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Re: Integration of exp(-az)*sigma(t-to)*dz ??
Posted:
Dec 6, 1996 1:03 PM
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In article <57jp6o$d42@brachio.zrz.TU-Berlin.DE>, esfahani@mikro.ee.tu-berlin.de (Farzad Esfahani) writes:
|>
|>
|> Hello you all,
|> Altough this may not be the right newsgroup I'll send to this group too
|> perhaps someone can help me.
|> I have to solve the following integration, who can help me or give me
|> some hints?
|>
|> I(t) =
|> A * Integral(from z=zo until oo) exp(-a*z) * sigma(t - to) dz
|>
|> Where:
|> zo is the lower and oo (infinite) is the upper integral limit
|> A, zo and a are constants
|> to = (z - zo)^2 / D, with D as a constant
|> sigma is a step function and is defined: sigma = 0 for t<to
|> sigma = 1 for t>=to
|>
|>
|> My actual problem is to find a way to integrate the step function sigma.
|> Is there any possiblities to approximate it to a mathematical function.
|> Or is there any rule for its integration that I am not aware of it.
|> I have already looked in some Math books but I didn't find it.
|>
|>
|> PS.: Please send me your responses also via e-mail, since I don't read this
|> newsgroup everyday.
|>
|>
|> Thanks in advance
|>
|> Farzad
|>
If I understand your notation correctly, your integral reduces to
I(t) = A * (exp(-a*(z0+sqrt(tD)))-exp(-az0)) for t>=0 and I(t)=0 otherwise
since t-t0<0 if (z-z0)^2/D>t, i.e. z>z0+sqrt(tD) and sigma(u)=0 for u<0 and 1
otherwise. right?
peter
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