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I was looking at a paper by Takao Suyama on GT theory, and I couldn't figure out how he derived his formula (3.59): $$\frac{1}{\pi}\int_a^bdx\frac{1}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{|(x-a)(x-b)|}}\frac{\log(e^{-t_1}x)}{2}=\frac{1}{2}\log\left[\frac{e^{-t_1}}{2\sqrt{ab}+a+b}\left(z+\sqrt{ab}-\sqrt{(z-a)(z-b)}\right)^2\right],$$ where $0<a<b$, $[a,b]\subset\mathbb{R}$, $t_1\in\mathbb{R}$, and $z\in\mathbb{C}\setminus[a,b]$. It's a physics paper, but my question is just how would one do the integral?

I tried expanding everything as a power series, and then using the fact that the integral of $x^n\log x$ is known, but then I couldn't figure out how to resum the resulting series, so I'm a bit confused how one would solve this integral.

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Okay, so I think I may have found the answer myself. So, really, the absolute value symbol is a trick. You can get rid of it by pulling out an $i$, and then you have $$\mathcal{I}:=\frac{1}{2\pi i}\int_a^b dx \frac{\log(x e^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}}$$ What you have to do is take a dumbbell contour around taking a clockwise "dumbbell" contour $\mathcal{C}$ around the region $[a,b]$ such that we integrate over a region $[a+i0,b+i0]$ from left to right and a region $[a-i0,b-i0]$ from right to left. Since there is a branch cut along $[a,b]$, we have a sign flip as we cross the branch cut, and hence we have \begin{equation} 2\mathcal{I}= \frac{1}{2\pi i}\int_{\mathcal{C}}dx \frac{\log(x e^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}} \end{equation} We now deform the contour to infinity, making note of the branch cut from the log and the pole at $x=z$. Then we pick up a residue at $z$ \begin{equation} 2\pi i\text{Res}_z\left(\frac{1}{2\pi i}\frac{\log (xe^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}}\right)=\log (z e^{-t_1}) \end{equation} There is also a contribution from integrating along the log branch cut that can be easily done. Summing all these together and doing a painful amount of simplification yields \begin{equation} \mathcal{I}=\frac{1}{2}\log\left[\frac{e^{-t_1}}{2\sqrt{ab}+a+b}\left(z+\sqrt{ab}-\sqrt{(z-a)(z-b)}\right)^2\right] \end{equation} as desired.

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