How to find the power series of $\sqrt{1+x^4}$?

Question

The complete question is to find the integral from $0$ to $1$ of $$\sqrt{1+x^4}$$

I am unsure of how to find the power series of this equation in order to do that. I haven't dealt with square root power series equations yet and any help would be appreciated. Thank you!

Answer 1

In order to find the power series of $\sqrt{1+x}$, you use the binomial series:$$\sqrt{1+x}=(1+x)^{\frac12}=\sum_{n=0}^\infty\binom{\frac12}nx^n$$and therefore$$\sqrt{1+x^4}=\sum_{n=0}^\infty\binom{\frac12}nx^{4n}.$$

Answer 2

$$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n\tag{1} $$ for any $x\in[-1,1)$ is a standard result. By applying termwise integration one gets $$ \sqrt{1-x} = \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{x^n}{1-2n}\tag{2} $$ for any $x\in[-1,1]$. By replacing $x$ with $-x^4$ one gets $$ \sqrt{1+x^4} = \sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{x^{4n}}{1-2n}\tag{3}$$ and finally $$ \int_{0}^{1}\sqrt{1+x^4}\,dx=\sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{1}{(1-2n)(4n+1)}.\tag{4} $$ By partial fraction decomposition, the RHS of $(4)$ only depends on $\sqrt{2}$ and $$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{(-1)^n}{(4n+1)}=\frac{\Gamma^2\left(\tfrac{1}{4}\right)}{8\sqrt{\pi}}.\tag{5}$$ The last equality can be derived through Euler's Beta function. Summarizing, $$ \int_{0}^{1}\sqrt{1+x^4}\,dx = \phantom{}_2 F_1\left(-\tfrac{1}{2},\tfrac{1}{4};\tfrac{5}{4};-1\right)= \frac{\sqrt{2}}{3}+\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}.\tag{6}$$ A numerical computation is extremely simple via $\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}=\frac{\pi}{3\,\text{AGM}(\sqrt{2},2)}$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}\root{1 + x^{4}}\dd x & \,\,\,\stackrel{x^{4}\ \mapsto\ x}{=}\,\,\, {1 \over 4}\int_{0}^{1}x^{-3/4}\,\pars{1 + x}^{1/2}\,\dd x \\[5mm] & = {1 \over 4}\int_{0}^{1}x^{-3/4}\,\pars{1 - x}^{0}\, \bracks{\vphantom{\Large A}1 - \pars{-1}x}^{1/2}\,\dd x \end{align} The integral is an Euler Type Hypergeometric Function. The above expression is reduced to \begin{align} &\overbrace{{1 \over 4}\bracks{\mrm{B}\pars{{1 \over 4},1}}}^{\ds{=\ 1}}\,\,\,\, \mbox{}_{2}\mrm{F}_{1}\pars{-\,{1 \over 2};{1 \over 4}; {5 \over 4},-1} \\[5mm] = &\ \bbx{\mbox{}_{2}\mrm{F}_{1}\pars{-\,{1 \over 2};{1 \over 4}; {5 \over 4},-1}} \approx 1.0894 \end{align}

$\ds{\mrm{B}}$ is the Beta Function.