Maths: the Laplace Transform of the convolution of two functions

The Laplace Transform of z(t) = x(t) * y(t) is Z(s) = X(s) . Y(s).
In the demonstration of this, I fail to see why we are allowed to
separate a double integral in two independent integrals.
z(t) = x(t) *
y(t) = Integral from -inf to +inf [x(u) . y(t-u) du]
Now let us calculate the two-sided(? - the one that goes from -inf to
+inf) Laplace Transform. It is called «bilatère» in French. What is
the correct English adjective?
Z(s) = Integ from -inf to +inf{Integ from -inf to +inf
[x(u).y(t-u).du] . exp(-s.t).dt}
I move the closing bracket to the end
= Integ from -inf to +inf{Integ from -inf to +inf [x(u).y(t-u).du
. exp(-s.t).dt]}
I add two exponentials which have 1 as a product
= Integ from -inf to +inf{Integ from -inf to +inf
[x(u).exp(-u.t).y(t-u).du.exp(-s.t).exp(u.t) dt]}
Now, I simplify a bit and I move «du» to the left
= Integ from -inf to +inf{Integ from -inf to +inf
[x(u).exp(-u.t).y(t-u).du . exp[-s(t-u)].dt]}
= Integ from -inf to +inf{Integ from -inf to +inf
[x(u).exp(-u.t).du . y(t-u).exp[-s(t-u)].dt]}
On the left side, we see an integral with respect to u variable.
On the right side, if we are allowed to make dt = d(t-u), we can
substitute v for t-u, and we can separate the integrals.
Why are we allowed to make dt = d(t-u)? This assumes that u is a
constant while u sweeps from -inf to +inf!
This message does not deal with an engineering problem but with a
maths problem. Could you suggest a more suitable newsgroup?
Writing this message without the usual mathetical notation was
difficult. Reading it is difficult. Is there a software tool that
permits a better rendering of mathematical expressions with only the
ASCII character set?
Thank you very much for your enlightenment.
Reply to
Jean Castonguay
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None that I know of, but I have seen some useful conventions. For example, "Integ from -inf to +inf{Integ from -inf to +inf [x(u).exp(-u.t).y(t-u).du]" becomes
+inf Integral[x(u).exp(-u.t).y(t-u).du], which is more like what one is -inf
accustomed to. Long fractions can be expressed the same way:
a_0.x^2 + a_1.x +a_2 -------------------- b_0.x^2 + b_1.x + b_2
(Underscore indicates subscript.)
Reply to
Jerry Avins
One, because u is independent of t; thus d(t-u)/dt = dt/dt - du/dt. Since u is independent of t, du/dt = 0. The other reason that this works where it wouldn't otherwise is because in sweeping t from -inf to +inf we sweep out any particular value that u may take at any moment.
There are math newsgroups -- the two newsreaders that I have used will both search on a term in the newsgroup title -- perhaps you can search for it?
Either use the notation that Jerry suggested, or use LaTeX notation (which is quite popular on the math newsgroups).
Reply to
Tim Wescott
A couple of (perhaps relevant) references might be Fubini's theorem, and Tonelli's theorem. More generally the "analysis" branch of mathematics should be able to justify such operations rigorously (things like interchanging the order of integration, exchanging indefinite integrals and infinite series, convergence of infinite series etc. etc.)
You will need fairly good math. skills My old textbook (probably out of date and out of print now!) was: "Real Analysis" by H. L. Royden
Reply to
a + b ------ c + d created by Andy´s ASCII-Circuit v1.22.310103 Beta
formatting link

much simpler. Each string in the text box gets placed with the mouse ans the auto line does the bar.
|\| -|-\ | >- -|+/ |/| created by Andy´s ASCII-Circuit v1.22.310103 Beta
formatting link
Reply to
I've found that alt.math.undergrad harbors the best help.
Reply to
That looks good to me! I'll have a go at it.
Reply to
Jerry Avins

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