If forced to construct a clean proof from scratch off the top of my head, I'd use a Möbius transformation preserving the upper half-plane (i.e., a hyperbolic isometry) to send $p = (x_1, y_1)$ and $q = (x_2, y_2)$ to $(0, 1)$ and a point $(0, y)$ respectively, then perform the easy length integral along the vertical axis.
– Andrew D. HwangFeb 11 '17 at 13:47
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The proposed duplicate is regarding the equivalence of two different expressions for the distance function, not the derivation from the Riemannian metric.
– Anthony CarapetisFeb 12 '17 at 09:42
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@AndrewD.Hwang I think this question is not duplicate with Expression of the Hyperbolic Distance in the Upper Half Plane . Because I want to know how to prove it . In the answers of Expression of the Hyperbolic Distance in the Upper Half Plane, just prove two expression are equal.
– Enhao LanFeb 12 '17 at 13:15
@lanse7pty: Will Jagy's answer to the linked question gives a good strategy, but I agree that the accepted linked answer does not address your question, and have voted to re-open.
– Andrew D. HwangFeb 12 '17 at 14:15
The linked question (as well as my answer to it) is about the form, not the substance, of the distance formula. This question is not a duplicate, and I have also voted to re-open.
– BlueFeb 12 '17 at 18:06