Let $\Gamma(a,x)$ denote the incomplete Gamma function. Consider the function
$$
f(x)=x^{-x}\Gamma(x,x)-\Gamma(0,x)=x^{-x}\Gamma(x,x)-E_1(x),
$$
where $E_1(x)$ denotes the exponential integral. I would like to estimate the rate of convergence of this function to $0$ as $x\to0+$.
From this paper for $\Gamma(0,x)$ we have $\Gamma(0,x)=-\ln(x)-\gamma+O(x)$, as $x\to0+$. Here and onward $\gamma=0.577…$ is the Euler constant.
Moreover, from the same paper for $\Gamma(x,x)$ with $|x|<1$ we have
$$
\Gamma(x,x)<e^{-x}\dfrac{(x+b_x)^x-x^x}{xb_x},
$$
where $b_x=(\Gamma(x+1))^{1/(x-1)}$, so $b_x\to 1$ as $x\to 0+$.
Thus we have
$$
f(x)\le e^{-x}\dfrac{(1+b_x/x)^{x}-1}{x b_x}+\ln(x)+\gamma+O(x),
$$
and this is where I got stuck: the upperbound seems to be too crude.
Would appreciate any help.
Best Answer
Using the relation between $\gamma(s,z)$ and Kummer's confluent hypergeometric function, we have $$\Gamma(x,x)=\Gamma(x)-x^{x-1}\, e^{-x}\,\sum_{n=0}^\infty \frac {x^n}{(x+1)_n }$$ Then $$f(x)=x^{-x}\Gamma(x)-\Bigg[\frac {e^{-x}} x\,\sum_{n=0}^\infty \frac {x^n}{(x+1)_n }+\Gamma(0,x)\Bigg]$$ Expanded as series around $x=0$ $$\sum_{n=0}^\infty \frac {x^n}{(x+1)_n }=1+x-\frac{x^2}{2}+\frac{5 x^3}{12}-\frac{7 x^4}{18}+O\left(x^5\right)$$ $$\Gamma(0,x)=-\log (x)-\gamma +x-\frac{x^2}{4}+\frac{x^3}{18}-\frac{x^4}{96}+O\left(x^5\right)$$ $$\frac {e^{-x}} x\,\sum_{n=0}^\infty \frac {x^n}{(x+1)_n }+\Gamma(0,x)=\frac{1}{x}-\log (x)-\gamma +x^2-\frac{9 x^3}{8}+\frac{467 x^4}{432}+O\left(x^5\right)$$
Combining all pieces, close to $x=0$, $$f(x)=\sum_{n=1}^\infty P_n\,x^n$$ the $P_n$ being polynomials of degree $(n+1)$ in $L=\log(x)$
$$\left( \begin{array}{cc} n & P_n \\ 1 & \frac{6 \gamma ^2+\pi ^2}{12} +\gamma L+\frac{1}{2} L^2\\ 2 & -\frac{12+2 \gamma ^3+\gamma \pi ^2-2 \psi ^{(2)}(1)}{12} -\frac{6 \gamma ^2+\pi ^2}{12} L-\frac{\gamma }{2}L^2-\frac{1}{6} L^3 \\ \end{array} \right)$$
This is not bad; computing the norm $$\Phi_p=\int_0^{\frac{1}{5}}\Bigg[f(x)-\sum_{n=1}^p P_n\,x^n\Bigg]^2\,dx$$ $$\left( \begin{array}{cc} p & \Phi_p \\ 1 & 1.27921\times 10^{-5} \\ 2 & 5.98123\times 10^{-6} \\ 3 & 3.19626\times 10^{-8} \\ 4 & 2.57442\times 10^{-9} \\ 5 & 5.86460\times 10^{-11} \\ \end{array} \right)$$