Laplace transform of a product of two functions

Question

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.

Where did i go wrong? Is there an article i can read about this formula?

Answer 1

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 $F(s) = \large\frac{2s+1}{s^2+9}$

Answer 2

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.