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

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 $$ \int_0^1 \frac{\mathrm{d}x}{\sqrt{\Gamma(x)}} < \frac{1}{\sqrt{1-\mathrm{e}^{-1}}} \int_0^1 \sqrt{x} \mathrm{d} x = \frac{2}{3} \frac{1}{\sqrt{1-\mathrm{e}^{-1}}} \approx 0.838511 $$

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.

Let us compute a simpler integral first. We replace the square root in the denominator by a first power. We have : \begin{eqnarray} I=\int\limits_0^1 \frac{1}{\Gamma(x)} dx = \lim_{n\rightarrow \infty} \int\limits_0^1 \frac{x^{(n+1)}}{n! n^x} d x = \lim_{n\rightarrow \infty} \frac{(-1)^{n+1}}{n!} \int\limits_0^1 \frac{(-x)_{(n+1)}}{ n^x} d x =\\ \lim_{n\rightarrow \infty} \frac{(-1)^{n+1}}{n!} \sum\limits_{k=0}^{n+1} \frac{s(n+1,k) (-1)^k}{(\log(n))^{k+1}} \cdot (k! - \Gamma(k+1,\log(n))) =\\ \lim_{n\rightarrow \infty} \frac{1}{n! \log(n)^2} \sum\limits_{k=1}^{n+1} \frac{1}{(k-1)!} \left. \frac{d^{k-1} x^{(n)}}{d x^{k-1}} \frac{1}{(\log(n))^{k-1}} \cdot (k! - \Gamma(k+1,\log(n))) \right|_{x=1} =\\ \lim_{n\rightarrow \infty} \frac{1}{n! \log(n)^2} \int\limits_0^{\log(n)} t \left(1+\frac{t}{\log(n)}\right)^{(n)} e^{-t} dt=\\ \lim_{n\rightarrow \infty} \int\limits_0^{\log(n)} t \binom{n+\frac{t}{\log(n)}}{n}\frac{1}{\log(n)^2} e^{-t} dt = \lim_{n\rightarrow \infty} \frac{\int\limits_0^n t \binom{e^n + \frac{t}{n}}{e^n} e^{-t} dt}{n^2} = 0.54123573432867053015.. \end{eqnarray} Here $s(n,k)$ are the Stirling numbers of the first kind. Now, it is easy to see that the limit over $n$ exists. Firstly note that if $n>1$ we have: \begin{equation} \log\left[\binom{e^n+\frac{t}{n}}{e^n} \cdot e^{-t}\right] = \sum\limits_{l=1}^\infty \frac{(-1)^{l+1}}{l} t^l \left(\frac{H_{e^n}^{(l)}}{n^l} - 1_{l=1}\right) \end{equation} where $H_n^{(l)}$ are the generalized harmonic numbers. Exponentiating both sides of the above equation and inserting the result into the integrand in the last expression above and doing the elementary integrals over powers of t we get: \begin{eqnarray} I= \frac{1}{2} &+& \frac{1}{1!} \sum\limits_{l=1}^\infty (-1)^{l+1} \cdot \frac{1}{l+2} \cdot \frac{1}{l} (H^{(l)}_{e^n} - 1_{l=1} n^l) + \\ &+&\frac{1}{2!} \sum\limits_{l=2}^\infty (-1)^{l+2} \cdot \frac{1}{l+2} \cdot \sum\limits_{\stackrel{l_1+l_2=l}{l_1,l_2 \ge 1}} \prod_{\xi=1}^2 \frac{1}{l_\xi}(H^{(l_\xi)}_{e^n} - 1_{l_\xi=1} n^{l_\xi}) + \\ &+&\frac{1}{3!} \sum\limits_{l=3}^\infty (-1)^{l+3} \cdot \frac{1}{l+2} \cdot \sum\limits_{\stackrel{l_1+l_2+l_3=l}{l_1,l_2,l_3 \ge 1}} \prod_{\xi=1}^3 \frac{1}{l_\xi}(H^{(l_\xi)}_{e^n} - 1_{l_\xi=1} n^{l_\xi}) + \\ &+&\cdots \end{eqnarray} Now, since \begin{equation} \frac{1}{l} \left(H^{(l)}_{e^n} - 1_{l=1} n^l\right) \stackrel{\rightarrow}{n\rightarrow \infty} \left\{ \begin{array}{rr} \gamma & \mbox{if $l=1$} \\ \frac{\zeta(l)}{l} & \mbox{otherwise} \end{array} \right. \end{equation} and since all the series above converge we can conclude that we have finished the calculation.

Now the next step would be to consider negative integer powers of the Gamma function in the integrand.