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?
$\endgroup$ 12 Answers
$\begingroup$Hint. If $s>|a|$ then $\exp((a-s)b)$ tends to $0$, not $\infty$, if $b\to\infty$.
$\endgroup$ $\begingroup$I'm not quite sure if I did it right but I think I did. Just correct me if I'm wrong.
