Approximation of ln(x) in the vicinity of x=0

Solution 1:

Since $$\lim_{x\to 0} \, x\log(x)=\lim_{x\to 0} \, (x^x-1)$$ you could use,as an approximation, $$x^2 \log(x) \sim x(x^x-1)$$

To give an idea, consider the norm $$\Phi=\int_{0}^{\frac 1{10}}\Big[x^2 \log(x) -x(x^x-1)\Big]^2\,dx=1.14\times 10^{-7}$$

Edit

Graphically, for $0 \leq x \leq 1$ $$x^2 \log(x) \sim (x-1) x \sqrt x$$ So, consider the norm $$\Phi(a)=\int_0^a \Big[x^2 \log(x) -k_a (x-1) x \sqrt x\Big]^2\,dx$$ which is explicit. Now, solve, for $k$, $\Phi'(a)=0$. This gives $$k_a=\frac{40 \sqrt{a} (-162 a+891 a \log (a)-1089 \log (a)+242)}{3267 \left(10 a^2-24a+15\right)}$$

If you want the best approximation for $a=\frac 1{100}$, $$k_{\frac 1{100}}=\frac{40 (24038+216018 \log (10))}{48224187}\sim \frac{298}{689}$$ and $$\Phi \left(\frac{1}{100}\right)=2.38 \times 10^{-12}$$

Update

We can do a bit better using the previous similarity $$x^2 \log(x) \sim (x-1) x \sqrt x$$ and use as an approximation $$x^2 \log(x) \sim \sum_{k=0}^p A_k^{(p)} x^{\frac{k+3}2}$$

This would give $$ A_0^{(1)}=\frac{8}{81} \sqrt{a} \,(9 \log (a)-2)\implies \Phi \left(\frac{1}{100}\right)=2.27 \times 10^{-12}$$ $$ A_0^{(2)}=-\frac{8 \sqrt{a}}{5} \qquad A_1^{(2)}=\log (a)+\frac{71}{45}\implies \Phi \left(\frac{1}{100}\right)=9.88 \times 10^{-15}$$