How to find this nice limit: $I=\lim_{t\to0^{+}}\lim_{x\to+\infty}f(x,t)$ Find the value:

$$I=\lim_{t\to0^{+}}\lim_{x\to+\infty}\dfrac{\displaystyle\int_{0}^{\sqrt{t}}dx\int_{x^2}^{t}\sin{y^2}dy}{\left[\left(\dfrac{2}{\pi}\arctan{\dfrac{x}{t^2}}\right)^x-1\right]\arctan{t^{\frac{3}{2}}}}$$

I spent some hours doing it, but have failed.
Thank you!
 A: One observes
$$lim_{x\to\infty}\left(\frac{2}{\pi}\arctan\frac{x}{t^2}\right)^x = e^{lim_{x\to\infty}x\left(\ln\arctan\frac{x}{t^2}-\ln\frac{\pi}{2}\right)}.$$
In computing that limit I find:
$$lim_{x\to\infty}x\left(\ln\arctan\frac{x}{t^2}-\ln\frac{\pi}{2}\right) = lim_{x\to\infty}\frac{\ln\arctan\frac{x}{t^2}-\ln\frac{\pi}{2}}{\frac{1}{x}} =\\
= lim_{x\to\infty}\frac{\frac{1}{\arctan\frac{x}{t^2}}\frac{1}{1+\frac{x^2}{t^4}}\frac{1}{t^2}}{-\frac{1}{x^2}} =
-lim_{x\to\infty}\frac{x^2t^2}{\arctan\frac{x}{t^2}(t^4+x^2)} =
-\frac{2t^2}{\pi}$$
We can therefore find the original limit as
$$\lim_{t\to 0^+}\frac{\sqrt{t}\int_\infty^t\sin(y^2)dy}{\left(e^{-\frac{2t^2}{\pi}}-1\right)\arctan(t^{3/2})} =\\
= \lim_{t\to 0^+}\frac{\frac{1}{2\sqrt{t}}\int_\infty^t\sin(y^2)dy + \sqrt{t}\sin(t^2)}{-\frac{4t}{\pi}e^{-\frac{2t^2}{\pi}}\arctan(t^{3/2}) + \left(e^{-\frac{2t^2}{\pi}}-1\right)\frac{1}{1+t^3}\frac{3}{2}\sqrt{t}} =\\
= \lim_{t\to 0^+}\frac{\frac{1}{2}\int_\infty^t\sin(y^2)dy + t\sin(t^2)}{-\frac{4t^{3/2}}{\pi}e^{-\frac{2t^2}{\pi}}\arctan(t^{3/2}) + \left(e^{-\frac{2t^2}{\pi}}-1\right)\frac{1}{1+t^3}\frac{3}{2}t}$$
and every term but the first one in the numerator tends to zero, so unless I've made some mistake (at this point it's more than likely--so let me know) the limit diverges.
