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Section 10.15 : Power Series and Functions
5. Give a power series representation for the integral of the following function.
\[h\left( x \right) = \frac{{{x^4}}}{{9 + {x^2}}}\]Show All Steps Hide All Steps
Hint : Integrating this function seems like (potentially) a lot of work, not to mention determining a power series representation of the result. It’s a good think that we know how to integrate power series.
First let’s notice that we can quickly find a power series representation for this function. Here is that work.
\[h\left( x \right) = \frac{{{x^4}}}{9}\frac{1}{{1 - \left( { - \frac{1}{9}{x^2}} \right)}} = \frac{{{x^4}}}{9}\sum\limits_{n = 0}^\infty {{{\left( { - \frac{1}{9}{x^2}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{9}{x^4}{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^n}\,{x^{2n}}} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 4}}} \] Show Step 2Now, we know how to integrate power series and we know that the integral of the power series representation of a function is the power series representation of the integral of the function.
Therefore,
\[\int{{h\left( x \right)\,dx}} = \int{{\sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 4}}} \,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{c + \sum\limits_{n = 0}^\infty {\frac{1}{{2n + 5}}{{\left( { - 1} \right)}^n}{{\left( {\frac{1}{9}} \right)}^{n + 1}}\,{x^{2n + 5}}} }}\]Remember that to integrate a power series all we need to do is integrate the term of the power series and we can’t forget to add on the “+c” since we’re doing an indefinite integral.