Summing the series $ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^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: $$ \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}=\frac{2\pi\Gamma(n-1/3)}{\Gamma(2/3)3^{3/2}n!}$$
(Previous post)
?
 A: 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
$$
\log(1-q)=-\sum\limits_{n=1}^\infty\frac{q^n}{n}\qquad-1<q<1
$$
With $q=(1+x^3)^{-1}$ we get
$$
\sum\limits_{n=1}^\infty\frac{1}{n(1+x^3)^n}=
-\log\left(1-\frac{1}{1+x^3}\right)=
\log\left(\frac{1+x^3}{x^3}\right)
$$
Hence
$$
\sum_{n=1}^{\infty} \int_0^{\infty} \frac{dx}{n(1+x^3)^n}=
\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx
$$
Let's proceed to calculation of the last integral. For the first we use integration by parts
$$
\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx=
x\log\left(1+\frac{1}{x^3}\right)\biggr|_0^\infty-
\int_0^{\infty}x\frac{d}{dx}\log\left(1+\frac{1}{x^3}\right)dx=
3\int_0^{\infty}\frac{1}{1+x^3}dx
$$
Now lets make substitution $u=x^{-1}$, then we get
$$
\int_0^{\infty}\frac{1}{1+x^3}dx=\int_0^{\infty}\frac{u}{1+u^3}dx
$$
Hence,
$$
\int_0^{\infty}\frac{1}{1+x^3}dx=
\frac{1}{2}\left(\int_0^{\infty}\frac{1}{1+x^3}dx+\int_0^{\infty}\frac{x}{1+x^3}dx\right)=
\frac{1}{2}\int_0^{\infty}\frac{1}{1-x+x^2}dx=
$$
$$
\frac{1}{2}\int_0^{\infty}\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}d x=
\frac{1}{2}\frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\Biggl|_0^\infty=\frac{2\pi}{3\sqrt{3}}
$$
Finally,
$$
\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx=3\int_0^{\infty}\frac{1}{1+x^3}dx=\frac{2\pi}{\sqrt{3}}
$$
A: 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, $I=\sqrt3\cdot\left[\arctan z\right]_{-1/\sqrt3}^{+\infty}=\displaystyle\sqrt3\left(\frac\pi2+\frac\pi6\right)=\frac{2\pi}{\sqrt3}$.
