I'm starting my masters in theoretical condensed matter physics soon, and I encountered this integral in a paper I have to read before starting my research.
The paper presented the answer as: \begin{align} & \mathcal P \frac1\pi {\LARGE \int}_0^\infty \frac{\omega' \tanh \left(\dfrac{\hbar\omega'}{2k_BT}\right)}{\omega^2-\omega'^2} \mathrm d\omega' \\[2mm] = & \ \mathrm{Re}\Psi \left(\frac12+\frac{\hbar\omega}{2\pi ik_BT}\right)- \ln\left(\frac{\hbar\omega}{k_BT}\right) \end{align}
(I left the exact constants to hopefully make the question clearer, but for simplicity, $a\equiv\dfrac{\hbar\omega}{2k_BT}$ in the title).
I found *this thread about a similar enough integral to get the portion with the Digamma function, but it leaves me with a different $\ln$ term.
*Edited to fix wrong link address
UPDATE So I've tried a couple different approaches:
- J.G.'s tip towards the Sokhotski–Plemelj theorem led me to an expression missing the argument inside the Digamma function.
- Integrating by parts led me to the following expression: $$-\frac{1}{2}{\LARGE[}\ln(a^2-x^2)\tanh(x)\Big|_{x=0}^\infty-\int_0^\infty\frac{\ln(a^2-x^2)}{\cosh^2(x)}\mathrm{d}x{\LARGE]}$$ The integral on the right gives (using Syvatoslav's method linked above): $\frac{\ln\pi}{2}+\mathrm{Re}\Psi(\frac12+\frac{a}{i\pi})$
So my two remaining problems are:
- As Syvatoslav (hey! totally glossed over you commenting) pointed out below, the left term diverges. However, the paper this calculation is from doesn't really mention why and how to treat it. There's just a term of the form $-\ln(2a)$.
- The term $\frac{\ln\pi}{2}$ isn't in the final expression given at all.
I don't know if these problems are can be solved mathematically or only physically from the system's properties, if the latter than I'll ask a new question in the physics forum.
