Related but not necessary to know: here
Looking at the temperature distribution in an infinitely long cylinder of metal with insulated sides and initial temperature distribution $f(x)= \left\{\begin{align}0,\quad|x|\lt L \\ C,\quad|x| \gt L \end{align} \right.$
$C$ is constant.
Now I want to workout $B_n$ for the fourier series, and I thought that I would want:
$$B_n = \frac2L\int_L^\infty C\sin\left(\frac{n\pi x}{L}\right) dx$$
But perhaps I haven't setup the integral correctly. Thank you for listening.
I based my choice off of the general form of the solution to the heat equation: $u(x,t)=\sum \limits_{n=1}^\infty B_n e^{({-n\pi C/L})^2} \sin\left(\frac{n\pi x}{L}\right)$ With $B_n = \frac2L \int_0^L \sin\left(\frac{n\pi x}{L}\right) f(x) dx$
It would seem you have a number of concepts confused, and I would personally suggest you refer to a textbook on this matter.
You want to use $$\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$$
which helps you obtain your standard $F''+p^2 F =0$ and $\dot{G} + c^2p^2 G=0$, if you can't get this you are in trouble.
$f(x)= \left\{\begin{align}0,\quad|x|\lt L \\ C,\quad|x| \gt L \end{align} \right.$
$C, L\lt x$ and $x\lt -L$
$u(x,t) = \frac{C}{2c\sqrt{\pi t}}\int_L^\infty$
(CURRENTLY EDITTING)