Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)


share|cite|improve this question
Hint: interchange sum and integral. – Robert Israel Jun 3 '12 at 21:08
up vote 4 down vote accepted

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.