How do you find the power series for $f(x)=int tln(1-t)dt$ from [0,x] and determine its radius of convergence?

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Answer 1 · 2017-01-23T16:50:28

$f(x) = -sum_(n=0)^oo x^(n+3)/((n+3)(n+1))$ with radius of convergence $R=1$ .

We have:

$f(x) = int_0^x tln(1-t)dt$

Focus on the function:

$ln(1-x)$

We know that:

$ln (1-x) = -int_0^x (dt)/(1-t)$

where the integrand function is the sum of the geometric series:

$sum_(n=0)^oo t^n = 1/(1-t)$

then we can integrate term by term, and we have:

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

and mutiplying by x term by term:

$xln(1-x) = - sum_(n=0)^oo x^(n+2)/(n+1)$

We can substitute this expression in the original integral and integrate again term by term:

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

To determine the radius of convergence we can then use the ratio test:

$abs (a_(n+1)/a_n) = abs (frac (x^(n+4)/((n+4)(n+2))) (x^(n+3)/((n+3)(n+1)))) = abs (x) ((n+3)(n+1))/((n+4)(n+2))$

$lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)$

and the series is absolutely convergent for $abs(x) <1$ which means the radius of convergence is $R=1$

How do you find the power series for f(x)=int tln(1-t)dtf(x)=int tln(1-t)dt from [0,x] and determine its radius of convergence?