I want to compute the following integral:
Integrate[-(λ/π) (1 - (wn - w)^2 / Ω0^2 Log[1 + Ω0^2/(wn - w)^2] +
Ω0^2/(2 (wn + w)^2)) ϕ/(w^2 + ρ^2),
{w, Ω, Infinity},
Assumptions -> {λ > 0, wn > 0, Ω0 > 0, ϕ > 0, ρ > 0, Ω > wn}]
$$ - \int \limits_\Omega^\infty \mathrm{d} \omega \frac{\lambda}{\pi} \left[ 1 - \frac{(\omega_n - \omega)^2}{\Omega_0^2} \log \left( 1 + \frac{\Omega_0^2}{(\omega_n - \omega)^2} \right) + \frac{\Omega_0^2}{2 (\omega_n + \omega)^2} \right] \frac{\phi}{\omega^2 + \rho^2} $$
The answer, after several minutes, is ComplexInfinity
. Why is that so? The first function in the bracket (1-...log(...)) is very similar to a Lorentzian with width Omega0
, it's asymptotics is 1/w^2
. The second function is just asymptotic version of the first, while wn, Omega >> Omega0
taking into an account. Last part is just multiplicative Lorentzian factor with a width rho, which is positive (in assumptions). So this function is 1) bounded 2) asymptotic o(1/w^4). Why is the answer ComplexInfinity
?
Bonus: is this integral doable analytically? If yes, how do I force Mathematica to show the result? If not, what should I do to get answer when I'm interested in this setup: 0 < wn < Omega
, Omega >> Omega0
, that log part is actually quite important, as the integral goes through full shape of the function (remember, 0 < wn < Omega, so integral catches all details of that function)
Bonus2: someone edited one of my previous questions so that symbols like π, etc appeared as they should: π
. How did he do that? :O
NIntegrate
instead. $\endgroup$