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

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.)

• Are you certain this has an answer? Mathematica can't find an explicit formula for this integral. Numerically, it is about $0.694654$. Feb 18, 2013 at 1:35
• The chance of an explicit formula is slim-to-none. Feb 18, 2013 at 2:12
• It is known that $\Gamma(x)>1$ for $0<x<1$, so the fact that the integral has value smaller than 1 is trivial. Feb 18, 2013 at 21:59
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: $$\log\left[\binom{e^n+\frac{t}{n}}{e^n} \cdot e^{-t}\right] = \sum\limits_{l=1}^\infty \frac{(-1)^{l+1}}{l} t^l \left(\frac{H_{e^n}^{(l)}}{n^l} - 1_{l=1}\right)$$ 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 $$\frac{1}{l} \left(H^{(l)}_{e^n} - 1_{l=1} n^l\right) \stackrel{\rightarrow}{n\rightarrow \infty} \left\{ \begin{array}{rr} \gamma & \mbox{if l=1} \\ \frac{\zeta(l)}{l} & \mbox{otherwise} \end{array} \right.$$ and since all the series above converge we can conclude that we have finished the calculation.