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Apologies in advance for the large number of parameters. I have run into the integral $$ \frac{\zeta + \eta}{\rho^{2}}\int_0^{\infty}\frac{\omega^2q^2}{\left[\,\omega^{2} - \left(\lambda + 2\mu\right)q^{2}/\rho\,\right]^{\, 2} + \left[\,\left(\zeta + \eta\right)\omega q^{2}/\rho\,\right]^{\, 2}}\, \mathrm{d}\omega $$ where $\omega, q, \zeta, \eta, \rho \in \mathbb{R}$ and $\lambda, \mu \in \mathbb{C}$.

With some difficulty, I was able to perform the integral and the result is $$\frac{\pi}{2\rho},$$ which wasn't apparent until the final step in my derivation. Notably for me, it is $q$-independent.

Is there a clever substitution that makes it clear that this is $q$-independent? (and $\zeta,\eta,\lambda,\mu$-independent). If the result of a definite integral is independent of a parameter, is it always possible to find a substitution that renders the integrand independent of that parameter? (I am not sure if this question is sensible).

S. Thornton
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    Interesting question ! But as far as I know it is good practice to ask in one post one question, and create other posts for related questions – Tortar Jul 27 '20 at 18:49
  • Thanks for the tip! I have pared down the questions. – S. Thornton Jul 27 '20 at 19:45
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    What happens when $q=0$? Anyway, you can reduce the number of parameters since $\lambda,\mu$ only appears as $\lambda'=\lambda+2\mu$ and $\zeta,\eta$ only appears as $\zeta'=\zeta+\eta$, so you want to show $$ \frac{\zeta'}{\rho}\int_0^{\infty}\frac{\omega^2q^2}{(\omega^2 - \lambda'q^2/\rho)^2 + (\zeta'\omega q^2/\rho)^2}, \mathrm{d}\omega=2\pi. $$which looks similar to this previous question. – user10354138 Jul 28 '20 at 01:35

1 Answers1

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I have found a satisfactory way to convince myself of this. First, note the integrand only depends on $\omega^2$ so I can integrate over the real line and take half. Define $$a^2\equiv \frac{\left(\lambda+2\mu\right)q^2}{\rho},\;\;b\equiv\frac{\left(\zeta+\eta\right)q^2}{\rho}.$$ The integral becomes $$\frac{b}{2\rho}\int_{-\infty}^{\infty}\frac{\omega^2}{\left(\omega^2-a^2\right)^2+\left(b\omega\right)^2}\mathrm{d}\omega.$$ Substitute $$\omega'=\frac{\omega}{b}$$ to find the integral is equal to $$\frac{1}{2\rho}\int_{-\infty}^{\infty}\frac{\omega'^2}{\left(\omega'^2-\frac{a^2}{b^2}\right)^2+\omega'^2}\mathrm{d}\omega'.$$ I came up with this substitution by guessing it'd be a scaling by some power of $b$ and matching powers so that it factors out of the denominator. To see the integral is independent of $a^2/b^2$ is equivalent to this post.

S. Thornton
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