# How do you find the power series for f(x)=int t^2/(1+t^2)dt from [0,x] and determine its radius of convergence?

Mar 3, 2017

${\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 3} / \left(2 n + 3\right)$

with radius of convergence $R = 1$.

#### Explanation:

Write the integral as:

${\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt} = {\int}_{0}^{x} \frac{{t}^{2} + 1 - 1}{1 + {t}^{2}} \mathrm{dt} = {\int}_{0}^{x} \left(1 - \frac{1}{1 + {t}^{2}}\right) \mathrm{dt}$

Using the linearity of integrals:

${\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt} = {\int}_{0}^{x} \mathrm{dt} - {\int}_{0}^{x} \frac{1}{1 + {t}^{2}} \mathrm{dt} = x - {\int}_{0}^{x} \frac{1}{1 + {t}^{2}} \mathrm{dt}$

Now consider the integrand function: it is in the form of the sum of a geometric series of ratio $- {t}^{2}$, so:

$\frac{1}{1 + {t}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- {t}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{2 n}$

and the series is convergent for $- {t}^{2} \in \left[- 1 , 1\right)$ that is for $t \in \left[- 1 , 1\right]$.
Within this interval we can therefore integrate term by term and we have:

${\int}_{0}^{x} \frac{1}{1 + {t}^{2}} \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\int}_{0}^{x} {t}^{2 n} \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right)$

Substituting this in the expression above we have:

${\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt} = x - {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right)$

Now note that the first term of the series is just $x$:

$x - {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right) = \cancel{x} - \cancel{x} + {x}^{3} / 3 - {x}^{5} / 5 + \ldots = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 3} / \left(2 n + 3\right)$

and we can conclude that:

${\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 3} / \left(2 n + 3\right)$

with radius of convergence $R = 1$.

Mar 3, 2017

$= {\sum}_{1}^{\infty} \frac{{\left(- 1\right)}^{n - 1}}{2 n + 1} {x}^{2 n + 1}$

#### Explanation:

You could have a go at this using the FTC, but it's gonna be easier just doing the integration and adding in some pre-determined Maclaurin Series (I know cos it started doing it the other way and it's a real drag).

${\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt}$

$= {\int}_{0}^{x} \frac{1 + {t}^{2} - 1}{1 + {t}^{2}} \mathrm{dt}$

$= {\int}_{0}^{x} 1 - \frac{1}{1 + {t}^{2}} \mathrm{dt}$

This is a trivial integral so I'll post the result:

$= x - \arctan x$

The power series for $\arctan x$, again from a look up because I assume you can get there yourself, is:

$\arctan x = {\sum}_{n = 0}^{\infty} \frac{{\left(- 1\right)}^{n}}{2 n + 1} {x}^{2 n + 1}$

$= x - \frac{1}{3} {x}^{3} + \frac{1}{5} {x}^{5} - \frac{1}{7} {x}^{7} + \frac{1}{9} {x}^{9} - . .$

It follows that:

$= x - \arctan x = \frac{1}{3} {x}^{3} - \frac{1}{5} {x}^{5} + \frac{1}{7} {x}^{7} - \frac{1}{9} {x}^{9} - . .$

$= {\sum}_{\textcolor{red}{n = 1}}^{\infty} \frac{{\left(- 1\right)}^{n - 1}}{2 n + 1} {x}^{2 n + 1}$

The ratio test confirms convergence at $\left\mid x \right\mid < 1$

If we go at it using the FTC , ie using this:

$\frac{d}{\mathrm{dx}} \left({\int}_{0}^{x} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt}\right) = {x}^{2} / \left(1 + {x}^{2}\right)$

...then we can start to generate the terms from the fundamental definition/idea, which is that:

f(x) = sum_(n=0)^(oo) (f^(n)(0))/(n!) x^n .

• $f \left(0\right) = {\int}_{0}^{0} {t}^{2} / \left(1 + {t}^{2}\right) \mathrm{dt} = 0$
• $f ' \left(x\right) = {x}^{2} / \left(1 + {x}^{2}\right)$
• $f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} / \left(1 + {x}^{2}\right)\right)$