation of variables, I got
$$V(x, y) = Ce^{-kx}\sin (ky)$$
Where $k = \dfrac{n\pi}{a},\ n \in \Bbb Z_{+}$
Since there is no single value of $n$ for which $V$ satisfy condition $4$, we write
$$V(x,y) = \sum_{n = 1}^{\infty}C_n\exp\left({-\dfrac{n\pi}{a}x}\right)\sin \left(\dfrac{n\pi}{a}y\right) \tag 1$$
Then,
$$V_0(y) = V(0,y) = \sum_{n = 1}^{\infty}C_n\sin \left(\dfrac{n\pi}{a}y\right)$$ $$\int_0^a V_0(y)\sin \left(\dfrac{n^\prime\pi}{a}y\right) dy = \sum_{n = 1}^{\infty}C_n \int_0^a\sin \left(\dfrac{n^\prime\pi}{a}y\right)\sin\left(\dfrac{n\pi}{a}y\right) dy = \dfrac a2C_{n^\prime} \tag 2$$
Where $n^\prime$ is some +ve integer.
But then the book claims that potential is given by $(1)$ for $$C_n = \dfrac2a\int_0^a V_0(y)\sin \left(\dfrac{n \pi}{a}y\right) dy, n \in \Bbb Z$$
I did not understand this, we solved $C_n$ from one specific $n^\prime$ not for all $n \in \Bbb Z_+$. How are we saying that $2$ is true for all $n$ ?
One more small thing, why did we integrate from $0$ to $a$ not something else ?