Fourier sine series in solution to 1D Heat Equation I have reduced my solution of a 1D heat equation boundary value problem to the following:
$$W(z, t) = \sum_{n=1}^\infty b_n \sin(\lambda_n z) e^{-\lambda_n^2 \alpha t}$$
To get the coefficients $b_n$, I apply the initial condition that: $W(z, 0) = T_0$, which gives the Fourier Sine Series:
$$\sum_{n=1}^\infty b_n \sin(\lambda_n z) = T_0$$
My question is how to obtain the coefficients $b_n$ for my problem here using the integral formula for the Fourier Sine Series? Namely, if
$$f(x) = \sum_{n = 1}^\infty b_n \sin(nx)$$
Then:
$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)$$
The argument in the sine function for my problem is $\lambda_n$ = (some function of n, and not explicitly equal to $n$ as in the integral formula above. Is there a way that I am suppose to transform the argument such that the formula can be applied? Thanks kindly in advance,
 A: Hint 
1) Multiply both sides of your second equation by ${\sin {\lambda _m}z}$ and integrate from $a$ to $b$.
2) Use this property of your sin functions called orthogonality
$$\eqalign{
  & \int\limits_a^b {\sin {\lambda _n}z\sin {\lambda _m}z{\rm{d}}z}  = {\delta _{mn}}\int\limits_a^b {{{\sin }^2}{\lambda _n}z{\rm{d}}z}   \cr 
  & {\delta _{mn}} = \left\{ {\matrix{
   1 & {m = n}  \cr 
   0 & {m \ne n}  \cr 
 } } \right. \cr} $$
where $a \le z \le b$ is your domain of interest.
A: Your formula isn't precise. 
To get the coefficients of
$$f(x)=\sum b_n X_n(\lambda_n x),\quad b\in[a,b]$$
you have to evaluate
$$b_n = \frac{1}{b-a} \int_{a}^{b} f(x) X_n(\lambda_n x)dx$$
where $X_n$ is an eigenfunction that belongs to the eigenvalue $\lambda_n$, which you have found by solving Storm Leuvil equation (in your example $X_n(x)=\sin(\lambda_n x)$).
You can figure it out by yourself using H.R.'s hint.
Note that 
$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} T_0 \sin(\lambda_n x)dx=0$$
which suggest that you have a mistake, for example in finding eigenfunctions\value .
