Show that this integral of a hyperbolic equals a certain equation 
Show that $$ \int_0^\infty e^{-ax}\sinh bx \,\mathrm dx = \frac{b}{a^2 - b^2}\,,\quad|b| < a $$ by expanding $\sinh bx$ in a Taylor series at $x = 0$ and integrating term by term.

The instructions confuse me somewhat. Expanding $\sinh bx$ in a Taylor series at $x = 0$ is the McLaurin series right? If this is so, I get this relationship:
$$\sinh bx =  \sum_{n = 0}^\infty \frac{{(bx)}^{2n+1}}{{(2n+1)}!}$$
Integrating this series term by term (skipping some steps) I find a new series:
$$\sum_{n=0}^\infty b^{n+1}\frac{x^n}{n!} = b^{n+1}e^x$$ 
 A: We have $$\int_{0}^{\infty}e^{-ax}\sinh\left(bx\right)dx=\sum_{n\geq0}\frac{b^{2n+1}}{\left(2n+1\right)!}\int_{0}^{\infty}e^{-ax}x^{2n+1}
 $$ hence, since $0\leq\left|b\right|<a
 $, if we put $$ ax=u
 $$ we have $$=\frac{b}{a^{2}}\sum_{n\geq0}\frac{b^{2n}}{a^{2n}\left(2n+1\right)!}\int_{0}^{\infty}e^{-u}u^{2n+1}du=\frac{b}{a^{2}}\sum_{n\geq0}\frac{b^{2n}}{a^{2n}\left(2n+1\right)!}\Gamma\left(2n+2\right)=$$ $$=\frac{b}{a^{2}}\sum_{n\geq0}\left(\frac{b}{a}\right)^{2n}=\frac{b}{a^{2}-b^{2}}.$$ 
A: Hints:
The key to solving this problem is the Gamma integral (for integer powers),
$$\int_0^\infty x^ne^{-x}dx=n!$$
The Taylor expansion is a sum of powers that you will multiply by $e^{-x}$ (after rescaling of $x$) and integrate, and in the end obtain an ordinary series.


After rescaling and integration, the general term $\dfrac{\alpha^{2k+1}x^{2k+1}}{(2k+1)!}$ (with $\alpha:=\dfrac ba$), yields $\dfrac{\alpha^{2k+1}(2k+1)!}{(2k+1)!}=\alpha(\alpha^2)^k$, hence the sum $\dfrac\alpha{1-\alpha^2}$. 

A: Notice 
Using Laplace transform of $\sinh(bx)$, we get  $$\int_{0}^{\infty}e^{-ax}\sin h(bx)dx$$$$=L[\sin h(bx)]_{s=a}$$
$$=\left[\frac{b}{s^2-b^2}\right]_{s=a}$$
$$=\left[\frac{b}{a^2-b^2}\right]$$$$=\color {red}{\frac{b}{a^2-b^2}}$$ 
$\forall \ \ |b|<a$
