How could I find the sum of the series
$$ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}$$
With: $$ \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}=\frac{2\pi\Gamma(n-1/3)}{\Gamma(2/3)3^{3/2}n!}$$
?
How could I find the sum of the series
$$ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}$$
With: $$ \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}=\frac{2\pi\Gamma(n-1/3)}{\Gamma(2/3)3^{3/2}n!}$$
?
Since integrals are taken over positive measurable functions we can interchange integration and summation
$$
\sum_{n=1}^{\infty} \int_0^{\infty} \frac{dx}{n(1+x^3)^n}=
\int_0^{\infty} \sum_{n=1}^{\infty}\left(\frac{1}{n(1+x^3)^n}\right)dx
$$
Consider the following Taylor expansion
$$
\log(1-q)=-\sum\limits_{n=1}^\infty\frac{q^n}{n}\qquad-1
Let us compute $I=\displaystyle\int_0^{+\infty}\log\left(\frac{1+x^3}{x^3}\right)\mathrm dx$.
The integration by parts with the functions $u(x)=\displaystyle\log\left(\frac{1+x^3}{x^3}\right)$ and $v'(x)=1$ yields $u'(x)=\displaystyle\frac{-3}{x(1+x^3)}$ and $v(x)=x$ hence $I=3\displaystyle\int_0^{+\infty}\frac{\mathrm dx}{1+x^3}$.
The change of variables $t=\displaystyle\frac1x$ yields $I=3\displaystyle\int_0^{+\infty}\frac{t\mathrm dt}{1+t^3}$.
Summing these yields $I=\displaystyle\frac32\int_0^{+\infty}\frac{(1+x)\mathrm dx}{1+x^3}=\frac32\int_0^{+\infty}\frac{\mathrm dx}{x^2-x+1}$.
The change of variables $2x-1=\sqrt3z$ yields $I=\displaystyle\frac32\cdot\frac2{\sqrt3}\int_{-1/\sqrt3}^{+\infty}\frac{\mathrm dz}{z^2+1}$, that is, $I=\sqrt3\cdot\left[\arctan z\right]_{-1/\sqrt3}^{+\infty}=\displaystyle\sqrt3\left(\frac\pi2+\frac\pi6\right)=\frac{2\pi}{\sqrt3}$.