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I happen to have the heat kernel on the two-dimensional hyperbolic space and I need to take partial derivatives in order to check that it satisfies the heat equation as expected. The problem is I can not apply the Leibniz formula because I get zero in the denominator. The function is $$P_2(x,t)=\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int_x^\infty\frac{se^{-s^2/4t}ds}{\sqrt{\cosh(s)-\cosh(x)}}.$$ I would be very grateful if you could help me take the partial derivative $\frac{\partial P_2}{\partial x}$. I need it symbolically, not numerically, because I want to use it in the heat equation.

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    $\begingroup$ How about substituting $s = r x$ and using the dominated convergence theorem? $\endgroup$ Nov 15, 2021 at 11:24

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A partial integration can remove the singularity: $$P_2(x,t)=\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int_x^\infty\frac{se^{-s^2/4t}ds}{\sqrt{\cosh s -\cosh x }}=$$ $$\qquad =\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int_x^\infty\frac{2\sqrt{s-x}\,s e^{-s^2/4t}}{\sqrt{\cosh s -\cosh x}}\left(\frac{d}{ds}\sqrt{s-x}\right)\,ds$$ $$\qquad=-\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int_x^\infty\sqrt{s-x}\left(\frac{d}{ds}\frac{2\sqrt{s-x}\,s e^{-s^2/4t}}{\sqrt{\cosh s -\cosh x}}\right)\,ds$$ $$\qquad=\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int_x^\infty e^{-s^2/4 t}\frac{ \left(s^3-s^2 x-3 s t+2 t x\right) (\cosh s-\cosh x)+s t (s-x) \sinh s}{t (\cosh s-\cosh x)^{3/2}}\,ds.$$ In the final expression the integrand vanishes$^\ast$ as $(s-x)^{1/2}$ when $s\rightarrow x$, so there are no contributions from the integration bounds when we differentiate the integral with respect to $x$.


$^\ast$ The numerator expands around $s=x$ as $$\left(s^3-s^2 x-3 s t+2 t x\right) (\cosh s-\cosh x)+s t (s-x) \sinh s$$ $$\qquad=-xt(s-x)\sinh s+xt(s-x)\sinh s+{\cal O}(s-x)^2={\cal O}(s-x)^2.$$ The denominator is of order $(s-x)^{3/2}$, so the ratio is of order $(s-x)^{1/2}$.

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  • $\begingroup$ Thank you for your answer. However I fail to see why the integrand vanishes as $(s-x)^{1/2}$. Could you please give me more details on that? $\endgroup$
    – MathqA
    Nov 15, 2021 at 17:13
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    $\begingroup$ sure, I have added the calculation. $\endgroup$ Nov 15, 2021 at 18:08
  • $\begingroup$ Ok! Thank you a lot. Now I see it. It was a really helpful answer! $\endgroup$
    – MathqA
    Nov 15, 2021 at 18:43

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