How do you find the power series for #f(x)=int e^(t^3)# from [0,x] and determine its radius of convergence?

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Answer 1 · 2017-01-09T10:57:14

#int_0^x e^(t^3) =sum_(n=0)^oo x^(3n+1)/((3n+1)n!)# with #R=oo#

Let's start with the MacLaurin expansion of #e^x#:

#e^x = sum_(n=0)^oo x^n/(n!)#

and substitute #x=t^3#

#e^(t^3) = sum_(n=0)^oo t^(3n)/(n!)#

This series converges absolutely in all #RR# so the radius of convergence is #R=oo#.

We can then integrate term by term and obtain a series with the same radius of convergence:

#int_0^x e^(t^3) = int_0^x sum_(n=0)^oo t^(3n)/(n!) = sum_(n=0)^oo int_0^xt^(3n)/(n!)= sum_(n=0)^oo 1/(n!)[t^(3n+1)/(3n+1)]_0^x#

Finally:
#int_0^x e^(t^3) =sum_(n=0)^oo x^(3n+1)/((3n+1)n!)#