The paper is entitled "Hyperkahler metrics and Supersymmetry", published in Communications in Mathematical Physics, in 1987. I am a little stuck on a contour integral. My question is related to eq. (2.13). Let
$\eta = z - \zeta x -\zeta^2 \bar{z}$
$G(\eta, \zeta) = \frac{1}{\zeta^2} \eta \ln \eta$
In the paper, they wrote, in the caption of fig. 1, using $r^2 = x^2+4z\bar{z}$:
$\eta = -z(\zeta-\zeta_+)(\zeta-\zeta_-)$, with
$\zeta_{\pm} = -(1/2z)(x \pm r)$.
But I think the $z$'s in the last 2 equations should be $\bar{z}$ instead.
We are interested in the contour integral:
$F(x,z,\bar{z}) = \operatorname{Re} \oint_C G(\eta(\zeta), \zeta)$
where $C$ is a contour going once around $\zeta_+$ counterclockwise, and once around $\zeta_-$ clockwise, and "missing" the origin. The answer is given by eq. (2.13), namely:
$F = r - x\ln(x+r) + \frac{1}{2}x\ln(4z\bar{z})$
I am just not getting the same answer. Moreover, the logarithm is multivalued, with its values differing by a multiple of $2\pi i$. Is the contour chosen in a smart way so that the multivaluedness of the logarithm in some sense cancels out? I would appreciate if someone could give some brief hints on how to perform the contour integral above.
You are correct that there is a typo exchanging $z$ and $\bar z$. The figure eight contour is independent of the multivaluedness of the logarithm--it cancels out between the two parts. Indeed, this is a general way to write definite integrals as contour integrals: for any function $f(u)$ (assuming appropriate singularity structure), the integral $$\int_v^w f(u)~du = \frac1{2\pi i} \int_C du~ f(u) \ln[(u-v)(u-w)]$$ where C is the figure eight contour; you can understand this because if you draw the branch cut between $v$ and $w$, then $$f(u)\ln[(u-v)(u-w)]|_+ -f(u)\ln[(u-v)(u-w)]|_- = 2\pi\, i\, f(u)$$ where the subscripts $\pm$ refer to being above and below the branch cut.
Another way to understand the contour is to visualize the multi-sheeted surface of $\ln[(u-v)(u-w)]$ as a parking garage with two spiral ramps going between the levels. Then you can see that the figure eight contour takes you up one ramp and down the next one so the integral is unambiguous.
I hope this helps, Martin Rocek