Limit Bessel Gaussian

Question

I am able to prove that: $$\lim_{\varepsilon \to 0}\int^{\infty}_{0}\varepsilon^{-1}\left(J_{\frac{3}{2}}\left(\frac{r}{\varepsilon}\right)\right)^{2}\exp{(-r^{2})}r dr<\infty.$$

But I am unable to have any information about the limit. Does this limit or $\liminf$ have an explicit form?

Thank you very much.

Answer 1

Using this result, we have that $$L(\varepsilon) = \int_{0}^{\infty}\varepsilon^{-1}\left(J_{3/2}(r/\varepsilon)\right)^{2}\exp(-r^{2})r \ dr = \frac{1}{2\varepsilon}\exp\left(-\frac{1}{2\varepsilon^{2}}\right)I_{3/2}\left(\frac{1}{2\varepsilon^{2}}\right)$$ where $I_{\nu}$ is the modified Bessel function of the first kind. Using this result, we have that $$I_{3/2}\left(\frac{1}{2\varepsilon^{2}}\right) = \frac{\left(\frac{1}{2}\frac{1}{2\varepsilon^{2}}\right)^{3/2}}{\sqrt{\pi}\Gamma(2)}\int_{-1}^{1}(1-t^{2})e^{-\frac{1}{2\varepsilon^{2}}t} \ dt = \frac{1}{8\varepsilon^{3}\sqrt{\pi}}\int_{-1}^{1}(1-t^{2})e^{-\frac{t}{2\varepsilon^{2}}} \ dt$$ $$=\frac{1}{8\varepsilon^{3}\sqrt{\pi}}\left[(16\varepsilon^{6} + 8\varepsilon^{4})e^{-1/2\varepsilon^{2}} - (16\varepsilon^{6} - 8\varepsilon^{4})e^{1/2\varepsilon^{2}}\right]$$ $$=\frac{(2\varepsilon^{3} + \varepsilon)e^{-1/2\varepsilon^{2}} - (2\varepsilon^{3} - \varepsilon)e^{1/2\varepsilon^{2}}}{\sqrt{\pi}}$$ Hence, $$L(\varepsilon) = \frac{(2\varepsilon^{2} + 1)e^{-1/\varepsilon^{2}} - (2\varepsilon^{2} - 1)}{2\sqrt{\pi}}$$ Therefore, $$\lim_{\varepsilon \to 0}L(\varepsilon) = \frac{1}{2\sqrt{\pi}}$$

Answer 2

$$I=\int\varepsilon^{-1}\Bigg(J_{\frac{3}{2}}\left(\frac{r}{\varepsilon}\right)\Bigg)^{2}\exp{(-r^{2})}\,r\, dr$$

Using, as @GEdgar commented $$J_{\frac{3}{2}}(x)=\sqrt{\frac{2}{\pi x}} \left(\frac{\sin (x)}{x}-\cos (x)\right)$$

$$I=\frac 2 \pi \varepsilon \int e^{-\varepsilon ^2 x^2} \left(\frac{\sin (x)}{x}-\cos (x)\right)^2\,dx$$ Expanding, assuming that $ \varepsilon>0$ and using Euler representation of the sine and cosine functions $$4\pi\,I=2 \sqrt{\pi } \left(1-2 \epsilon ^2\right) \text{erf}(x \epsilon )-\frac{8 \epsilon e^{-x^2 \epsilon ^2} \sin ^2(x)}{x}+$$ $$i \sqrt{\pi } e^{-\frac{1}{\epsilon ^2}} \left(2 \epsilon ^2+1\right) \left(\text{erfi}\left(\frac{1}{\epsilon }-i x \epsilon \right)-\text{erfi}\left(\frac{1}{\epsilon }+i x \epsilon \right)\right)$$

Using the bounds

$$K=\int_0^\infty\varepsilon^{-1}\Bigg(J_{\frac{3}{2}}\left(\frac{r}{\varepsilon}\right)\Bigg)^{2}\exp{(-r^{2})}\,r\, dr$$ $$K=\frac{1+2 \epsilon ^2+e^{\frac{1}{\epsilon ^2}} \left(1-2 \epsilon ^2\right)}{2 \sqrt{\pi }}$$ and then the limit.