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?

2 Answers
Jul 25, 2017

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).

Jul 25, 2017

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))