Also note if you use the residue theorem there is a different path
depending on whether t>0 or t<0. This means when you calculate the
residue you only use one pole (phi+ or phi-) depending on the path. The
path chosen makes sure that the part of it not on the real axis does not
contribute to the integral.
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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