Integrating over $(0,1)$, and letting $t=\dfrac1x$ , followed by $u=t-1$, we are left with evaluating
$\displaystyle\int_0^\infty\frac{e^{-au^2}}{u+1}~du~=~\frac{\pi\cdot\text{erfi}\Big(\sqrt a\Big)-\text{Ei}(a)}{2~e^a}~$ for $a=50$, where erfi stands for the imaginary
Integrating over $(0,1)$, and letting $t=\dfrac1x$ , followed by $u=t-1$, we are left with evaluating
$\displaystyle\int_0^\infty\frac{e^{-au^2}}{u+1}~du~=~\frac{\pi\cdot\text{erfi}\Big(\sqrt a\Big)-\text{Ei}(a)}{2~e^a}~$ for $a=50$, where erfi stands for the imaginary
error function, and Ei represents the exponential integral. $~$ See Liouville's theorem and the
Risch algorithm for more information.