How to find the Limit of this Exponential function I am trying to take the following limit
$$
\lim_{t\to\infty} \frac{1}{t}\exp \left(\frac{1}{2} + \frac{1}{2}\int_0^\infty \operatorname{erfc} \left(\frac{\sqrt{\beta}}{2^{1/4} t^\beta} x + \frac{t^\beta}{2^{3/4} \sqrt{\beta}}\right) \,\mathrm{d}x\right)$$
where 
$\operatorname{erfc} \left(z\right)$ is the complementary error function given by
\begin{equation}
\operatorname{erfc}(z) = \frac{2}{\sqrt{\pi}}\int_z^{\infty} \mathrm{e}^{-t^2} \,dt 
\end{equation}
and $0 \le \beta \le 1$.
I am not quite sure how to proceed. The integral in the numerator should make the numerator tend to infinity, if I am not mistaken, which would suggest LHospital's rule can be applied? Is that a, correct assumption ?
Thank you.
 A: Assuming $A,B>0$,
$$\begin{eqnarray*}\int_{0}^{+\infty}\text{erfc}(Ax+B)\,dx &=& \frac{2}{A\sqrt{\pi}}\int_{0}^{+\infty}\int_{x+B}^{+\infty}e^{-t^2}\,dt\,dx\\[0.2cm]&=&\frac{1}{A\sqrt{\pi}}\left(e^{-B^2}-B\sqrt{\pi}\,\text{erfc}(B)\right) \end{eqnarray*}$$
hence $\text{erfc}(B)\leq \frac{1}{B\sqrt{\pi}}\,e^{-B^2}$ and, by the Cauchy-Schwarz inequality:
$$ \int_{0}^{+\infty}\text{erfc}(Ax+B)\,dx \leq \sqrt{\int_{0}^{+\infty}\frac{dx}{\pi(Ax+B)^2}\int_{0}^{+\infty}e^{-2(Ax+B)^2}\,dx} $$
so:
$$ \int_{0}^{+\infty}\text{erfc}(Ax+B)\,dx \leq \sqrt{\frac{1}{AB\pi}\cdot\frac{\sqrt{\pi}}{2A}\,\text{erfc}(B)}\leq \frac{1}{AB\sqrt{2\pi}} e^{-B^2/2}.$$
In our case $A=\frac{\sqrt{\beta}}{2^{1/4}t^\beta}$ and $B=\frac{t^\beta}{2^{3/4}\sqrt{\beta}}$, so $AB=\frac{1}{2}$.
The previous inequality hence gives that the integral in our limit converges to zero very fast as $t\to +\infty$, like $\exp(-t^{2\beta})$, so the original limit is trivially zero (unless $\beta=0$, but in such a case I wonder how $\frac{1}{\sqrt{\beta}}$ is defined).
A: Put $$\frac{\sqrt{\beta}}{2^{1/4}t^{\beta}}x+\frac{t^{\beta}}{2^{3/4}\sqrt{\beta}}=u,
 $$ then we have $$F\left(t,\beta\right)=\int_{0}^{\infty}\textrm{erfc}\left(\frac{\sqrt{\beta}}{2^{1/4}t^{\beta}}x+\frac{t^{\beta}}{2^{3/4}\sqrt{\beta}}\right)dx=\frac{\sqrt{\beta}}{2^{1/4}t^{\beta}}\int_{t^{\beta}/(2^{3/4}\sqrt{\beta})}^{\infty}\textrm{erfc}\left(u\right)du
 $$ $$=\frac{\sqrt{\beta}}{2^{1/4}t^{\beta}}\left(-\frac{t^{\beta}}{2^{3/4}\sqrt{\beta}}\textrm{erfc}\left(\frac{t^{\beta}}{2^{3/4}\sqrt{\beta}}\right)+\frac{\exp\left(-\left(\frac{t^{\beta}}{2^{3/4}\sqrt{\beta}}\right)^{2}\right)}{\sqrt{\pi}}\right)
 $$ because $$\int\textrm{erfc}\left(x\right)dx=x\textrm{erfc}\left(x\right)-\frac{e^{-x^{2}}}{\sqrt{\pi}}.
 $$ Now since $$\lim_{x\rightarrow\infty}\textrm{erfc}\left(x\right)=0
 $$ we have, if $0<\beta\leq1
 $ $$\lim_{t\rightarrow\infty}F\left(t,\beta\right)=0
 $$ so $$\lim_{t\rightarrow\infty}\frac{1}{t}\exp\left(\frac{1}{2}+\frac{1}{2}F\left(t,\beta\right)\right)=0.
 $$ If $\beta\rightarrow0
 $ we note that $$ \lim_{\beta\rightarrow0^{+}}F\left(t,\beta\right)=0
 $$ so $$\lim_{t\rightarrow\infty}\frac{1}{t}\exp\left(\frac{1}{2}+\frac{1}{2}\lim_{\beta\rightarrow0^{+}}F\left(t,\beta\right)\right)=0.
 $$
