Get ready Part 2

$\begin{aligned} I & = \int_0^\infty t^{-\frac 12} e^{-a(t+t^{-1})} dt & \small \color{#3D99F6} \text{Let }x^2 = t \implies 2x \ dx = dt \\ & = \int_0^\infty 2 e^{-a\left(x^2+\frac 1{x^2}\right)} dx \\ & = 2 \int_0^\infty e^{-a\left(x-\frac 1x\right)^2-2a} dx \\ & = \frac 2{e^{2a}} \int_0^\infty e^{-a\left(x-\frac 1x\right)^2} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f \left(\frac 1x\right)}{x^2} \ dx \\ & = \frac 1{e^{2a}} \int_0^\infty \left(1+\frac 1{x^2}\right) e^{-a\left(x-\frac 1x\right)^2} dx & \small \color{#3D99F6} \text{Let }u = x - \frac 1x \implies du = \left(1 + \frac 1{x^2} \right) dx \\ & = \frac 1{e^{2a}} \int_{-\infty}^\infty e^{-au^2} du & \small \color{#3D99F6} \text{Since the integrand is even,} \\ & = \frac 2{e^{2a}} \int_0^\infty e^{-au^2} du & \small \color{#3D99F6} \text{Error function erf }(z) = \frac 2{\sqrt \pi} \int_0^z e^{-t^2} dt \\ & = \frac {\color{#3D99F6}\text{erf }(\infty)}{e^{2a}}\sqrt{\frac \pi a} = \frac {\color{#3D99F6}1}{e^{2a}}\sqrt{\frac \pi a} & \small \color{#3D99F6} \text{erf }(\infty) = 1 \text{ and put }a = 1985 \\ & = 2.82228 \times 10^{-1726} \end{aligned}$

Therefore, $10^{1726}I \approx \boxed{2.8223}$ .