How fast does this function converge to zero?

Question

Consider the function given by $$f:(0,\infty)\rightarrow \mathbb{R}, t\mapsto \int\limits_{-\delta}^{\delta}x^{2k}\frac{1}{\sqrt{4\pi t}}e^{-\frac{x^2}{4t}}dx,$$

where $k\in\mathbb{N}$ and $\delta >0$. I know the function will converge to zero for $t\downarrow 0$, i.e. $\lim\limits_{t\downarrow 0}f(t)=0$. But I would like to know how fast the convergence is.

Is it for instance possible to show some estimate like: $f(t)\leq C\cdot t^{N(k)}$ for some number $N(k)$ or to show that $\lim\limits_{t\downarrow 0}t^{-\alpha}f(t)=0$, for some $\alpha >0 $? If this is possible, which number $N(k)$ or $\alpha$ would be the best possible?

Best regards

Answer 1

If you replaced $\delta$ by $\infty$, the integral would be $(4t)^k \Gamma(k+1/2)/\sqrt{\pi}$.

Answer 2

With $u = t/\sqrt t$:

$$ \int\limits_{-\delta}^{\delta}x^{2k}\frac{1}{\sqrt{4\pi t}} e^{-\frac{x^2}{4t}}dx = \int\limits_{-\delta/\sqrt t}^{\delta/\sqrt t} t^k u^{2k}\frac{1}{\sqrt{4\pi }} e^{-\frac{u^2}{4}}du \le t^k \int\limits_{\Bbb R} u^{2k}\frac{1}{\sqrt{4\pi }} e^{-\frac{u^2}{4}}du $$ so $N(k) = k$ is okay.