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As in wiki,the distance function is

enter image description here

I don't know how to get this formula, I mean how to prove it.

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    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. – 2017-02-11
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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. – 2017-02-12
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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. – 2017-02-12
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    @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. – 2017-02-12
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    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. – 2017-02-12

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