Question

To evaluate the indefinite integral as a power series and the radius of convergence

\(f(t) = \int {{x^2}} \ln (1 + x)dx\)

Short answer

The power series representation for the function \(\int {{x^2}} \ln (1 + x)\) is \(\left( {\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{n + 3}}}}{{n(n + 3)}}} } \right) + C\) and the radius of convergence is \(R = 1\).

Step-by-step solution

Concept of Geometric Series

Geometric Series

The sum of the geometric series with initial term \(a\) and common ratio \(r\) is \(\sum\limits_{n = 0}^\infty a {r^n} = \frac{a}{{1 - r}}\)

Calculation of the indefinite integral\(f(t) = \int {{x^2}} \ln (1 + x)dx\)

Let

\(f(t) = \int {{x^2}} \ln (1 + x)dx\)

Modify the function as follows:

$\(f(t) = \int {\left[ {{x^2}\int {\frac{1}{{1 + x}}} dx} \right]} dx\)

By geometric series the function \(\frac{1}{{1 + x}}\) can be written as follows:

\(\begin{aligned}\frac{1}{{1 + x}} &= \frac{1}{{1 - ( - x)}}\\\frac{1}{{1 + x}} &= \frac{a}{{1 - r}}\end{aligned}\)

Therefore, \(f(t)\) is a sum of a geometric series with initial term \(a = 1\) and \(r = - x\).

Expansion of the expression\(f(t) = \sum\limits_{n = 0}^\infty   a {r^n}\)

The expansion of series by using geometric series.

\(\begin{aligned}f(t) &= \sum\limits_{n = 0}^\infty a {r^n}\\f(t) &= \sum\limits_{n = 0}^\infty {(1)} {( - x)^n}\\f(t) &= \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} {x^n}\\f(x) &= \int {\left[ {{x^2}\int {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} } {x^n}dx} \right]} dx\\f(x) &= \int {\left[ {{x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{n + 1}}}}{{n + 1}}} } \right]} dx\end{aligned}\)

Therefore, the power series representation as follows:

\({x^2}\ln (1 + x) = \int {\left[ {{x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{n + 1}}}}{{n + 1}}} } \right]} dx\)

Integrate the expression\({x^2}\ln (1 + x) = \int {\left[ {{x^2}\sum\limits_{n = 0}^\infty   {\frac{{{{( - 1)}^n}{x^{n + 1}}}}{{n + 1}}} } \right]} dx\)

\(\int {{x^2}} \ln (1 + x) = \int {\left[ {{x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{n + 1}}}}{{n + 1}}} } \right]} dx\)

\(\int {{x^2}} \ln (1 + x) = \int {\left[ {\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{n + 3}}}}{{n + 1}}} } \right]} dx\)

\(\int {{x^2}} \ln (1 + x) = \left[ {\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{n + 4}}}}{{(n + 1)(n + 4)}}} } \right] + C\)

Replace \(n\) by \(n + 1\) in the summation part

\(\int {{x^2}} \ln (1 + x) = \left[ {\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{n + 3}}}}{{n(n + 3)}}} } \right] + C\)

Therefore, the power series representation is. \(\int {{x^2}} \ln (1 + x) = \left[ {\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{n + 3}}}}{{n(n + 3)}}} } \right] + C\).

Calculation of the radius of convergence

Ratio test:

If \(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\), then the series \(\sum\limits_{n = 1}^\infty {{a_n}} \) is absolutely convergent

Let

\({a_n} = \frac{{{x^{n + 3}}}}{{n(n + 3)}}\).

Then,

\({a_{n + 1}} = \frac{{{x^{n + 4}}}}{{(n + 1)(n + 4)}}\).

Obtain \(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\) to apply the Ratio test.

\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{\frac{{{x^{n + 4}}}}{{(n + 1)(n + 4)}}}}{{\frac{{{x^{n + 3}}}}{{n(n + 3)}}}}} \right|\)

Take \(\mathop {\lim }\limits_{n \to \infty } \) on both sides.

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{x^{n + 4}}}}{{(n + 1)(n + 4)}}}}{{\frac{{{x^{n + 3}}}}{{m(n + 3)}}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{n + 4}}}}{{(n + 1)(n + 4)}} \cdot \frac{{n(n + 3)}}{{{x^{n + 3}}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^n}{x^4}}}{{\left( {{n^2} + 5n + 4} \right)}} \cdot \frac{{{n^2} + 3n}}{{{x^n}{x^3}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } |x|\frac{{{n^2} + 3n}}{{\left( {{n^2} + 5n + 4} \right)}}\end{aligned}\)

Simplification of the expression\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\)

Simplify the terms.

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } |x|\frac{{{n^2} + 3n}}{{\left( {{n^2} + 5n + 4} \right)}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } |x|\frac{{{n^2}\left( {1 + \frac{3}{n}} \right)}}{{{n^2}\left( {1 + \frac{5}{n} + \frac{4}{{{n^2}}}} \right)}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } |x|\frac{{1 + \frac{3}{n}}}{{1 + \frac{5}{n} + \frac{4}{{{n^2}}}}}\end{aligned}\)

Apply the limit and simplify the terms as shown below.

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } |x|\frac{{1 + \frac{3}{n}}}{{1 + \frac{5}{n} + \frac{4}{{{n^2}}}}} &= |x|\frac{{1 + 0}}{{1 + 0 + 0}}\\\mathop {\lim }\limits_{n \to \infty } |x|\frac{{1 + \frac{3}{n}}}{{1 + \frac{5}{n} + \frac{4}{{{n^2}}}}} &= |x|\end{aligned}\)

Therefore, by the ration test \(\int {{x^2}} \ln (1 + x) = \left[ {\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{n + 3}}}}{{n(n + 3)}}} } \right] + C\) converges as \(|x| < 1\).

The series converges when the limit is less than\(1\)

\(\begin{aligned}|x| < 1\\x < 1\end{aligned}\)

Therefore, the radius of convergence is \(R = 1\).