# Laplace transform of a product of two functions

I have read questions and answers about this topic and i am still confused, using this formula we can calculate the Laplace transform of a product of two functions:

$$L[a_{(t)} b_{(t)}]={{1}\over{2 i \pi}} \int_{\sigma -i \infty}^{\sigma +i \infty}A_{(z)}B_{(s-z)}dz$$

But when i test this formula on an example i get wrong result. My example is: a(t)=t , b(t)=e^(-t)

So the correct answer should be

$$L[te^{-t}]={{1}\over{(s+1)^2}}$$

But when i substitute:

$$L[t]={{1}\over{s^2}}$$

$$L[e^{-t}]={{1}\over{s+1}}$$

into the above formula, i get:

$$L[te^{-t}]={{1}\over{2 \pi i}}\int_{\sigma-i \infty}^{\sigma + i \infty} {{1}\over{z^2}}{{1}\over{s-z+1}}dz=\sum res$$

Residue at z=0 is 1/(s+1)^2 and residue at z=s+1 is also 1/(s+1)^2 so the result i get using this formula is twice the correct result.

I think that the problem is that you're using both residues. But the proper definition of the inverse Laplace transform involves a "dummy" constant $\sigma>0$. This excludes the residue in z=0 and, hence, you don't have the undesired 2 factor.
The integral you've given calculates the inverse Laplace transform. To go from the time-domain to the Laplace-domain (i.e. perform the Laplace Transform), you shouldn't need any contour integration. The (unilateral) Laplace transform is given by $$F(s) = \int_{0}^{\infty} {f(t)e^{-st}dt}$$
Unless you specifically want to practice integration, these are usually given in tables for common functions (like $f(t)=t$ and $f(t) = e^{-t}$). To look them up, google "table of laplace transforms".
EDIT: I might've misread your question. If you're indeed trying the inverse Laplace transform, then the integral you're doing is known as a Bromwhich integral. See this discussion: Inverse Laplace transform of fraction $F(s) = \large\frac{2s+1}{s^2+9}$
• Sorry, my question was not clear enough. If I have two functions a(t) and b(t), and I know only their corresponding Laplace transforms A(s) and B(s), but not functions themselves, how can i calculate the Laplace transform of a(t)*b(t) ? One way is to find a(t) and b(t) using inverse Laplace transform and then find the LT of the product of results And the other way is to use the formula: $$L[a_{(t)} b_{(t)}]={{1}\over{2 i \pi}} \int_{\sigma -i \infty}^{\sigma +i \infty}A_{(z)}B_{(s-z)}dz$$ and i am getting wrong result using this formula – Dusan Krantic Jan 10 '15 at 17:00