How do I find the laplace transform of a product?

Question

How do I find the laplace transform of a product? Specifically $e^{5t}\cos{t}$?

Answer 1

There is not a general formula for the Laplace transform of a product. However, the integral $$\int_0^\infty \cos(t)e^{(5-s)t}\,dt$$ is by definition the Lapalace transform of $e^{5t}\cos t$, and is not hard to solve by parts (note that $s$ must be greater than $5$).

Answer 2

In general, the only neat formula involving products is for the convolution product: ${\cal L}\{f \ast g\} = {\cal L}\{f\} {\cal L}\{g\}.$

However, one can prove that ${\cal L}\{e^{at}f\}(s) = {\cal L}\{f\}(s-a)$, a frequency shifting formula. Since ${\cal L}\{\cos t\}(s) = \frac{s}{s^2+1}$, it follows that $${\cal L }\{e^{5t}\cos t\}(s) = \frac{s-5}{s^2-10s + 26}.$$

Answer 3

There is such a formula, but you need some complex analysis.

If $f(t)$ and $g(t)$ are exponentially bounded, for any real numbers $a$ and $b$ such that all singularities of $F = {\mathcal L}(f)$ and $G = \mathcal L(g)$ are to the left of $\text{Re}(s) = a$ and $\text{Re}(s) = b$ respectively, then for $\text{Re}(s) > a+b$,

$$\mathcal L(f g)(s) = (2 \pi i)^{-1} \int_C F(p) G(s-p) \; dp$$

where $C$ is the straight line $\{a + it:-\infty < t < \infty\}$.

Basically this comes from the fact that the Laplace transform at imaginary $s$ is a Fourier transform, and the Fourier transform of a product is a convolution.

Of course it's much easier to do the case $f(t) = e^{5t}$, $g(t) = \cos(t)$ directly, but here it is anyway in that case: $F(s) = \dfrac{1}{s-5}$, $a > 5$, $G(s) = \dfrac{s}{s^2+1}$, $b > 0$, so you integrate $\dfrac{1}{2\pi i}\dfrac{s-p}{((s-p)^2+1)(p-5)}$ on the vertical line $\text{Re}(p) = a$ where $\text{Re}(s) > a+b$. You can close up the contour using a big semicircle and use residues: the only pole inside the contour is at $p=5$ where the residue is $(s-5)/((s-5)^2+1)$ (the other poles are at $p = s \pm i$ which are outside the contour). Thus the answer is $(s-5)/((s-5)^2+1)$.