I got difficulty when I try to plot I(x) for $m=1$ and $t=0.2$. The questions is how to get the asymptotic form of the following integral?
$I(x,t)=\int_{0}^{\infty} \frac{f(y)}{2 \sqrt{\pi t}} \Biggl[ \mathrm{sin} \left( {\frac{ (x-y)^2}{4t}}+{\frac{\pi}{4}} \right)-\mathrm{sin} \left( {\frac{ (x+y)^2}{4t}}+{\frac{\pi}{4}} \right)\Biggr] \mathrm{d} y+\int_{0}^{\infty} f(y) \int_{0}^{t} \Biggl[ \mathrm{sin} \left( {\frac{ (x-y)^2}{8r}}+{\frac{\pi}{4}} \right)+\mathrm{sin} \left( {\frac{ (x+y)^2}{8r}}+{\frac{\pi}{4}} \right)\Biggr]~\Biggl[ \mathrm{e}^{ {-(x+y)^2}/{8r}} \frac{(x+y-4mr)}{4 \pi r \sqrt{r(t-r)}} +\dfrac{m^2\sqrt{2}}{\sqrt{\pi (t-r)}}\mathrm{e}^{ {m(x+y)+2 m^2r}} ~\text{Erfc} \biggl(\dfrac{x+y+4mr}{2\sqrt{2r}}\biggr) \Biggr]\mathrm{d} r~ \mathrm{d} y. $
where $f(y)=\mathrm{sin}^2(y)$, $\pi<y<2 \pi$. (Or we can consider $f(y)=\mathrm{exp}(-(y-4.68)^2/0.4)$, if the piecewise function might cause a problem.) I need the asymptotic form, because I would like to see the energy changing with the formula
$E(t)=\frac{1}{2} \int^{\infty}_{0} (I_{t}^2+I_{xx}^2)\mathrm{d}x.$
I checked the book of A. Erdélyi and N. Bleistein.However I could not apply to my problem.
Thanks!