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These are replies submitted by BasSPe

@Axel Vogt Normally you would take the integral from -inf to inf yes. By using notions of symmtery you can already half the integral and use 0 to inf and replace (1/2pi) by (1/pi). Next, when finding the dynamic response of structures you are only interested in the range of frequencies in which the first few eigenfrequencies are located. These first few have a way much larger contribution to the dynamic response and so by knowing the range of the first few eigenfrequencies you can cut of the integral.

The pole at omega=10 is because the excitation has the exact same frequency.  

@mmcdara Thanks for your reply!

I eventually found my solutions as follows

# write the input for the fourier transform

uL_tilde := Re(eval(u_tilde, x = L)*exp(omega*t*I));

# numerically solve the integral
lower_limit := 0.001;
upper_limit := 40;
n_steps := 400;
step_omega := (upper_limit - lower_limit)/n_steps;
sol := step_omega*(subs(omega = lower_limit, uL_tilde) + subs(omega = upper_limit, uL_tilde) + sum(subs(omega = i*step_omega + lower_limit, uL_tilde), i = 1 .. n_steps))/Pi;

So I did more or less the same as you suggested and wrote the fourier the fourier transform myself and kind of numerically solved the integral for uL(x,t). The runtime with these step sizes was very do-able :)

Thanks again for the advise 

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