Maybe Frullani can help

Notice the integral can be written in the form $\int_0^\infty \frac{\frac1{\Gamma(x+\frac12)}-\frac1{\Gamma(ax+\frac12)}}{x}dx$ Then using Frullani's integral, the integral is equal to $\left( \frac1{\Gamma(\frac12)} - \frac1{\Gamma(\infty)} \right) \ln \left(\frac a1\right) = \frac1{\sqrt\pi} \ln (a)$ Hence, for $I(a)=1$ we require $\frac1{\sqrt\pi}\ln(a)=1$ , or $\boxed{a=e^{\sqrt\pi}\approx5.88527}$