# Answer ComplexInfinity for a clearly convergent integral

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

• Do you have good reason to believe that this integral can be done analytically? If not, then set values for your parameters and try NIntegrate instead. – march Mar 17 '17 at 20:25
• I need it analytically. – user16320 Mar 17 '17 at 20:29

Mathematica knows how to do the indefinite integral analytically.

indefInt = 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)]


Then you can take the result, and insert the limits following the fundamental theorem.

But, Mathematica fails to produce a finite upper limit with Limit[indefInt, w->∞]. Therefore, use

ser = Normal[Series[indefInt, {w, ∞, 0}]]


Unfortunately, the leading term still contains Log[1/w], and (Log[1/w])^2. In this case, you need take the limit more carefully.

Apply the transformation Log[1/w] -> -L, and construct the series for large L:

ser2 = Series[ser/.Log[1/w] -> -L, {L, ∞, 0}]


Unfortunately, the output is of the form $(\text{stuff})L + (\text{const}) + \mathcal{O}(L^{-1})$. Fortunately, however, the $(\text{stuff})$ vanishes under your conditions, so the limiting value is given by $(\text{const})$, which you can extract by copy and pasting from output of Series:

upperlimit = 1/(2 π) λ ϕ * (-((π (2 wn^4 + 2 ρ^4 - ρ^2 Ω0^2 + wn^2 (4 ρ^2 + Ω0^2)))/(2 ρ (wn^2 + ρ^2)^2)) + 1/Ω0^2 (-((I (wn - I ρ)^2 (π^2 + 3 Log[wn - I ρ]^2 + 12 Log[wn - I ρ] Log[-(1/(-wn + I ρ))] + 12 Log[-(1/(-wn + I ρ))]^2))/(3 ρ)) + (I (wn + I ρ)^2 (π^2 + 3 Log[wn + I ρ]^2))/(3 ρ) + (I (wn - I ρ)^2 (π^2 + 3 Log[wn - I (ρ - Ω0)]^2 + 12 Log[wn - I (ρ - Ω0)] Log[-(1/(-wn + I (ρ - Ω0)))] + 12 Log[-(1/(-wn + I (ρ - Ω0)))]^2))/(6 ρ) - (I (wn + I ρ)^2 (π^2 + 3 Log[wn + I (ρ - Ω0)]^2))/(6 ρ) + (I (wn - I ρ)^2 (π^2 + 3 Log[wn - I (ρ + Ω0)]^2 + 12 Log[wn - I (ρ + Ω0)] Log[-(1/(-wn + I (ρ + Ω0)))] + 12 Log[-(1/(-wn + I (ρ + Ω0)))]^2))/(6 ρ) - (I (wn + I ρ)^2 (π^2 + 3 Log[wn + I (ρ + Ω0)]^2))/(6 ρ)))


The lower limit is just lowerlimit = indefInt/.w->Ω. And so the value of your definite integral is

upperlimit - lowerlimit


I hope that helps.

• This is nice. When I run Simplify with my assumptions on (stuff), I get zero. Which is nice, but when I run Simplify for just Floor[Arg[-wn/(wn-i rho)]/2pi], I don't get -1 as expected, but I get Cases. That's why I am confused that Mathematica Simplifies that horrible (stuff) to zero right away. Thank you, although quite messy, it gets job done in a clear, feasible way! – user16320 Mar 18 '17 at 17:30