Thank you brombo, I'll take a closer look at the file you send me!
Op maandag 6 februari 2023 om 22:29:25 UTC+1 schreef brombo:
> I cleaned things up here is what the notebook looks like (see attached
> html) -
>
>
> 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:
> [image: Schermafbeelding 2023-02-06 163521.jpg]
>
> Kind regards,
> Tom van Woudenberg
> Delft University of Technology
>
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