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Take a look at the over 111000 values for the hypergeometric function. This shows that many results can be derived from using this genius function. Now for the arc length formula derivation. An arc length function for the arc length of $x^n$ would be messy, so we want something like this using the main arc length formula, not for others like this or this one, but rather for y=f(x). Series expansion for the hypergeometric function. There may be a small typo in here:

$$\mathrm {S(n,x)=\text{Arc length with respect to x}(\text{monomial power function})=}$$ $$\mathrm{\int\sqrt{1+x^n}dx=\int\sqrt{1+\left(y’(x)\right)^2}\implies y=\pm\int x^{n/2}dx=c\pm \frac{2x^{\frac n2+1}}{n+2}\implies \text{S(n,x)=[Arc length with respect to x]}\left(c\pm \frac{2x^{\frac n2+1}}{n+2}\right)=}$$ $$\mathrm{x \ _2F_1\left(-\frac12, \frac1n,\frac1n+1,-x^n\right)=\left\{\sum_{k=0}^\infty \frac{(-1)^k\left(-\frac12\right)_k\left(\frac1n\right)_k x^{kn+1}}{\left(\frac1n+1\right)_kk!},|-x^n|\le 1\ \text{or} \ \frac{Γ\left(\frac1n +\frac12\right)Γ\left(\frac1n+1\right)x^{\frac n2}}{Γ\left(\frac1n\right)Γ\left(\frac1n+\frac32\right)}\sum_{k=0}^\infty \frac{(-1)^k\left(-\frac12\right)_k\left(-\frac1n-\frac12\right)_k}{\left(\frac12-\frac1n\right)_kk! x^{kn-1}} +C\sum a_n (0)_k,|-x^n|>1,n\not\in -\frac{2}{2\Bbb Z+1} \right\}=}$$ $$\mathrm{\left\{-\frac1{2\sqrt\pi}\sum_{k=0}^\infty \frac{(-1)^kΓ\left(k-\frac12\right) x^{kn+1}}{(kn+1)k!},|-x^n|\le1\ \text{or}\ \frac{x^{\frac n2+1}}{\sqrt\pi}\sum_{k=0}^\infty \frac{(-1)^k Γ\left(k-\frac12\right)}{(2kn-n-2)k!x^{kn}},\quad |-x^n|>1,n\not\in -\frac{2}{2z+1},z\in\Bbb Z \right\}}$$

This large expression would be useful for seeing the sum version of the series. This brings up the question of if this is the correct summation expansion using the main definition of the hypergeometric function. If there was a mistake somewhere, I would like to know where and have it corrected in an answer. Assuming this form is correct using the linked “main definition”, then what is another representation of this S(n,x) function?

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  • $\begingroup$ I am having a little trouble following your post. Could you please explain what exactly your question is in simple terms for slow folks like myself. Are you just looking for alternative forms of $S$? $\endgroup$ Jun 30, 2021 at 19:09
  • $\begingroup$ @AaronHendrickson Yes, essentially the beginning is trying to derive a series for the arc length of $c\pm \frac{2x^{\frac n2+1}}{n+2}$ which boils down to $\int \sqrt{x^n+1}dx$. I just want to verify my results which are at the bottom of the large MathJax block of text. If this series is correct, then I thought I could rewrite “S” in terms of other functions. This would make it easier to understand, at least for me. Does this help? To find the series, I used Wolfram Mathworld. $\endgroup$ Jun 30, 2021 at 19:20

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Your hypergeometric representation for $S$ is indeed correct. Another way to obtain this is to write $$ \begin{align} S &=\int(1+x^n)^{1/2}\,\mathrm dx\\ &\overset{u=x^n}{=}\frac{1}{n}\int u^{1/n-1}(1+u)^{1/2}\,\mathrm du\\ &=\frac{1}{n}\int_0^u z^{1/n-1}(1+z)^{1/2}\,\mathrm dz\\ &\overset{z=ut}{=}u^{1/n}\frac{1}{n}\int_0^1 \frac{t^{1/n-1}(1-t)^{(1+1/n)-1/n-1}}{(1-(-u)t)^{-1/2}}\,\mathrm dt, \end{align} $$ and then use DLMF 15.6.1 to identify the remaining integral as that of the hypergeometric function $F(-1/2,1/n;1+1/n;-u)$. Then, reintroducing $u=x^n$ gives $$ S=x\,{_2F_1}\biggl({-\frac{1}{2},\frac{1}{n}\atop 1+\frac{1}{n}};-x^n\biggr). $$ As for a general procedure for reducing this to simpler functions you are in good company. Of course you have the list of reduction formulae available through Wolfram functions, which will give you results in terms of more elementary functions for specific values of $n$. As for other potentially interesting forms you can use differential formula in the linked question above to show $y(z)=F(-1/2,1/n;1+1/n;z)$ satisfies $$ zy^\prime+\tfrac{1}{n}y=\tfrac{1}{n}\sqrt{1-z},\quad y(0)=1. $$ If $n=1$ this gives $$ zy^\prime+y=\sqrt{1-z}\implies \partial_z(zy)=\sqrt{1-z}, $$ which immediately lends to finding a simpler expression for $S$.

