0
$\begingroup$

I want to integrate

$$ \int_{-\infty}^{\infty} dw \hspace{0.51em}\frac{w^{n} e^{i wt}}{(w^{2}-\Omega^{2})^2 +(\gamma w)^2} $$

I did so using contour integration because it has the poles:

$$w=\mp \frac{i\gamma}{2} \pm \sqrt{\Omega^{2}-\frac{\gamma^{2}}{4}}$$

resulting in:

$$2^{1-n} i \pi \left(\frac{e^{-\frac{1}{2} \tau \left(\gamma +i \sqrt{4\Omega ^2-\gamma ^2}\right)} \left(-\sqrt{4 \Omega ^2-\gamma^2}+i \gamma \right)^n}{\gamma \left(\sqrt{4 \Omega ^2-\gamma^2}-i \gamma \right)+4 i \Omega ^2}-\frac{e^{\frac{1}{2} i \tau \left(\sqrt{4 \Omega ^2-\gamma ^2}+i \gamma \right)}\left(\sqrt{4 \Omega ^2-\gamma ^2}+i \gamma \right)^n}{\gamma \left(\sqrt{4 \Omega^2-\gamma ^2}+i \gamma \right)-4 i \Omega^2}\right)$$

I decided that I wanted to check with the numerical results because I hadn't done contour integration in a while, for that I did the following, I chose $ n = 3, \gamma = 3, \Omega = 4 $, and plotted for $ \tau $

f[n_, γ_, Ω_, τ_] := I 2^(1 - n) Pi((E^(-(1/2) τ (γ + I Sqrt[
 -γ^2 + 4 Ω^2])) (I γ - Sqrt[-γ^2 
 + 4 Ω^2])^n)/(4 I Ω^2 + γ (-I γ + Sqrt[-γ^2 
 + 4 Ω^2])) - (E^(1/2 I τ (I γ + Sqrt[-γ^2 + 4 Ω^2])) 
(I γ + Sqrt[-γ^2 + 4 Ω^2])^n)/(-4 I Ω^2 + γ 
(I γ + Sqrt[-γ^2 + 4 Ω^2])))
Plot[f[3, 3, 4, τ], {τ, 0, 5}] 

and then tried the same with numerical integration

int[n_, γ_, Ω_, τ_] := γ w^n Exp[I w τ]/((w^2 - 
Ω^2)^2 + (γ w)^2)
fau[τ_] := NIntegrate[Im[int[3, 3, 4, τ]], {w, -Infinity, Infinity}]
Plot[fau[τ], {τ, 0, 5}]
fau[τ_] := NIntegrate[Re[int[3, 3, 4, τ]], {w, -Infinity, Infinity}]
Plot[fau[τ], {τ, 0, 5}]

However the obtained results are quite different, so my main question is is it possible that numerical algorithms suffer with these kind of integrals making them divergent when they're not?

$\endgroup$
12
  • $\begingroup$ For $n = 3$, your integrand scales asymptotically as $e^{iwt}/w$. I wouldn't expect that to be convergent. $\endgroup$ Aug 3, 2021 at 13:50
  • $\begingroup$ Also, is the $\Omega$ in the denominator of your integrand supposed to be squared? $\endgroup$ Aug 3, 2021 at 13:55
  • $\begingroup$ it is supposed to be squared sorry about that $\endgroup$ Aug 3, 2021 at 13:56
  • $\begingroup$ I mean I maybe deeply confused but $$\int_{-\infty}^{\infty} e^{iwt}/w dw $$ converges, it's proportional to the sign function (inverse fourier transform of 1/w ) $\endgroup$ Aug 3, 2021 at 13:58
  • 1
    $\begingroup$ It is conditionally convergent, behaving like the alternating harmonic series (-1)^n/n. I would not b e surprised if NIntegrate has troubles with it. $\endgroup$ Aug 3, 2021 at 14:06

1 Answer 1

1
$\begingroup$

You incorrectly found the poles. See the output of

ToRadicals[Solve[((w^2 - \[CapitalOmega]^2)^2 + (\[Gamma] w)^2) == 0, w, 
Assumptions -> \[CapitalOmega] > 0 && \[Gamma] > 0], 
Assumptions -> \[CapitalOmega] > 0 && \[Gamma] > 0]

Here is its part

$$ w\to \fbox{$i \left( \begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{2} \left(-\gamma -\sqrt{\gamma ^2-4 \Omega ^2}\right) & \gamma >0\land 0<\Omega \leq \frac{\gamma }{2} \\ \frac{1}{2} \left(\sqrt{\gamma ^2-4 \Omega ^2}-\gamma \right) & \gamma >0\land \Omega =\frac{\gamma }{2} \\ \text{Root}\left[\Omega ^4+\text{$\#$1}^4+\left(2 \Omega ^2-\gamma ^2\right) \text{$\#$1}^2\&,1\right] & \text{True} \\ \end{array} \\ \end{array} \right)\text{ if }\gamma >2 \Omega $}$$

The symbolic integration

n = 3; \[CapitalOmega] = 4; \[Gamma] = 3; \[Tau] = 1; 
Integrate[w^n Exp[I *
 w *\[Tau]]/((w^2 - \[CapitalOmega]^2)^2 + (\[Gamma] w)^2), {w,-\Infinity, Infinity}]

(E^(-I Sqrt[ 1/2 (23 - 3 I Sqrt[55])]) (-23 + 3 I Sqrt[55] + (23 + 3 I Sqrt[55]) E^( 1/2 I (Sqrt[46 - 6 I Sqrt[55]] + Sqrt[ 46 + 6 I Sqrt[55]]))) \[Pi])/(6 Sqrt[55])

N[%]

8.88178*10^-16 - 0.980392 I

is in accordance with

NIntegrate[w^n Exp[I * w *\[Tau]]/((w^2 - \[CapitalOmega]^2)^2 + (\[Gamma] w)^2),
{w, \-Infinity, Infinity}, AccuracyGoal -> 4, PrecisionGoal -> 4, 
 Method -> "ExtrapolatingOscillatory"]

0. - 0.980372 I

$\endgroup$
4
  • $\begingroup$ I did have a typo because I wrote $\frac{\gamma}{4}$ instead of $\frac{\gamma}{2}$ but it was right in the code, my bad just edited the question $\endgroup$ Aug 3, 2021 at 16:08
  • $\begingroup$ @GerardoSuares: Your mistakes are more serious than a typo. $\endgroup$
    – user64494
    Aug 3, 2021 at 16:14
  • $\begingroup$ @GerardoSuares: I think four (even three) parameters are too many for Integrate in this case. $\endgroup$
    – user64494
    Aug 3, 2021 at 16:17
  • $\begingroup$ The integral was found by residue integration rather than using Integrate, and actually no, it was just the typo once I added AccuracyGoal -> 4, PrecisionGoal -> 4, Method -> "ExtrapolatingOscillatory" to NIntegrate the answers agree N[f[3, 3, 4, 1]] gives -2.94118 I while NIntegrate gives -2.94112 I (Including the gamma meaning 3*- 0.980372 I) would you mind telling me how you knew you should include those? and thanks I already marked your answer as correct $\endgroup$ Aug 3, 2021 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.