Sign problem, Laplace transform of sin(at)

Question

I have a problem in my integration by parts but I can't find it:

$$L(\sin(\alpha t)) = \int_0^{\infty}\sin(\alpha t)e^{-st}dt$$

$$= -\frac{1}{\alpha}\left[\cos(\alpha t)e^{-st}\right]_0^{\infty}-\frac{s}{\alpha}\int_0^{\infty}\cos(\alpha t)e^{-st}dt$$

$$= -\frac{1}{\alpha}\left[\cos(\alpha t)e^{-st}\right]_0^{\infty}-\frac{s}{\alpha^2}\left(\left[\sin(\alpha t)e^{-st}\right]_0^{\infty}+s\int_0^{\infty}\sin(\alpha t)e^{-st}dt\right)$$

$$\Leftrightarrow L(\sin(\alpha t)) = \frac{\alpha}{\alpha^2+s^2}\left[e^{-st}\left(-\cos(\alpha t) -\frac{s}{\alpha}\sin(\alpha t)\right)\right]_0^{\infty}$$

$$ $$ $$ = -\frac{\alpha}{\alpha^2+s^2}$$

but it should be :

$$L(sin(\alpha t)) = \frac{\alpha}{\alpha^2+s^2}$$

Where is my mistake ?

Answer 1

When you calculate the integral, you have to remember that it is the upper limit minus the lower limit. It is not plus.

Answer 2

Your signs are perfectly fine until the point where you evaluate at $\infty$ and subtract the evaluation at $0$. The first gives nothing, the second gives the negative of what you've written