I happen to have the heat kernel on the twodimensional 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.
1 Answer
A partial integration can remove the singularity: $$P_2(x,t)=\frac{\sqrt{2}e^{t/4}}{(4\pi t)^{3/2}}\int_x^\infty\frac{se^{s^2/4t}ds}{\sqrt{\cosh s \cosh x }}=$$ $$\qquad =\frac{\sqrt{2}e^{t/4}}{(4\pi t)^{3/2}}\int_x^\infty\frac{2\sqrt{sx}\,s e^{s^2/4t}}{\sqrt{\cosh s \cosh x}}\left(\frac{d}{ds}\sqrt{sx}\right)\,ds$$ $$\qquad=\frac{\sqrt{2}e^{t/4}}{(4\pi t)^{3/2}}\int_x^\infty\sqrt{sx}\left(\frac{d}{ds}\frac{2\sqrt{sx}\,s e^{s^2/4t}}{\sqrt{\cosh s \cosh x}}\right)\,ds$$ $$\qquad=\frac{\sqrt{2}e^{t/4}}{(4\pi t)^{3/2}}\int_x^\infty e^{s^2/4 t}\frac{ \left(s^3s^2 x3 s t+2 t x\right) (\cosh s\cosh x)+s t (sx) \sinh s}{t (\cosh s\cosh x)^{3/2}}\,ds.$$ In the final expression the integrand vanishes$^\ast$ as $(sx)^{1/2}$ 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 $$\left(s^3s^2 x3 s t+2 t x\right) (\cosh s\cosh x)+s t (sx) \sinh s$$ $$\qquad=xt(sx)\sinh s+xt(sx)\sinh s+{\cal O}(sx)^2={\cal O}(sx)^2.$$ The denominator is of order $(sx)^{3/2}$, so the ratio is of order $(sx)^{1/2}$. }

$\begingroup$ Thank you for your answer. However I fail to see why the integrand vanishes as $(sx)^{1/2}$. Could you please give me more details on that? $\endgroup$– MathqANov 15, 2021 at 17:13

1$\begingroup$ 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! $\endgroup$– MathqANov 15, 2021 at 18:43