You can use residues but I made a mistake. You have to do a partial
fraction decomposition of the reciprocal quadratic factor. Then both
poles of your integrand lie in the lower half plane so that the inverse
transform is zero for t<0. For t>0 you have calculate the residue of
each component of the partial fraction.
On 2/6/23 10:36 AM, 'Tom van Woudenberg' via sympy wrote:
Hi there,
When trying to solve a integral as part of a manual inverse fourier
transform, SymPy return the unevaluated integral. Does anybody know if
SymPy is able to solve this integral with some help? It would be good
enough if I'd be able to obtain the result for specific values of t.
import sympy as sp
phi,t = sp.symbols('phi,t',real=True)
sp.I*(1 -
sp.exp(4*sp.I*sp.pi*phi))*sp.exp(-8*sp.I*sp.pi*phi)/(2*sp.pi*phi*(-4*sp.pi**2*phi**2
+ 1.5*sp.I*sp.pi*phi + 4))
solution_numeric = 1 / sp.pi *
sp.integrate(sp.re(solution_in_frequency_domain_numeric*sp.exp(sp.I*2*phi*t)),(phi,0,4))
print(solution_numeric)
returns:
(Integral(sin(4*pi*phi)*re(exp(2*I*phi*t)/(-4*pi**2*phi**2*exp(8*I*pi*phi)
+ 1.5*I*pi*phi*exp(8*I*pi*phi) + 4*exp(8*I*pi*phi))), (phi, 0, 4)) +
Integral(cos(4*pi*phi)*im(exp(2*I*phi*t)/(-4*pi**2*phi**2*exp(8*I*pi*phi)
+ 1.5*I*pi*phi*exp(8*I*pi*phi) + 4*exp(8*I*pi*phi))), (phi, 0, 4)) +
Integral(-im(exp(2*I*phi*t)/(-4*pi**2*phi**2*exp(8*I*pi*phi) +
1.5*I*pi*phi*exp(8*I*pi*phi) + 4*exp(8*I*pi*phi))), (phi, 0,
4)))/(2*pi**2*phi)
Plotting the result for t,0,15 should give this result according to Maple:
Schermafbeelding 2023-02-06 163521.jpg
Kind regards,
Tom van Woudenberg
Delft University of Technology
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