Question

How do you find a power series converging to `f(x)=int arcsint/t dt` from [0,x] and determine the radius of convergence?

Answer 1

`f(x) = x+x^3/18+(3x^5)/200+(5x^7)/784+(35x^9)/10458 + ...`

Explanation:

Let us start by using a well known series that is easily shown using a Binomial Series expansion

`1/sqrt(1-x^2) = 1+x^2/2+(3x^4)/8+(5x^6)/16+(35x^8)/128 + ...`

Then we can form the Maclaurin Series for arcsinx# via integration, as:

`arcsinx = int_0^x 1/sqrt(1-t^2) \ dt`

# " " = int_0^x {1+t^2/2+(3t^4)/8+(5t^6)/16+(35t^8)/128 + ... } \ dt #

`" " = [t+t^3/6+(3t^5)/40+(5t^7)/112+(35t^9)/1162 + ... ]_0^x`

`" " = x+x^3/6+(3x^5)/40+(5x^7)/112+(35x^9)/1162 + ...`

So now we can re-write `f(x)` in terms of the integral of a power series:

`f(x) = int_0^x \ arcsint/t \ dt`
`" " = int_0^x \ {t+t^3/6+(3t^5)/40+(5t^7)/112+(35t^9)/1162 + ...}/t \ dt`

`" " = int_0^x \ 1/t \ {t+t^3/6+(3t^5)/40+(5t^7)/112+(35t^9)/1162 + ...} \ dt`

`" " = int_0^x \ {1+t^2/6+(3t^4)/40+(5t^6)/112+(35t^8)/1162 + ...} \ dt`

`" " = [t+t^3/18+(3t^5)/200+(5t^7)/784+(35t^9)/10458 + ...]_0^x`

`" " = x+x^3/18+(3x^5)/200+(5x^7)/784+(35x^9)/10458 + ...`