On rational bounds for the gamma function

AbstractIn the article, we prove that the double inequality x2+p0x+p0<Γ(x+1)<x2+9/5x+9/5$$ \frac{x^{2}+p_{0}}{x+p_{0}}< \Gamma(x+1)< \frac{x^{2}+9/5}{x+9/5} $$ holds for all x∈(0,1)$x\in(0, 1)$, we present the best possible constants λ and μ such that λ(x2+9/5)x+9/5≤Γ(x+1)≤μ(x2+p0)x+p0$$ \frac{\lambda(x^{2}+9/5)}{x+9/5}\leq\Gamma(x+1)\leq\frac{\mu (x^{2}+p_{0})}{x+p_{0}} $$ for all x∈(0,1)$x\in(0, 1)$, and we find the value of x∗$x^{\ast}$ in the interval (0,1)$(0, 1)$ such that Γ(x+1)>(x2+1/γ)/(x+1/γ)$\Gamma(x+1)>(x^{2}+1/\gamma)/(x+1/\gamma)$ for x∈(0,x∗)$x\in(0, x^{\ast})$ and Γ(x+1)<(x2+1/γ)/(x+1/γ)$\Gamma(x+1)<(x^{2}+1/\gamma)/(x+1/\gamma )$ for x∈(x∗,1)$x\in(x^{\ast}, 1)$, where Γ(x)$\Gamma(x)$ is the classical gamma function, γ=limn→∞(∑k=1n1/k−logn)=0.577…$\gamma=\lim_{n\rightarrow\infty}(\sum_{k=1}^{n}1/k-\log n)=0.577\ldots$ is Euler-Mascheroni constant and p0=γ/(1−γ)=1.365…$p_{0}=\gamma/(1-\gamma )=1.365\ldots$ .