Prove the laplace transform of $\sinh(at)$?

Question

My problem today is as above. Here is what I have done:

Use integral definition of laplace transform to get

$$\int_0^\infty \sinh(at)\exp(-st)dt$$

$$= \lim_{b \to \infty}\int_0^b \sinh(at)\exp(-st)dt$$

By the definition $\sinh(at) = \frac{1}{2}(\exp(at)-\exp(-at))$

We can write

$$= \lim_{b \to \infty}\frac{1}{2}\int_0^b (\exp(at)-\exp(-at))\exp(-st)dt$$

$$= \lim_{b \to \infty}\frac{1}{2}\int_0^b \exp((a-s)t) - \exp((-a-s)t)dt$$

$$\lim_{b \to \infty}\frac{1}{2}\left[\left(\frac{1}{a-s}\exp((a-s)b) - \frac{1}{-a-s}\exp((-a-s)b)\right) - \left(\frac{1}{a-s} - \frac{1}{-a-s}\right)\right]$$

But taking the limit as $b \to \infty$ yields infinity terms. What have I done wrong here?

Answer 1

Hint. If $s>|a|$ then $\exp((a-s)b)$ tends to $0$, not $\infty$, if $b\to\infty$.