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

1 Answer
Jan 23, 2017

#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#