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

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