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Evaluate the indefinite integral as a power series. integral tan^-1(x)/(x) dx. What is the radius of convergence?

This problem aims to familiarize us with the power series of an indefinite integral. This question requires the understanding of fundamental calculus, which includes indefinite integrals, Power series, and the radius of convergence.

Now, Indefinite integrals are mostly normal integrals but are expressed without higher and lower limits on the integrand, the expression $\int f(x)$ is employed to represent the function as an antiderivative of the function.

Whereas a power series is an indefinite series of the form $ \sum_{n= 0}^{ \infty } a_n  (x – c)^{n} $ where $a_n$ symbolizes the coefficient of the $n^{th}$ duration and $c$ represents a constant. Such power series are helpful in mathematical analysis, and are transformed into Taylor series to solve infinitely differentiable expressions.

Expert Answer

If we expand the expression $tan^{-1}x$ into an indefinite summation, we get something as follows:

\[x – \dfrac{x^3}{3} + \dfrac{x^5}{5} – \dfrac{x^7}{7} + \dfrac{x^9}{9} \space ….. \]

The given integral can be written as a power series:

\[\int \dfrac{tan^{-1}x}{x} dx = \int \dfrac{1}{x} \left( x – \dfrac{x^3}{3} + \dfrac{x^5}{5}  – \dfrac{x^7}{7} + \dfrac{x^9}{9} \space …. \right) dx\]

\[= \int \left( 1 – \dfrac{x^2}{3} + \dfrac{x^4}{5} – \dfrac{x^6}{7} + \dfrac{x^8}{9} \space …. \right) dx\]

By solving the integral:

\[=x – \dfrac{x^3}{9} + \dfrac{x^5}{25} – \dfrac{x^7}{49} + \dfrac{x^9}{81} \space ….\]

This above sequence can be written in the form of:

\[=\sum_{n= 1}^{ \infty } \dfrac{ (-1) ^ {n-1} x^{2n -1}} {( 2n -1 )^2 }\]

Which is the required power series.

The radius of convergence is given as:

\[R = lim_{n \rightarrow \infty } \left| \dfrac{a_n}{a_{n+1}} \right|\]

Here, we have:

\[a_n = \dfrac{ (-1) ^ {n-1} } {( 2n -1 )^2 }\]

\[a_{n+1} = \dfrac{ (-1) ^ {n} } {( 2n +1 )^2 }\]

Therefore:

\[R = lim_{n \rightarrow \infty } \left| \dfrac{(  -1)^{n-1}}{ (2n -1 )^2 } \times  \dfrac { (2n +1 )^2 } {(  -1)^{n}} \right|\]

\[R = lim_{n \rightarrow \infty } \left| \dfrac{(2n+1)^{2}}{ (2n -1 )^2 } \right|\]

\[R = lim_{n \rightarrow \infty } \left| \dfrac{4n^2 \left( 1 + \dfrac{1}{2n} \right)^2 }{ 4n^2 \left( 1 – \dfrac{1}{2n} \right)^2 } \right|\]

\[R = lim_{n \rightarrow \infty } \left| \dfrac{ \left( 1 + \dfrac{1}{2n} \right)^2 }{ \left( 1 – \dfrac{1}{2n} \right)^2 } \right|\]

Therefore, the radius of convergence is $R = 1$.

Numerical Result

Indefinite integral as a power series is $ \sum_{n= 1}^{ \infty } \dfrac{ (-1) ^ {n-1} x^{2n -1}} {( 2n -1 )^2 } $.

Radius of convergence is $ R =1 $.

Example

Using the Power Series, evaluate the given integral $ \int \dfrac{x}{1+x^3} dx $.

The given integral can be written as a power series as follows:

\[ = \sum_{n= 0}^{ \infty } (-1) ^ {n}  x^{3n +1} \]

The series converges when $|-x^3| < 1$ or $|x| < 1$ , so for this particular power series $R = 1$.

Now, we integrate:

\[ \int \dfrac{x}{1+x^3} dx = \int \sum_{n= 0}^{ \infty } (-1) ^ {n}  x^{3n +1} dt \]

Indefinite integral as a power series comes out to be:

\[ = C + \sum_{n= 0}^{ \infty } (-1) ^ {n} \dfrac{ x^{3n +2}}{( 3n +2 ) } \]

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