# 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: $$\int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}=\frac{2\pi\Gamma(n-1/3)}{\Gamma(2/3)3^{3/2}n!}$$

?

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Hint: interchange sum and integral. – Robert Israel Jun 3 '12 at 21:08

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, $$\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx=3\int_0^{\infty}\frac{1}{1+x^3}dx=\frac{2\pi}{\sqrt{3}}$$

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is it always ok to interchange sum and integral without knowing the nature of the integrant ? – Theorem Jun 3 '12 at 21:23
@Ananda Well, not always. But if the series and its derivative is uniformly convergent, then we're cool. – Pedro Tamaroff Jun 3 '12 at 21:56
Thank you very much for your clear answers! – Chon Jun 4 '12 at 20:00
@Chon, not at all. – Norbert Jun 4 '12 at 20:11
@AlexNelson There is no need in this because I consider my integrals as Lebesgue integrals. Hence I can apply dominated convergence theorem in the first line. The second equality holds if $x\neq 0$. But since I'm talking about Lebesge integral I can forget about this points (becase it is of measure zero) – Norbert Aug 13 '13 at 20:36

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, $I=\sqrt3\cdot\left[\arctan z\right]_{-1/\sqrt3}^{+\infty}=\displaystyle\sqrt3\left(\frac\pi2+\frac\pi6\right)=\frac{2\pi}{\sqrt3}$.

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