# Expanding the integrand gives a different result

I integrated this term in Mathematica:

$$\int_{-\infty}^{\infty} d\omega\cdot \sin(s\cdot \omega)\cdot \frac{1}{e^{\beta\cdot \hbar\cdot \omega}-1}\cdot \frac{\omega}{\omega^{2}+\gamma^{2}}$$

The code in Mathematica:

Integrate[Sin[s*ω]*(1/(Exp[β*ℏ*ω] -1))*ω/(ω^2 + γ^2), {ω, -Infinity, Infinity},
Assumptions -> {ℏ > 0, s >= 0, γ > 0, β > 0}]


The result is:

$$\frac{-\pi}{2}\cdot e^{-s\cdot \gamma}$$

If I suppose that $\beta$ is small and expand the exponential in the integrand, and keep the lowest term the result would be zero. However, the result for the unexpanded term does not depend on $\beta$. Why does expanding the exponential lead to an incorrect answer?

• If you expand the exponential in $\beta$, doesn't that whole term become $\frac{1}{\beta \hbar \omega}$? I don't obviously see why the integral of $\frac{\sin(s\omega)}{\omega^{2} + \gamma^{2}}$ is zero. – Jerry Schirmer Jun 10 '14 at 16:03
• @JerrySchirmer the integrand is odd. – Harald Hanche-Olsen Jun 10 '14 at 16:04
• @JerrySchirmer. Yes it does. After that the integral is odd and the limit of integration is over a symmetric rage of negative and positive values so the result would be zero. – MOON Jun 10 '14 at 16:05
• why would you expect to get the correct result from your approximation procedure? It seems to me that you cannot use the dominated convergence theorem, for example. – Harald Hanche-Olsen Jun 10 '14 at 16:05
• Oh, duh. Yes. Shouldn't math in the morning. – Jerry Schirmer Jun 10 '14 at 16:07

The integrand is not analytic at $\beta = 0$ so expanding around $\beta = 0$ does not make sense.
• The integrand has a simple pole at $\beta=0$, so expanding could make sense. I think the problem here is global, not local. – Harald Hanche-Olsen Jun 10 '14 at 16:28