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?

Write the integral as:

#int_0^x t^2/(1+t^2)dt = int_0^x (t^2+1-1)/(1+t^2)dt = int_0^x (1-1/(1+t^2))dt #

Using the linearity of integrals:

#int_0^x t^2/(1+t^2)dt = int_0^x dt -int_0^x1/(1+t^2)dt = x- int_0^x1/(1+t^2)dt#

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

#1/(1+t^2) = sum_(n=0)^oo (-t^2)^n = sum_(n=0)^oo (-1)^n t^(2n)#

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

#int_0^x1/(1+t^2)dt = sum_(n=0)^oo (-1)^n int_0^xt^(2n)dt =sum_(n=0)^oo (-1)^nx^(2n+1)/(2n+1)#

Substituting this in the expression above we have:

#int_0^x t^2/(1+t^2)dt = x - sum_(n=0)^oo (-1)^nx^(2n+1)/(2n+1)#

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

#x - sum_(n=0)^oo (-1)^nx^(2n+1)/(2n+1) = cancelx - cancelx +x^3/3 -x^5/5+... = sum_(n=0)^oo (-1)^n x^(2n+3)/(2n+3)#

and we can conclude that:

#int_0^x t^2/(1+t^2)dt = sum_(n=0)^oo (-1)^n x^(2n+3)/(2n+3)#

with radius of convergence #R=1#.

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).

If we start with:

#int_0^x t^2/(1+t^2)dt #

#= int_0^x (1 + t^2 - 1)/(1+t^2)dt #

#= int_0^x 1 - 1/(1+t^2)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)^(oo) ((-1)^n)/(2n+1) x^(2n+1)#

#= x - 1/3 x^3 + 1/5 x^5 - 1/7 x^7 + 1/9 x^9 - ..#

It follows that:

#= x - arctan x = 1/3 x^3 - 1/5 x^5 + 1/7 x^7 - 1/9 x^9 - ..#

#= sum_(color(red)(n=1))^(oo) ((-1)^(n-1))/(2n+1) x^(2n+1)#

The ratio test confirms convergence at #abs x < 1#

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

#d/dx( int_0^x t^2/(1+t^2)dt ) = x^2/(1+x^2)#

...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 #.

So we start with:

• #f(0) = int_0^0 t^2/(1+t^2)dt = 0 #

After that, using the FTC and differentiation:

• #f' (x) = x^2/(1+x^2)#

• #f'' (x) = d/dx( x^2/(1+x^2))#

And you can loop through that and it's really, really boring and long-winded.