Question

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

Answer 1

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

Explanation:

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`