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$Solution$ $by$ $Sergio$ $Esteban$

interchanging the sum and integral justified by fubini's theorem

$\Omega =\sum_{i=1}^\infty\sum_{k=2}^\infty\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn^2}{k^n(k^nm+k^mn)} \frac{1}{(i-k+2)^q}\int_{0}^1\frac{x^j+x^{j-1}+...+1}{x^{j+2}+x^{j+1}+...+1}dx$ = $=\sum_{k=2}^\infty\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn^2}{k^n(k^nm+k^mn)}\sum_{i=1}^\infty \frac{1}{(i)^q}\int_{0}^1\frac{1-x^{j+1}}{1-x^{j+3}}dx$

i) $\sum_{k=2}^\infty\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn^2}{k^n(k^nm+k^mn)}=\frac{\zeta(2)+2\zeta(3)+\zeta(4)}{2}$, This is $LIMIT-285$ that we can find on the official RMM website

ii) $\sum_{i=1}^\infty \frac{1}{(i)^q}=\zeta(q)$ by the riemman zeta function definition

iii) by notable ratios or "cocientes notables"

$I_{n}=\int_{0}^1\frac{x^j+x^{j-1}+...+1}{x^{j+2}+x^{j+1}+...+1}dx = \int_{0}^1\frac{1-x^{j+1}}{1-x^{j+3}}dx$, by a power series expansion of the integrand and then integrate term-by-term to arrive at an infinite sum:

the integrand is $\frac{1-x^{j+1}}{1-x^{j+3}}=(1-x^{j+1})=(1+x^{j+3}+x^{2(j+3)}+x^{3(j+3)}+...)$,then $I_{n}=\int_{0}^1(1-x^{j+1})\sum_{k=0}^\infty x^{k(j+3)}dx=\sum_{k=0}^\infty\int_{0}^1(x^{k(j+3)}-x^{k(j+3)+(j+1)})dx=$

$=\sum_{k=0}^\infty (\frac{1}{k(j+3)+1}-\frac{1}{k(j+3)+(j+1)+1})=\frac{1}{j+3} \sum_{k=1}^\infty (\frac{1}{k-\frac{j+2}{j+3}}-\frac{1}{k-\frac{1}{j+3}})=$

=$\frac{1}{j+3}${${\psi(1-\frac{1}{j+3})- \psi(1-\frac{j+2}{j+3})}$}=$\frac{1}{j+3}${${\psi(1-\frac{1}{j+3})- \psi(\frac{1}{j+3})}$}, by Euler´s reflection formula

$=\frac{\pi}{j+3} cot (\frac{\pi}{j+3})=\frac{\pi}{j+3} tan(\frac{\pi(j+1)}{j+3})$

by i),ii) and iii)

$\therefore \Omega= \frac{\pi}{2(j+3)} tan(\frac{\pi(j+1)}{j+3}) (\zeta(2)+2\zeta(3)+\zeta(4))\zeta(q)$

Thu, 28 Nov 2019 22:25 GMT