How do you find the power series for $f(x)=int sinh(t^2)$ from [0,x] and determine its radius of convergence?

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Answer 1 · 2018-02-03T14:03:30

$int_0^x sinh t^2 dt = sum_(n=0)^oo t^(4n+3)/((4n+3)(2n+1)!)$

with radius of convergence $R=oo$ .

Start from the MacLaurin series of $sinh x$ :

$sinhx = sum_(n=0)^oo x^(2n+1)/((2n+1)!)$

Using the ratio test:

$lim_(n->oo) abs((a_(n+1))/(a_n)) = lim_(n->oo) abs( ((x^(2(n+1)+1)/((2(n+1)+1)!))/(x^(2n+1)/((2n+1)!)))$

$lim_(n->oo) abs((a_(n+1))/(a_n)) = lim_(n->oo) abs( x^(2n+3)/x^(2n+1)) ((2n+1)!)/((2n+3)!)$

$lim_(n->oo) abs((a_(n+1))/(a_n)) = x^2 lim_(n->oo) 1/((2n+3)(2n+2)) = 0$

we can see that the series is absolutely convergent for every $x in RR$ .

Substituting $x=t^2$ we have:

$sinh t^2 = sum_(n=0)^oo (t^2)^(2n+1)/((2n+1)!) = sum_(n=0)^oo t^(4n+2)/((2n+1)!)$

and as the series has radius of convergence $R=oo$ we can integrate term by term and obtain a series that still has $R=oo$ :

$int_0^x sinh t^2 dt = sum_(n=0)^oo int_0^x t^(4n+2)/((2n+1)!)dt$

$int_0^x sinh t^2 dt = sum_(n=0)^oo t^(4n+3)/((4n+3)(2n+1)!)$

How do you find the power series for f(x)=int sinh(t^2)f(x)=int sinh(t^2) from [0,x] and determine its radius of convergence?