Question

Find power series expansion and give its interval of validity for `int_0^x t^(-1)ln(1+2t)dt` ?

I have a problem with this.First will i find expansion for inside intergal,and later i will integrate it?

Answer 1

`int_0^x t^(-1)ln(1+2t)dt = sum_(n=0)^oo (-1)^n2^(n+1)x^(n+1)/(n+1)^2dt`

for `abs x <1/2`

Explanation:

Start from the MacLaurin expansion of `ln(1+x)`:

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

that has radius of convergence `R=1`.

Let `x=2t`:

`ln(1+2t) = sum_(n=0)^oo (-1)^n(2t)^(n+1)/(n+1) = sum_(n=0)^oo (-1)^n2^(n+1)t^(n+1)/(n+1)`

converging for `absx <1`, so `abs t < 1/2`.

Divide by `t` term by term:

`t^(-1)ln(1+2t) = sum_(n=0)^oo (-1)^n2^(n+1)t^(-1)t^(n+1)/(n+1) = sum_(n=0)^oo (-1)^n2^(n+1)t^n/(n+1)`

We can now integrate term by term obtaining a power series that has at least the same interval of convergence as the integrand:

`int_0^x t^(-1)ln(1+2t)dt = sum_(n=0)^oo (-1)^n2^(n+1)int_0^x t^n/(n+1)dt`

`int_0^x t^(-1)ln(1+2t)dt = sum_(n=0)^oo (-1)^n2^(n+1)x^(n+1)/(n+1)^2dt`