Evaluating $\int_0^1 \frac{1}{\sqrt{\Gamma(x)}} dx$

Question

What is the value of the following integral?

$$\int_0^1 \frac{1}{\sqrt{\Gamma(x)}} \,dx$$

Here $\Gamma(x)$ is Euler's gamma function.

EDIT: Can we improve the upper bound strictly smaller than $1$?

(Thanks for the hint about $\Gamma(x)$ being $> 1$ in the domain of integration.)

Answer 1

First $$ \int_0^1 \frac{\mathrm{d}x}{\sqrt{\Gamma(x)}} = \int_0^1 \frac{\sqrt{x} \mathrm{d} x}{\sqrt{\Gamma\left(1+x\right)}} $$ We now prove that for all $0<s<1$ $$ \Gamma\left(1+s\right) = \int_0^1 t^{s} \exp(-t) \mathrm{d} t + \int_1^\infty t^{s} \exp(-t) \mathrm{d} t > \int_0^1 t \exp(-t) \mathrm{d} t + \int_1^\infty \exp(-t) \mathrm{d} t = 1 - \mathrm{e}^{-1} $$ Thus $$ \frac{1}{\sqrt{\Gamma(1+x)}} < \frac{1}{\sqrt{1-\mathrm{e}^{-1}}} $$ giving $$ \int_0^1 \frac{\mathrm{d}x}{\sqrt{\Gamma(x)}} < \frac{1}{\sqrt{1-\mathrm{e}^{-1}}} \int_0^1 \sqrt{x} \mathrm{d} x = \frac{2}{3} \frac{1}{\sqrt{1-\mathrm{e}^{-1}}} \approx 0.838511 $$

Answer 2

You can get the bound $\sim 0.782..$ by using the series expansion: $$\frac{1}{\sqrt{\Gamma(x)}} < \sqrt{x} + \frac{\gamma}{2} x^{3/2}$$ And integrating: $$\int_0^1\frac{1}{\sqrt{\Gamma(x)}}dx < 2/3+\gamma/5\sim 0.782$$ Of course, using more even number of terms will lead to lower bounds. In fact, the first four terms give a result accurate to three digits.