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

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

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Answer 1 · 2018-05-27T23:54:56

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

for $abs x <1/2$

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:

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

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