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I want to get the expansion of the inverse of the hyperbolic sine $$f(x)=\sinh^{-1}(x)$$ with the interval of convergence $$|x|>1$$ I solved it by integrating the series of the derivative of f(x) but I could not get the constant of the integration. Here is my solution: $$f'(x)=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-1/2}=\frac{1}{x}(1+\frac{1}{x^2})^{-1/2}$$ $$f'(x)=\frac{1}{x}(1+\frac{-1}{2}\frac{1}{x^2}+\frac{-1}{2}\frac{-3}{2}\frac{1}{2!}\frac{1}{x^4}-...)$$ $$f'(x)=\frac{1}{x}+\frac{-1}{2}\frac{1}{x^3}+\frac{-1}{2}\frac{-3}{2}\frac{1}{2!}\frac{1}{x^5}-...$$ $$f(x)=\sinh^{-1}(x)=\ln(x)-\frac{1}{2}\frac{x^{-2}}{-2}+\frac{-1}{2}\frac{-3}{2}\frac{1}{2!}\frac{x^{-4}}{-4}-...+C$$ How to get C ? ( I thought of substituting by x=infinity or x=0 but this would not help!)

MCS
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    Wouldn't it be easier to integrate the original integral by parts? You know that $f'(x)=\frac{1}{\sqrt{x^2+1}}$ and $\frac{x^2+1}{2}$ is an antiderivative of $x$. – A.Γ. Dec 25 '17 at 21:37
  • Thanks for this hint .. But because I am studying a power series course , so the question in the course is to use the expansion to solve the integral. – MCS Dec 25 '17 at 22:26

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You may use the fact that $$ f(x)=\ln(x+\sqrt{x^2+1})=\ln\Big(x\cdot \left(1+\sqrt{1+x^{-2}}\right)\Big)=\ln x+\ln\Big(1+\sqrt{1+x^{-2}}\Big). $$ Compare with your expansion, remove $\ln x$ and set $x\to\infty$ to get $C=\ln 2$.

P.S. Compare also with Wolfram (see Eq. 8 and 9).

P.P.S. It is easier to solve the integral by parts (see my comment above).

A.Γ.
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