How to solve this simple Laplace integral?

Oct 2015
141
8
Greece
Check this Image. I tried to post it as an image but it takes the entire space of the post...

It seems I have forgotten a lot about solving integrals... But I really need to remember quickly how to solve this kind of fractional integrals since I haven't memorized the Laplace "trick" formulas (like 1 over Laplace is \(\displaystyle \frac{1}{s}\)) and tomorrow I have a test.
 
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skipjack

Forum Staff
Dec 2006
21,387
2,410
I can't see the image for some reason. Can you just type enough to explain what the integral is?
 
Oct 2015
141
8
Greece
By the way I finished the test today, and it was 90% of Fourier transformation exercises :p while in the past you had like one exercise only... I probably failed the test and the weird thing is that i knew how to solve the integrals...

The problem is that the textbook says that \(\displaystyle \mathcal{L}[1] = \frac{1}{s}\) but if I try to calculate it,

\(\displaystyle
\mathcal{L}[1] = \lim_{T->\infty} \int_{-T}^{T} 1 \cdot e^{-st} dt =
\lim_{T->\infty} [\frac{e^{-st}}{-s}]_{-T}^{T} =
\lim_{T->\infty} [\frac{e^{-sT}}{-s} - \frac{e^{sT}}{-s}] =
\frac{1}{s} \lim_{T->\infty} [-e^{-sT} + e^{sT}]
\)

I believe that \(\displaystyle \lim_{T->\infty} [-e^{-sT}] = 0 \) and \(\displaystyle \lim_{T->\infty} [e^{sT}] = \infty\) by trying to "see it" using the graph below.



But of course this is not true, because the limit of them both must be 1 in order to get \(\displaystyle \frac{1}{s}\)

This is a really great problem that is keeping me from calculating Laplace integrals in general.

I certainly need to review calculus all over again, but I have no time... I will try this summer though.... This limit is extremely easy and yet I don't remember how to solve it.

Offtopic: How did you put that image into a thumbnail? When I try to add a big image in a post usually it takes over the whole screen.
 
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