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#