This is my first question on mathstackexchange and I've tried my best to follow the guidelines. If something about my post is inadequate please feel free to let me know. Also, I do apologise for the length of this post but I wanted to give some additional context which might be useful.
I'm currently working on representing the function $e^{-t^\alpha}$, where $\alpha \in (0,1)$, as an integral average of the function $t^{\alpha/2}r^{\alpha/2} K_{\alpha/2}(tr)$, where $K_{\alpha/2}$ is the modified Bessel function of the second kind, i.e. an equality of the form $$ e^{-t^\alpha} =t^{\frac{\alpha}{2}} \int_0^\infty f(r)r^{\alpha/2}K_{\alpha/2}(tr)dr $$ The function that I have constructed which I think does the job is, up to positive constants, given by $$ f(r) := r^{1-\frac{\alpha}{2}} \Im\left[ \int_0^\infty J_{\alpha/2}(rs)s^{1-\frac{\alpha}{2}} e^{-(is)^\alpha} ds \right],$$ where $\Im$ denoted the imaginary part.
This can be seen by letting $-\mu = \nu = \frac{\alpha}{2}$, $\alpha = r$, and $\beta = t$ in the following identity $$ \int_0^\infty t^{\mu+\nu+1}K_\mu(\alpha t)J_\nu(\beta t) dt = \frac{(2\alpha)^\mu(2\beta)^{\nu}\Gamma(\mu + \nu +1)}{(\alpha^2 + \beta^2)^{\mu + \nu +1}}, $$ see for example page 410 in Watson's Treatise On The Theory Of Bessel Functions. The resulting integral then is $$ t^{\frac{\alpha}{2}} \int_0^\infty f(r)r^{\alpha/2}K_{\alpha/2}(tr)dr = \Im \left[ \int_0^\infty \frac{re^{-(ir)^\alpha}}{t^2+r^2} dr \right] $$ which, if I have not made some blunder, evaluates to the desired result upon applying the residue theorem to the function $g(r) := \frac{re^{-(ir)^\alpha}}{t^2+r^2}$ integrated over $(-\infty,\infty)$ and noting that $g(-r) = -\overline{g(r)}$ for $r>0$.
What I need for the application of this result is that both $f(r)$ and also $r^\alpha f(r)$ are integrable. My attempt goes like this:
Recall the relations \begin{align} \frac{d}{dz} z^{-\nu} J_\nu(z) &= -z^{-\nu}J_{\nu+1}(z)\\ \frac{d}{dz} z^{\nu+1} J_{\nu+1}(z) &=z^{\nu+1}J_\nu(z). \end{align} Substituting $u = \frac{s}{t}$ in the integral defining $f(r)$ we have \begin{align} r^\alpha f(r) &= r^{1+\frac{\alpha}{2}} \Im\left[ \int_0^\infty J_{\alpha/2}(rs)s^{1-\frac{\alpha}{2}} e^{-(is)^\alpha} ds \right] \\ &= r^{\alpha-1} \Im\left[ \int_0^\infty J_{\alpha/2}(u)u^{1-\frac{\alpha}{2}} e^{-(iu)^\alpha r^{-\alpha}} ds \right] \end{align} and integrating by parts using the first relation with $\nu = \frac{\alpha}{2}-1$ (the boundary value at $u=0$ vanishes because it's real-valued and we are taking the imaginary part) now gives $$ r^\alpha f(r) = r^{-1} \Im \left[ i^\alpha \int_0^\infty J_{\alpha/2-1}(u)u^{\frac{\alpha}{2}} e^{-(iu)^\alpha r^{-\alpha}} du\right] $$ and if we now use the second relation with $\nu = \frac{\alpha}{2}-1$ we then get $$ r^\alpha f(r) = r^{-1-\alpha} \Im \left[ i^{2\alpha} \int_0^\infty J_{\alpha/2}(u)u^{\frac{3\alpha}{2}-1} e^{-(iu)^\alpha r^{-\alpha}} \right]. $$ This integral can be shown to be uniformly bounded by introducing the Hankel functions of the first kind $H^{(1)}_\nu(z) = J_\nu(z) + iY_\nu(z)$ and rotating the path of integration by a small angle $\delta>0$ into the upper complex plane, as once this is done, the Hankel function decreases exponentially and thus we may even take the limit as $r \to \infty$ (this argument is adapted from Some Theorems on Stable Processes, R. M. Blumenthal and R. K. Getoor). If the above approach works, this shows that both $f(r)$ and $r^\alpha f(r)$ are absolutely integrable, and this would make me very happy. However, I have some doubts. This is mainly due to the following:
Suppose that $\alpha = \frac{1}{2}$ and consider $$ r^{-\frac{1}{2}}f(r) = \Im \left[ \int_0^\infty r^{\frac{1}{4}}J_{1/4}(rs)s^{1-\frac{1}{4}} e^{-(is)^{\frac{1}{2}}} ds \right]. $$ By (7) 4.14 in Tables Of Integral Transforms, The Bateman Manuscript Project we have the identity $$\mathcal{L}[t^\nu J_\nu(\alpha t)](\lambda) = \frac{2^\nu\Gamma(\nu+\frac{1}{2})}{\sqrt{\pi}}\frac{\alpha^\nu}{(\lambda^2 + \alpha^2)^{2\nu+\frac{1}{2}}} $$, where $\mathcal{L}$ denotes the Laplace transform. Applying this to $r^{-\frac{1}{2}}f(r)$ gives, up to a positive constant, $$ \mathcal{L}[r^{-\frac{1}{2}}f(r)](\lambda) = \Im \left[\int_0^\infty \frac{se^{-(is)^{\frac{1}{2}}}}{\lambda^2 + s^2}ds \right] = e^{-\sqrt{r}}$$. However, $e^{-\sqrt{r}}$ is the Laplace transform of the function $h(r) = \frac{1}{2\sqrt{\pi}}r^{-\frac{3}{2}}e^{-\frac{1}{4r}}$, see for example this stackexchange post and hence $f(r) = r^{-1}e^{-\frac{1}{4r}}$ up to a positive constant, which is definitely not integrable.
This now begs the question where my mistake is, and if this integrability even holds. I've spent almost two weeks racking my brain and I just can't find the mistake so I'd appreciate if someone could help me out. I (at least I think) have already proven that for the $\alpha \in (1,2)$ case, an opposite equality holds, namely that we can represent $t^{\alpha/2}r^{\alpha/2} K_{\alpha/2}(tr)$ in terms of $e^{-r^\alpha}$, but a definite statement regarding the $\alpha \in (0,1)$ case seems to elude me.
For those who are interested, the reason I'm even doing is is that these integral representations allow for a proof of statements of the form "If $\alpha \in (1,2)$, convergence in the pointwise definition of the Laplace operator to the power $\alpha$ as the generator of the fractional heat group (on the whole space) implies convergence in the pointwise definition of the harmonic extension definition of the same operator as outlined in the 2017 paper Ten equivalent characterisations of the fractional Laplace operator" and the corresponding $\alpha \in (0,1)$ case.
