Mind explosion!

$\int_{0}^{\infty}\frac{1}{1+t^2}dt=lim_{b\rightarrow{\infty}}\int_{0}^{b}\frac{1}{1+t^2}dt$ $=lim_{b\rightarrow{\infty}}\left [ arctan(b)-arctan(0)\right]=\frac{\pi}{2}$ It follows that the expression given above is equivalent to $\sqrt{2}\sqrt{\frac{\pi}{2}}=\sqrt{\pi}\Rightarrow\;a^{log_{\pi}2}=\pi^{log_{\pi}2}=\boxed{2}$