quick and easy integral question - Page 2

jabar · #11 10-18-2009, 01:01 PM

i think s+a must be minus in this case.

count_volta · #12 10-18-2009, 03:22 PM

Quote:

Originally Posted by LKjell

Something is indeed wrong because the inverse transformation of
$\frac{1}{s+a} => e^{-at}$
Hence

$\int_0^\infty e^{-at}e^{-st}dt$

What you have wrong is the limit. It is from 0 to infinity. If it was Fourier transformation then it is from minus infinity to plus infinity. And you set $\sigma = 0$

This is directly out of an example in the book. They work it out and get the answer. I just posted what they wrote.

I really don't think its wrong. Remember the above is only true if Re{s} < -a

LKjell · #13 10-18-2009, 10:53 PM

$\int e^{at}dt = \frac{1}{a}e^{at}$

$-\int_a^b = \int_b^a$

$\int_0^{-\infty} e^{-(s+a)t}dt = -\frac{1}{s+a}[e^{-(s+a)t}]_0^{-\infty}$

$\lim_{t\to -\infty}e^{-(s+a)t} = \infty, \qquad \sigma, a >0$

$\lim_{t\to 0}e^{-(s+a)t} = 1$

notxjack · #14 10-19-2009, 10:52 PM

Quote:

Originally Posted by LKjell

$\lim_{t\to -\infty}e^{-(s+a)t} = \infty, \qquad \sigma, a >0$

that integral has a domain (values of s+a) of convergence, so we can analyze it using a radius (values of s+a) of convergence, which is SOP for most bilateral laplace transforms.

Eduard Munteanu · #15 11-02-2009, 04:43 PM

The usual (unilateral) Laplace transform goes from 0 to +inf, while the bilateral Laplace transform goes from -inf to +inf.

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