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Let $\delta_+,\delta_-$ be two complex variables. Denote $e_i, i = 1,...,6$ be (multi-valued) holomorphic functions of $\delta_+,\delta_-$ defined as follows: $$ \begin{align} e_1 & = -2\left(1 - \frac{\delta_+}{12} - \frac{\delta_-}{36}\right), \\ e_2 &= -\left( 1 - \mathrm{i}\sqrt{\frac{\delta_+}{3}} - \frac{\delta_+}{36} - \frac{\delta_-}{12} \right),\\ e_3 &= -\left( 1 + \mathrm{i} \sqrt{\frac{\delta_+}{3}} - \frac{\delta_+}{36} - \frac{\delta_-}{12} \right)\\ e_4 &= \left( 1 + \mathrm{i}\sqrt{\frac{\delta_-}{3}} - \frac{\delta_+}{12} - \frac{\delta_-}{36} \right),\\ e_5 &= \left( 1 - \mathrm{i} \sqrt{\frac{\delta_-}{3}} - \frac{\delta_+}{12} - \frac{\delta_-}{36} \right),\\ e_6 &= 2\left(1 - \frac{\delta_+}{12} - \frac{\delta_-}{36}\right) \end{align} $$ Note that as $\delta_\pm \to 0$, $e_2$ and $e_3$ both converge to $-1$, and $e_1 \to -2$.

My question is how to obtain the asymptotic behavior of the following integration as $\delta_\pm\to0$: $$ \int_{e_1}^{e_2} \frac{3x^3 - xu}{\sqrt{(x - e_1)(x - e_2)(x - e_3)(x - e_4)(x - e_5)(x - e_6)}} \mathrm{d} x $$ where $u = c_0 + c_1(\delta_+ + \delta_-)$ for some contants $c_i$.

This question arised from and answered by pp.47--pp.48 of Klemm-Lerche-Theisen's paper Nonperturbative Effective Actions of N=2 Supersymmetric Gauge Theories. They write the lower terms as: $$ -3 - \frac{1}{12} \delta_+\left(\log\delta_+ - 2\log2 - 3 \log3 - 1\right) + \frac{\log2}{6} \delta_- $$ But they skip the details, and I don't know how to calculate it. I understand why there would be logarithmic singularities, since as $\delta_{\pm} \to 0$ the integration near $e_2$ is something like $$ \int^{e_2} \frac{\mathrm{d}x}{x-e_2}. $$

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  • $\begingroup$ Skimming the referenced paper, the authors do actually give quite a lot of detail. You might try to follow the steps they outline and exhibit exactly where you get stuck. Now, the result you exhibit doesn't in fact diverge logarithmically - since the point $x=e_2 $ has to be treated with a bit more care than in your final expression ... $\endgroup$ Commented Aug 29 at 2:49

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