Just need to know if my work for this is correct.
To start off, I sum over the Laplace Transform of $\sin(ax)$:$$\int_0^\infty\sin(ax)e^{-nx}dx=\frac{a}{n^2+a^2}$$
$$\sum_{n=1}^\infty\int_0^\infty\sin(ax)e^{-nx}dx=a\sum_{n=1}^\infty\frac{1}{n^2+a^2}=\frac{\pi\coth(\pi a)}{2}-\frac{1}{2a}$$
This is the series for the hyperbolic cotangent, which can be derived through Euler's Product for the sine function. Summing over the LHS I get:
$$\int_0^\infty\frac{\sin(ax)}{e^x-1}dx=\frac{\pi\coth(\pi a)}{2}-\frac{1}{2a}$$ I'm not sure if my interchange of sum and integral is justified, but continuing on I expand the sine into its Taylor Series: $$\int_0^\infty\frac{\sin(ax)}{e^x-1}dx=\int_0^\infty\sum_{n=0}^\infty \frac{(-1)^n (ax)^{2n+1}}{(e^x-1)(2n+1)!}dx=\sum_{n=0}^\infty\frac{(-1)^n a^{2n+1} }{(2n+1)!}\int_0^\infty\frac{x^{2n+1}}{e^x-1}dx=\sum_{n=1}^\infty(-1)^{n+1}\zeta(2n)a^{2n-1}=\frac{\pi\coth(\pi a)}{2}-\frac{1}{2a}$$ for $|a|<1$, and where I have employed the integral representation of the Zeta Function $$\zeta(n+1)n!=\int_{0}^\infty \frac{x^n}{e^x-1}dx$$
However, the Taylor Series expansion of this function is $$\frac{\pi\coth(\pi a)}{2}-\frac{1}{2a}=\sum_{n=1}^\infty\frac{2^{2n-1}B_{2n}\pi^{2n}}{(2n)!}a^{2n-1}$$ The coefficients in both polynomial expansions must be the same. This must mean that $$(-1)^{n+1}\zeta(2n)=\frac{2^{2n-1}B_{2n}\pi^{2n}}{(2n)!}$$ $$\zeta(2n)=(-1)^{n+1}\frac{2^{2n-1}B_{2n}\pi^{2n}}{(2n)!}$$ I'm not quite sure if I've made any mistakes so any help is appreciated.