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?

Question

Elijah Abram · Accepted Answer

#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#

Christopher Allers · Answer

undefined To find a power series converging to $ f(x) = \int_0^x t^{-2} \sinh(t^2) \, dt $ and determine the radius of convergence, you can integrate the power series representation for $ t^{-2} \sinh(t^2) $. Then, you'll obtain a power series representation for $ f(x) $. The radius of convergence will be the same as that of the original power series.

The power series representation for $ t^{-2} \sinh(t^2) $ is given by $ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} t^{2n-2} $.

After integrating, you'll get the power series representation for $ f(x) $.

To determine the radius of convergence, you can use the ratio test or another appropriate convergence test on the original power series representation for $ t^{-2} \sinh(t^2) $. This will provide you with the radius of convergence for $ f(x) $.