An integral and series to prove that $\log(5)>\frac{8}{5}$

Question

A relationship between $\log(5) \approx 1.6094$ and $\dfrac{3}{2}=1.5$ can be justified by the harmonic approximation $$\log(5) \approx H_2=1+\frac{1}{2}=\frac{3}{2}$$

that can be obtained by regrouping Lehmer's logarithm

$$\log(5) = \sum_{k=0}^\infty \left(\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}-\frac{4}{5k+5}\right)$$

symmetrically around the negative term

$$\log(5)=\frac{3}{2}+\sum_{k=1}^\infty \left( \frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2} \right)$$

The corresponding integral is $$\log(5)-\frac{3}{2}=\int_0^1 \frac{x^2(1-x)(1+3x+x^2)}{1+x+x^2+x^3+x^4}\:dx$$

(answer https://math.stackexchange.com/a/1656356/134791 by Olivier Oloa)

This is a direct proof that $\log(5)>\dfrac{3}{2}$ because the integrand is non-negative in $(0,1)$.

However, $\dfrac{8}{5}=1.6$ would be a closer approximation using small numbers, so the natural question is:

What are Dalzell-type integral and series for $\log(5)-\dfrac{8}{5}$?

Answer 1

Here is an initial solution.

Take two truncations of the series $$\sum_{k=1}^\infty \left(\frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2}\right)$$

that evaluate to $\log(5)-r_1$ and $\log(5)-r_2$ such that

$$r_1<\frac{8}{5}<r_2$$

and build a linear combination of the truncations

$$\alpha r_1+(1-\alpha)r_2=\frac{8}{5}$$

with $0<\alpha<1$.

Because of linearity, the resulting series will have positive terms by construction (when combined properly) and the corresponding integral has positive integrand.

$$\log(5)=\frac{8}{5}+\frac{1}{287} \int_0^1 \frac{x^7(1-x)(1+3x+x^2)(89+198x^{5})}{1+x+x^2+x^3+x^4} dx$$

An acceptable solution should have smaller degree in the numerator.

Answer 2

I'm not quite sure what you're looking for, but I get

$$ \log(5) = \frac{8}{5} -\frac{1}{5}\int_0^1 {\frac {x \left( 12\,{x}^{5}+10\,{x}^{4}+15\,{x}^{3}-5\,{x}^{2}- 7 \right) }{{x}^{4}+{x}^{3}+{x}^{2}+x+1}} \; dx $$