Computing a double integral with applications to prime numbers

Question

I was reading the preprint [1] which contains on p. 7 the following formula (for $4<s\le6$): $$ f_1(s)=\frac{2e^\gamma}{s}\left\{\log(s-1)+\int_4^s\int_3^t\frac{\log(u-2)}{u-1}du\,dt \right\} $$

and which concludes, on page 16, with $$ c=\frac{2e^\gamma}{3f_1(6)}=1.18751\ldots. $$

But evaluating the function numerically in PARI/GP

f1(s)=2*exp(Euler)/s*(log(s-1)+intnum(t=4,s,intnum(u=3,t,log(u-2)/(u-1))))
\p 200
2*exp(Euler)/3/f1(6)

gives me c = 0.8239599331..., a very different result. I tried it in Maxima

assume(t >= 4);
float(integrate(integrate(log(u-2)/(u-1),u,3,t),t,4,6));

and it gives 0.81786458079115, which corresponds to c = 1.4518071082, not very close to either result. (bfloat doesn't seem to work here -- it has trouble with the polylogarithms it seems.)

How can I reliable compute this double integral?

[1] Pin-Hung Kao, Almost-Prime Polynomials with Prime Arguments, arXiv:1606.03505 [math.NT]

Answer 1

It seems the authors might have used (I admit that the numbers do not match completely) $$ c=\frac{2e^\gamma}{3F_1(6)}\approx1.18787 $$ (where $F_1$ is also defined on page 7) instead of $$ c=\frac{2e^\gamma}{3f_1(6)}\approx 0.82396 $$ I don't have time or energy to understand which one is the correct one to put in the end. Mathematica calculates $$ F_1(s)=\frac{1}{s}\bigl(2e^\gamma(1+\frac{\pi^2}{12}+\log(s-2)\log(s-1)+\text{Li}_2(2-s)\bigr) $$ and $$ \begin{aligned} f_1(s)&=\frac{2}s e^\gamma\Bigl(s-4+\frac{\pi^2}{12}s+\frac{5}{2}(\log 2)^2 +2\log 2-3\log2\log 3+\log(s-1)\\ &\quad +\log(s-2)(2-s+(s-1)\log(s-1))\\ &\quad+\text{Li}_2(-1/2)+\text{Li}_2(1/4)+(s-1)\text{Li}_2(2-s)\Bigr) \end{aligned} $$

This might be a mistake, and I suggest that you contact the author and ask.

Answer 2

Hint:

Though it's tedious but the double integral is doable by subbing $u=1+x$ then repeating integration by parts several times. You might use the following relation \begin{equation} \operatorname{Li}_2(z)=\int_z^0\frac{\log(1-x)}{x}\ dx \end{equation}