Question

How do you find a power series converging to `f(x)=int ln(1+t^2) dt` from [0,x] and determine the radius of convergence?

Answer 1

Substitute the series for `ln(1+t^2)` and integrate term-by-term to get `x^3/3-x^5/10+x^7/21-x^9/36+cdots`. The radius of convergence is 1.

Explanation:

The Taylor series for `ln(1+x)` centered at `x=0` can be found by integrating the series for `1/(1+x)=1-x+x^2-x^3+x^4-x^5+cdots` term-by-term and using the fact that `ln(1)=0` to get `ln(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5-x^6/6+cdots`.

Next, replace `x` with `t^2` to get `ln(1+t^2)=t^2-t^4/2+t^6/3-t^8/4+t^10/5-t^12/6+cdots`.

Now integrate this term-by-term, from `t=0` to `t=x` to get

`\int_{0}^{x}ln(1+t^2)dt=x^3/3-x^5/10+x^7/21-x^9/36+cdots`

The original radius of convergence for `1/(1+x)` was 1, and that does not change upon substitution of `t^2` for `x` or upon either integration.

The interval of convergence does get "expanded" by one point, however. The interval of convergence for the final answer is `(-1,1]` (include 1 but do not include `-1`).

Answer 2

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

for `absx < 1`

Explanation:

Note that:

`d/dx (ln(1+x^2)) = (2x)/(1+x^2)`

Now:

`1/(1+x^2)`

can be expressed as the sum of a geometric series of ratio `(-x^2)` with `x^2 < 1`:

`1/(1+x^2) = sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^nx^(2n)`

and then:

`(2x)/(1+x^2) = 2xsum_(n=0)^oo (-1)^nx^(2n) = 2sum_(n=0)^oo (-1)^nx^(2n+1)` for `absx < 1`

Inside the interval of convergence we can integrate term by term to have a power series with the same radius of convergence:

`ln(1+x^2) = int_0^x (2t)/(1+t^2)dt = 2sum_(n=0)^oo int_0^x(-1)^nt^(2n+1)dt`

`ln(1+x^2) = 2 sum_(n=0)^oo (-1)^nx^(2n+2)/(2n+2) = sum_(n=0)^oo (-1)^nx^(2n+2)/(n+1)`

and integrating again:

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