Another form for $S$ that may be of interest to you is that terms of the incomplete beta function, which can be found directly with this relation.

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  • $\begingroup$ A few comments. I now have a proof for the Hypergeometric expression, but would it be possible to check the derived series expansion above also found here, the second terms looks to be 0 in the link? I would like to genuinely see this expressed in terms of the incomplete gamma function. Thanks again. $\endgroup$ Jun 30, 2021 at 20:06
  • $\begingroup$ If your goal is to search for potentially useful forms of $S$ in terms of other special functions, then I would not work directly from series expansions. Instead become very familiar with specialized values of ${_2F_1}$ and perhaps more importantly it's various properties so that you can get good at transforming hypergeometric expressions into other forms. $\endgroup$ Jun 30, 2021 at 20:30
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Example:

$$\mathrm{S(3,1)=\text{Arc length on unit interval of} \ \frac{2x^\frac52}5=\ _2F_1 \left(-\frac12,\frac13,\frac43,-1\right)=-\frac1{2\sqrt\pi}\sum_{k=0}^\infty\frac{(-1)^kΓ\left(k-\frac12\right)}{(3k+1)k!}=\frac16(1-\sqrt3i)B_{-1}\left(\frac13,\frac32\right)=1.11144797…}$$

I tried one for the |-$x^2$|>1 restriction series which was wrong, but still converged.

$$\mathrm{S(c,n,x_1,x_2)=\int_{x_1}^{x_2}\sqrt{cx^n+1}dx}$$

$$\mathrm{Arclength\left(C\pm x^n\right)\text{ with respect to x}=S\left(n^2,2(n-1),x\right)=\int\sqrt{\left(nx^{n-1}\right)^2+1}dx=x\ _2F_1\left(-\frac12,\frac1{2n-2},\frac1{2n-2}+1,-n^2x^{2n-2}\right) = \frac{\sqrt\pi x \left(\frac1{2n-2}\right)^{\left(-\frac12\right)}}2P_\frac12^{\left(\frac1{2(n-1)},-\frac32\right)}\left(2n^2x^{2(n-1)}+1 \right)=\frac{xB_{-n^2x^{2n-2}}\left(\frac1{2n-2},\frac32\right)}{2\sqrt[2n-2]{-n^2x^{2n-2}}(n-1)}}$$

$$\mathrm{n=-1:S(1,-4,1,2)=Arclength\left(\frac1x \right)\text { from 1 to 2}=\int_1^2 \sqrt{x^{-4}+1}dx= \frac{Γ^2\left(\frac34\right)}{\sqrt{\pi}}-\frac{\sqrt{17}} {2}-\frac{1+i}{2\sqrt2}B_{-16}\left(\frac34,\frac12\right)=1.3209039…}$$

$$\mathrm{n,x>0:S\left(n^2,2n-2,x\right)=\frac{B_{-n^2x^{2n-2}}\left(\frac1{2n-2},\frac32\right)}{2(n-1)\sqrt[n-1]{ni}}}$$

To be able to change coefficient of $\mathrm x^n:$

$$\mathrm{S(c,n,x)=Arclength\left(C\pm\frac{2\sqrt cx^{\frac n2+1}}{n+2}\right)\text{ with respect to x}=\int \sqrt{cx^n+1}dx=x\ _2F_1\left(-\frac12,\frac1n,\frac1n+1,-cx^2\right)=\frac{\sqrt\pi x \left(\frac1n\right)^{\left(-\frac12\right)}}{2}P_\frac12^{\left(\frac1n,-\frac32\right)}\left(2cx^n+1\right)=\frac{B_{-cx^n}\left(\frac1n,\frac32\right)}{\sqrt[n]{-c}n}}$$

$$\mathrm{n=i:S(1,2i-2,1,2)=Arclength\left(x^i\right) from \ 1\ to \ 2=\int_1^2 \sqrt{x^{2i-2}+1}dx=-\frac{1+i}4 B_{\frac14\left(cos(2ln(2))+isin(2ln(2))\right)}\left(-\frac{1+i}4,\frac32\right)-\frac{\sqrt\pi}2\left(-\frac{1+i}{4}\right)^{\left(-\frac12\right)}=.780647…+.168809…i}$$

Jacobi P function

Falling:$x^{(a)}$ and rising factorials

Incomplete Beta Function

Source for $\frac1x$ arc length and formula proof

Graph of arc length from 1 to 2 of $x^k$: enter image description here

Graph of arc length from 0 to 1 of $x^k,x\ge0$: enter image description here

$x^i$ arc length source. WA needs more time to compute

I know there are a few other series representations in the link and through more general functions here. Please correct me and give me feedback!

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