I am reading Wong's book on "Asymptotic Approximations of Integrals". On page 497, the book recalls (without proof) the following estimate: for all $\delta>0$ and $\nu>1$, $$ \int_\delta^\infty t^m e^{-\nu t^2} dt \le K_\delta e^{-\nu \delta^2}, $$ where $K_\delta$ is a constant independent of $\nu$.
May I know whether the above estimate is a well-known result? Could you provide a reference for it?
You can proceed by induction on $m$. Integrating by parts with $t^{m-1}$ and $te^{-νt^2}$ we get $$\int_δ^\infty t^m e^{-νt^2} dt = \left[\frac{-1}{2ν} t^{m-1} e^{-νt^2}\right]_δ^\infty + \frac{m-1}{2ν}\int_δ^\infty t^{m-2} e^{-νt^2} dt = \frac{1}{2ν}\left(δ^{m-1}e^{-νt^2} + (m-1)\int_δ^\infty t^{m-2} e^{-νt^2} dt\right).$$ Bounding $\frac{1}{2ν}\leq \frac{1}{2}$ (since $ν>1$) and applying the inductive hypothesis to the remaining integral gives you the result.
Performing the change of integration variables from $t$ to $s$ via $ t = \delta \sqrt {1 + s}$ gives $$ \int_\delta ^{ + \infty } {t^m {\rm e}^{ - \nu t^2 } {\rm d}t} = \frac{{\delta ^{m + 1} }}{2}{\rm e}^{ - \nu \delta ^2 } \int_0^{ + \infty } {(1 + s)^{(m - 1)/2} {\rm e}^{ - \nu \delta ^2 s} {\rm d}s} . $$ Since $\nu>1$, we can assert that $$ \int_0^{ + \infty } {(1 + s)^{(m - 1)/2} {\rm e}^{ - \nu \delta ^2 s} {\rm d}s} \le \int_0^{ + \infty } {(1 + s)^{(m - 1)/2} {\rm e}^{ - \delta ^2 s} {\rm d}s} . $$ Hence, the claim follows by letting $$ K_\delta = \frac{{\delta ^{m + 1} }}{2}\int_0^{ + \infty } {(1 + s)^{(m - 1)/2} {\rm e}^{ - \delta ^2 s} {\rm d}s} . $$
