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

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Answer 1 · 2017-02-08T20:00:01

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

with $R=oo$

Start from the MacLaurin series of $sinhx$ :

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

with radius of convergence $R=oo$

Substitute $x=t^2$

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

Multiply by $t^(-2)$ term by term:

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

As $R=oo$ we can integrate term by term on all $RR$ .

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

and the resulting series also have $R=oo$

How do you find a power series converging to f(x)=int t^-2 sinh (t^2) dtf(x)=int t^-2 sinh (t^2) dt from [0,x] and determine the radius of convergence?