Re: Laplace Transform with Branch Points
From: Maxim (ab_def_at_prontomail.com)
Date: 07/01/04
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Date: 1 Jul 2004 15:01:47 -0700
gtg655n@mail.gatech.edu (James) wrote in message news:<d27f5d81.0406300935.7ae2c115@posting.google.com>...
> Hi, I'm trying to compute the inverse Laplace transform of
> f(p)=1/(p*sqrt(p+1)). I've reasoned that there are branch points at p
> = -1 and infinity so I make a branch cut from -1 to negative infinity
> in the complex plane. I draw the semi circle on the left hand side
> and draw a straight line at Re(p) = alpha from negative infinity on
> the the imaginary axis to infinity on the imaginary axis. The semi
> circle is drawn such that the branch cut is not enclosed. I integrate
> on these boundaries and add them up. I have one pole at p=0, so I add
> the residue at this point.
>
> I end up with an integral of the form:
> 1/(2*pi*I)*int(exp(-u*t)/(u*sqrt(u-1))du, u=-1..inf)
> (where I^2=-1)
>
> I'm not 100% sure that this final integral is correct, but I think it
> is. I have had a lot of trouble with computing this integral. Any
> suggestions/corrections are greatly appreciated.
Let F(p)=e^(p*t)/(p*sqrt(p+1)); we need to find
f(t) = 1/(2*Pi*I)*Integrate(F(p), p = gamma-I*infinity ..
gamma+I*infinity).
Take gamma=1; we can transform the contour of integration to the
contour
p = -infinity-I*eps .. 1-I*eps .. 1+I*eps .. -infinity+I*eps
for a small eps>0, so we traverse the lower and then the upper bank of
the branch cut. The integral from -infinity-I*eps to -1-I*eps can be
approximated with the integral of -F(p) from -infinity to -1 because
of the branch cut of the square root; the integral from -1+I*eps to
-infinity+I*eps is approximated with the integral of F(p) from -1 to
-infinity. The integral along the remaining part of the contour can be
found by taking the residue of F(p) at 0, which equals 1. We obtain
f(t) = 1/(2*Pi*I)*(-2*Integrate(F(p), p = -infinity .. -1) + 2*Pi*I),
evaluating which is just a matter of technical computations. Making
the substitution u^2=-p-1, we get
f0(t) = Integrate(F(p), p = -infinity .. -1) =
2*I*Integrate(e^((-u^2-1)*t)/(u^2+1), u = 0 .. infinity).
Perhaps we can transform it to the error function directly with
another change of variables, but there is another way: differentiating
on parameter t, we get
f0'(t) = -2*I*Integrate(e^((-u^2-1)*t), u = 0 .. infinity) =
-2*I*e^(-t)*sqrt(Pi/t)/2 = -I*Pi*diff(erf(sqrt(t)), t),
where erf(t) = 2/sqrt(Pi)*Integrate(exp(-u^2), u = 0 .. t) and diff is
the derivative. Therefore,
f0(t) = -I*Pi*erf(sqrt(t)) + f0(0),
f0(0) = 2*I*Integrate(1/(u^2+1), u = 0 .. infinity) = I*Pi.
Putting it all together, we obtain
f(t) = 1/(2*Pi*I)*(-2*(-I*Pi*erf(sqrt(t)) + I*Pi) + 2*Pi*I) =
erf(sqrt(t)).
Maxim Rytin
m.r@inbox.ru
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