# Calculation/Verification of an integral kernel for $\operatorname{e}^{t\Delta}(1-\Delta)^{-\frac{1}{4}}$

Given the operator $T = \operatorname{e}^{t\Delta}(1-\Delta)^{-\frac{1}{4}} \colon L^p(\mathbb{R}^3) \to L^p(\mathbb{R}^3), p \in (1,\infty), t > 0$ I want to calculate its kernel $K_t$ in order to see $T$ as integral operator. Here, $\Delta$ is the Laplacian in $\mathbb{R}^3$. I know already how this kernel should look like but still fail in calculating/verifying it. Maybe someone can give me a hint?! I have the feeling a suitable substitution is needed.

This is what I want to get: $$K_t(x) = \frac{1}{\sqrt{\frac{3}{4} \pi^3}} \int_0^{\infty}s^{-\frac{3}{4}}\operatorname{e}^{-s}(s+t)^{-\frac{3}{2}} \operatorname{e}^{-\frac{|x|^2}{s+t}} \, ds,$$ where $x \in \mathbb{R}^3$ and $t > 0$.

Here is how far I got so far: I used the heat kernel in order to write out a representation for the heat semigroup $\operatorname{e}^{t\Delta}$ and I used that the the fractional power $(1-\Delta)^{-\frac{1}{4}}$ has the representation $$(1-\Delta)^{-\frac{1}{4}}g(y) = \frac{\sin(\frac{1}{4}\pi)}{\pi} \int_0^{\infty} s^{-\frac{1}{4}} (s+1-\Delta)^{-1}g(y) \, ds$$ and that $$(\lambda - \Delta)^{-1} = \int_0^{\infty} \operatorname{e}^{- \lambda r} \operatorname{e}^{r \Delta} \, dr$$ for $\lambda$ in the resolvent set of $\Delta$.

Alltogether I come to $$\operatorname{e}^{t\Delta}(1-\Delta)^{-\frac{1}{4}}g(x) = \frac{1}{(4\pi t)^{\frac{3}{2}}} \frac{1}{\sqrt{2}\pi} \int_{\mathbb{R}^3} \int_0^{\infty} s^{-\frac{1}{4}}\operatorname{e}^{-\frac{|x-y|^2}{4t}}(s+1-\Delta)^{-1}g(y)\, ds dy \\ = \frac{1}{(4\pi t)^{\frac{3}{2}}} \frac{1}{\sqrt{2}\pi} \int_{\mathbb{R}^3} \int_0^{\infty} s^{-\frac{1}{4}}\operatorname{e}^{-\frac{|x-y|^2}{4t}} \int_0^{\infty} \operatorname{e}^{-(s+1)r} \operatorname{e}^{r \Delta} g(y) \, dr ds dy \\ = \frac{1}{\Gamma(\frac{1}{4})}\frac{1}{(4\pi t)^{\frac{3}{2}}} \ \int_{\mathbb{R}^3} \int_0^{\infty} \operatorname{e}^{-\frac{|x-y|^2}{4t}} r^{-(\frac{3}{4})}\operatorname{e}^{-r} \operatorname{e}^{r \Delta} g(y) \, dr dy$$

where I performed the integration in the $s$ variable in the last step. From there I am not sure how to proceed. One option I tried is to use again the heat kernel:

$$\operatorname{e}^{t\Delta}(1-\Delta)^{-\frac{1}{4}}g(x) = \\ \frac{1}{\Gamma(\frac{1}{4})}\frac{1}{(4\pi t)^{\frac{3}{2}}}\frac{1}{(4\pi r)^{\frac{3}{2}}} \ \int_{\mathbb{R}^3} \int_0^{\infty} \operatorname{e}^{-\frac{|x-y|^2}{4t}} r^{-(\frac{3}{4})}\operatorname{e}^{-r} \int_{\mathbb{R}^3} \operatorname{e}^{-\frac{|y-w|^2}{4r}} g(w) \, dw dr dy$$

But here I am completely stuck. Some parts seem already to be like in the desired kernel. I am glad for any hint. I assume it is not so hard to get the result so I would prefer constructive hints instead of a complete calculation.

First I remark that $$\operatorname{e}^{t\Delta}(1-\Delta)^{-\frac{1}{4}} = \mathcal{F}^{-1} \operatorname{e}^{-t|\xi|^2}(1-|\xi|^2)^{-\frac{1}{4}}\mathcal{F},$$ where $\mathcal{F}$ denotes the Fourier transform. Further by the definition of the Fourier transform and its inverse, we see for $g \in \mathcal{S}(\mathbb{R}^3)$ that $$\mathcal{F}^{-1} \operatorname{e}^{-t|\xi|^2}(1-|\xi|^2)^{-\frac{1}{4}}\mathcal{F} g = \mathcal{F}^{-1} \operatorname{e}^{-t|\xi|^2}(1-|\xi|^2)^{-\frac{1}{4}} \ast g,$$ so that in order to calculate the kernel, we need to calculate $$k_t(x) = \mathcal{F}^{-1} \operatorname{e}^{-t|\xi|^2}(1-|\xi|^2)^{-\frac{1}{4}} (x).$$ We will thereto use the equalities, which I suggested to be useful in my question post, for $-|\xi|^2$ instead of $\Delta$. By doing so we get $$k_t(x) = \int_{\mathbb{R}^3}\operatorname{e}^{i\xi \cdot x} \operatorname{e}^{-t|\xi|^2}(1-|\xi|^2)^{-\frac{1}{4}} \, d\xi \\ = \frac{1}{\sqrt{2} \pi} \int_{\mathbb{R}^3}\operatorname{e}^{i\xi \cdot x} \int_0^{\infty}\int_0^{\infty} s^{-\frac{1}{4}}\operatorname{e}^{-r(s+1)} \operatorname{e}^{-(r+t)|\xi|^2} \, dr ds d\xi\\ = \frac{\Gamma(\frac{3}{4})}{\sqrt{2} \pi} \int_{\mathbb{R}^3}\operatorname{e}^{i\xi \cdot x} \int_0^{\infty} \operatorname{e}^{-r}r^{-\frac{3}{4}} \operatorname{e}^{-(r+t)|\xi|^2} \, dr d\xi\\ = \frac{\Gamma(\frac{3}{4})}{\sqrt{2} \pi} \int_{\mathbb{R}^3}\operatorname{e}^{iz \cdot \frac{x}{\sqrt{r+t}}} \int_0^{\infty} \operatorname{e}^{-r}r^{-\frac{3}{4}} \operatorname{e}^{-|z|^2} (r+t)^{-\frac{3}{2}} \, dr dz \\ =2 \Gamma(\frac{3}{4})\sqrt{\pi} \int_0^{\infty} \operatorname{e}^{-r}r^{-\frac{3}{4}} \operatorname{e}^{-\frac{|x|^2}{r+t}} (r+t)^{-\frac{3}{2}} \, dr.$$ We used the substitution $\sqrt{r+t}\xi = z.$ Further we used, that $(2\pi)^{-\frac{3}{2}}\operatorname{e}^{-z^2}$ is a fixed point of the Fourier transform. We also used the Gamma function $\Gamma(z) = \int_0^{\infty} t^{z-1} \operatorname{e}^{-t} \, dt$. However, somehow I end up with a different scalar factor in the kernel.