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.



Partial derivative of the heat kernel
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 $x$.




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$^\ast$ 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 $(s-x)^{1/2}$.
