Let $d\ge 3$ be an integer and $m,r>0$. Then how would one show that $$ \int_m^\infty (s^2-m^2)^{(d-3)/2}e^{-sr}ds \le \frac{C}{r^{(d-1)/2}} e^{-mr} $$ for somce constant $C$ which may depend on $m$ but not $r$.
Background. The LHS is basically the Fourier transform $G(x)$ of $1/(k^2+m^2)$ at $x$ where $|x|=r$ (you can check this by taking $d=3$). It is known that $G(x)$ is bounded by the RHS (or at least, I have seen papers mentioning this), but how would one prove the aforementioned inequality.
To make the story short, using quite obvious changes of variables, the integral is $$I=\frac{2^{\frac{d}{2}-1} \left(\frac{m}{r}\right)^{\frac{d-2}{2}}}{\sqrt{\pi }} \,\Gamma \left(\frac{d-1}{2}\right)\,K_{\frac{d-2}{2}}(m r)$$ At least, when $m$ is large,we have $$I=\frac{(2 m)^{\frac{d-3}{2}}\,\Gamma \left(\frac{d-1}{2}\right) e^{-mr}}{r^{\frac {d-1 } 2}}\Bigg[1+\frac{(d-1)(d-3)}{8 m r}+O\left(\frac{1}{m^2}\right) \Bigg]$$ If we neglect the high order terms, then you would like to show that $$\frac{(2 m)^{\frac{d-3}{2}}\,\Gamma \left(\frac{d-1}{2}\right) e^{-mr}}{r^{\frac {d-1 } 2}} < \frac C {r^{\frac {d-1 } 2}} e^{-mr}$$ that is to say $$(2 m)^{\frac{d-3}{2}}\,\Gamma \left(\frac{d-1}{2}\right) < C$$