2
$\begingroup$

I'm trying to prove a particular series is convergent, and I would like to use the Cauchy integral test for fun, even though it's not the most convenient. I need to evaluate,

$$\int_0^\infty dn \, \frac{x^n}{(3n+1)(3n+2)}$$

for $x\in\mathbb{R}$. Using Mathematica I have found the integral to be,

$$\frac{1}{3x^{2/3}} \left[ x^{1/3}\Gamma(0,-\log(x)/x)-\Gamma(0,-2\log(x)/x)\right]$$

providing $x<1$. So I just need to express the integral in terms of incomplete gamma functions, but I haven't found a substitution. Can someone offer a hint (and not a complete solution)?

$\endgroup$
2
  • $\begingroup$ But of course you know that existence of the integral is enough, and that is clear for $|x|\le 1$. And we have divergence otherwise. $\endgroup$ Commented Jan 29, 2015 at 8:16
  • $\begingroup$ @AndréNicolas Yes, I know, but I'd like to actually compute the integral. The only reason I mentioned the Cauchy test was to motivate the integral. $\endgroup$
    – JPhy
    Commented Jan 29, 2015 at 8:51

1 Answer 1

2
$\begingroup$

Hint. Recall that the incomplete gamma function may be defined as $$ \Gamma(s,a)=\int_a^{\infty} t^{s-1}e^{-t}dt,\quad a>0, s \in \mathbb{R}. \tag1 $$ Observe that, for $0<x<1$, we have $$ \begin{align} \int_0^\infty dn \,\frac{x^n}{3n+1}&=\int_0^\infty (3n+1)^{-1}x^n \:dn\\\\ &=x^{-\frac13}\int_{0}^\infty (3n+1)^{-1}x^{\frac13 (3n+1)} \:dn\\\\ &=\frac{1}{3}x^{-\frac13}\int_{1}^\infty v^{-1}x^{\frac13 v} \:dv,\quad (v=3n+1),\\\\ &=\frac{1}{3}x^{-\frac13}\int_{-\frac13 \ln x}^\infty t^{-1}e^{-t} \:dt,\quad t= \left(-\frac13 \ln x\right)\times v,\\\\ &=\frac{1}{3 x^{1/3}}\Gamma\left(0,-\frac13 \ln x\right), \end{align} $$ similarly $$ \begin{align} \int_0^\infty dn \,\frac{x^n}{3n+2}&=\frac{1}{3 x^{2/3}}\Gamma\left(0,-\frac23 \ln x\right). \end{align}$$ Then conclude by observing that, by partial fraction decomposition, you have $$\int_0^\infty dn \,\frac{x^n}{(3n+1)(3n+2)}=\int_0^\infty dn \,\frac{x^n}{3n+1}-\int_0^\infty dn \,\frac{x^n}{3n+2},$$ getting the announced result.

$\endgroup$
3
  • $\begingroup$ yo've been faster then me (+1) $\endgroup$
    – tired
    Commented Jan 29, 2015 at 11:27
  • $\begingroup$ @tired Thank you very much for your upvote. $\endgroup$ Commented Jan 29, 2015 at 11:28
  • $\begingroup$ Brilliant, thank you. +$\infty$. $\endgroup$
    – JPhy
    Commented Jan 29, 2015 at 11:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .