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I have to find an infinitite series expansion for the integral: $$\int \frac{x}{8+x^3} \, dx$$

First, I started by determining the Taylor series of the integrand $$\frac{x}{8+x^3}=\frac{x}{8} \cdot \frac{1}{1-(-(x/2)^3)} = \frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}$$

Then, I integrate $$\int \frac{x}{8+x^3} \, dx = -\frac{1}{8} \int x \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \, dx$$

But, I'm not sure how to continue.

Thank you for your help.

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  • $\begingroup$ @avid19 I know that $\int x^n \, dx = \frac{x^{n+1}}{n+1}$. But, how can I apply this to this product of $x \cdot \sum_{i=0}^{\infty} \left( - \frac{x}{2} \right)^{3i}$. $\endgroup$
    – hlapointe
    Nov 22, 2015 at 18:50
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    $\begingroup$ Just simplify the expression to $\sum \frac{(-1)^{3i}}{2^{3i}} x^{3i+1}$. Then you just have a sum of something that looks like $x^n$. Another hint, what is the integral of a sum? $\endgroup$
    – user223391
    Nov 22, 2015 at 18:52
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    $\begingroup$ You might need to be careful interchanging infinite sums and integrals because in general you can't. However in most case you'll deal with, it's okay and here it should come out okay. $\endgroup$
    – user223391
    Nov 22, 2015 at 18:58

2 Answers 2

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$$\frac{x}{8} \cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i}=-\frac{1}{4} (-\frac{x}{2})\cdot \sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i} \,=-\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,$$ $$\int -\frac{1}{4}\sum_{i=0}^{\infty} \left(-\frac{x}{2}\right)^{3i+1} \,dx=\frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{3i+2} \left(-\frac{x}{2}\right)^{3i+2} \,$$

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$$\int \frac{x}{8+x^3}dx$$ $$=\lim_{h\to 0} \,\ h\sum_{r=1}^{n} \frac{rh}{8+(rh)^3}$$ $$=\lim_{n\to\infty} \,\ \frac{1}{n}\sum_{r=1}^{n} \frac{\frac{r}{n}}{8+(\frac{r}{n})^3}$$ $$=\lim_{n\to\infty} \,\ \sum_{r=1}^{n} \frac{rn}{8n^3+r^3}$$

Is this the series what you were looking for?
Please confirm.

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  • $\begingroup$ @SchrodingersCat In the second line, the summation's upper bound $n$ has not been defined $\endgroup$
    – Vessel
    Nov 18, 2020 at 20:14

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