Who's up to the challenge? 52

Let us denote the integral by $\displaystyle \mathfrak{J}$

$\displaystyle \mathfrak{J}=\frac{1}{2}\int_{0}^{1} x\Gamma(\frac{x}{2})\Gamma(1-\frac{x}{2})dx$ , Since $0<\frac{x}{2}<1$ when $x\in(0,1)$ we have By the $\text{Reflection formulae}$ ,

$\displaystyle \mathfrak{J} = \frac{1}{2}\int_{0}^{1} \frac{\pi x}{2\sin(\frac{\pi x}{2})}dx$

Substitute $\displaystyle\frac{\pi x}{2}=t$ and we have $\displaystyle \mathfrak{J} = \frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \frac{t}{sint}dt$

We have the identity $\displaystyle G=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{t}{sint}dt$ which turns the integral into,

$\displaystyle \mathfrak{J} = \boxed{\frac{4G}{\pi}}$

we know by the numeric integration technique that the above integral can be evaluated as: 7/6 (by simp son's method of numeric integration) 7/6=(0.2915a)/b On multiplying normally we get the value of a/b=4 Thus we can conclude that a=4 and b=1

Hence a+b=4+1=5