I am trying to use Fourier's method to solve a problem.
$u(x,t) = \sum \limits_{n=1}^\infty B_ne^{-(n\pi C / L)^2 t}\sin\left(\frac{n\pi x}{L}\right), B_n=\frac2L\int_0^L \sin\left(\frac{n\pi x}{L}\right)f(x) \,dx$
Of the $1-D$ heat equation $t\gt0$ on $0\leq x\leq L$ with boundary $u(0,t)=0=u(L,t)$, $t\gt 0$ and initial condition $u(x,0)=f(x)$, $0\leq x \leq L$
Now first I want to calculate $B_n=\frac2L\int_0^L \sin\left(\frac{n\pi x}{L}\right)f(x) \,dx$, and so I look for $f(x)$, now $f(x)$ appears to be equal to $\sum \limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{L}\right)$, which means I have:
$B_n = \frac2L\int_0^L \sin\left(\frac{n\pi x}{L}\right)\left[\sum \limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{L}\right)\right] \,dx$
Is this correct, shall I work no the above expression, or is there something I should first note.
Don't look for it in your solution: look in the statement of the problem. Is it given there (like $f(x)=3\sin^2 x$ or something)? If there is no concrete initial condition, there cannot be an concrete solution: the formula that you have for $u$ at the beginning of the post is the answer.
Putting the series with $B_n$ into the formula for evaluating $B_n$ is akin to plugging a UPS into itself. All you can get from this is $B_n=B_n$.