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I have an exercise where there is the following given:

  • $f$ is a causal function.

  • $f$ is Laplace transformable:$\int_{0}^{\infty} f(t)e^{-zt} \, dt = L(z) $ with $Real(z)> -1$

I have to determine if the integral : $\int_{-\infty}^{\infty}f(t)$ is convergent or not. First thing that popped in my head was that if $\int_{-\infty}^{\infty}f(t)\leq \int_{0}^{\infty} f(t)e^{-zt} \, dt$ then of course it integral is convergent.

The problem is that I'm not sure if this is always true for all z where $Real(z)>-1$. If it's not true I have no idea on how to determine if the integral is convergent or not. Anyone who knows if my idea was good and how to prove it? If it's nog the right way to handle this problem how to do it then? Thanks in advance !

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1 Answer 1

up vote 2 down vote accepted

If you know that $\int_0^\infty f(t)\,e^{-zt}\,dt$ converges for $\operatorname{Re}(z) > -1$, you can just take $z=0$ to see that $\int_0^\infty f(t)\,dt$ converges, and since $f$ is causal, this is the same as $\int_{-\infty}^\infty f(t)\,dt$.

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Thanks for your answer ! –  tim_a Jan 26 '13 at 17:41

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