I have figured, hopefully, out the volume of the figure of $$(y-x^x)(y-x^{-x})=0\le x\le 1$$ about the line y=1.
This volume came out to be $$V=2π\int_0^1 (1-x)(x^{-x}-x^x)dx= \boxed{-2π\sum_{n=1}^∞\sum_{k=1}^{n-1} \binom{n-1}{k-1}\frac{(-1)^{k+n}+(-1)^n}{k^n}}$$$$=1.774473…$$
All of the context and work is here.
The question is from that the summand does not have the denominator of $n^n$ or another variable which would have made it quite difficult to put into another form except at the upper values for k.
A desired form for this integral is required because this would possibly create a closed form of this integral. In addition, this sum is a little hard to understand which is why an alternate form may make it easier to read.
What is another form of this integral and sum which can be simplified and possibly put in terms of special functions? A closed form would be great, but still optional. Please do not include another integral form as that would be redundant. Anything except integral representations works. Finally, please give me feedback and correct me please!
We'll show $$ V=-2\pi\sum_{n=1}^\infty\sum_{k=1}^{n-1}{n-1\choose k-1}\frac{(-1)^{k+n}+(-1)^n}{k^n} =4\pi\sum_{r=1}^\infty\left(\frac{1}{(2r)^{2r}}-\frac{1}{(2r+1)^{2r}}\right). $$ First observe that for odd $k$ the terms vanish because $(-1)^{k+n}+(-1)^n=0$. Hence we can restrict the sum to even $k=2r$: $$ V=-2\pi\sum_{n=1}^\infty\sum_{r=1}^{\left[\frac{n-1}{2}\right]}{n-1\choose 2r-1}\frac{2\,(-1)^n}{(2r)^n}. $$ Now, rearrange the summation $$1\le n,\quad 1\le r\le\frac{n-1}{2}\quad\Longleftrightarrow\quad1\le r,\quad2r+1\le n$$ to get $$ V=-4\pi\sum_{r=1}^\infty\sum_{n=2r+1}^\infty{n-1\choose 2r-1}\left(-\frac{1}{2r}\right)^n=-4\pi\sum_{r=1}^\infty\sum_{n=2r}^\infty{n\choose 2r-1}\left(-\frac{1}{2r}\right)^{n+1}.\qquad(*) $$ The next step will evaluate the inner sum to a closed expression. For doing that, we start with $$ \sum_{k=0}^\infty x^k=\frac{1}{1-x},\quad(|x|<1). $$ Differentiating $a$ times by $x$ gives $$ \sum_{k=0}^\infty k(k-1)\ldots(k-a+1)x^{k-a}=\frac{a!}{(1-x)^{a+1}},\quad(|x|<1). $$ Multiplying this identity with $\frac{x^a}{a!}$ and observing that the terms in the sum for $0\le k<a$ vanish, we get $$ \sum_{k=a}^\infty {k\choose a}x^{k}=\frac{x^a}{(1-x)^{a+1}},\quad(|x|<1). $$ That is almost what we need for $\sum_{n=2r}^\infty{n\choose 2r-1}\left(-\frac{1}{2r}\right)^{n+1}$. Multiplying with $x$ and subtracting the first term completes it: $$ \sum_{k=a+1}^\infty {k\choose a}x^{k+1} = \sum_{k=a}^\infty {k\choose a}x^{k+1}-x^{a+1}=\frac{x^{a+1}}{(1-x)^{a+1}}-x^{a+1},\quad(|x|<1). $$ Setting $a=2r-1$, $k=n$, $x=-\frac{1}{2r}$, we find that indeed $|x|\le\frac{1}{2}<1$, and $$ \sum_{n=2r}^\infty{n\choose 2r-1}\left(-\frac{1}{2r}\right)^{n+1}= \frac{\left(-\frac{1}{2r}\right)^{2r}}{\left(1+\frac{1}{2r}\right)^{2r}}- \left(-\frac{1}{2r}\right)^{2r}=\frac{1}{(2r+1)^{2r}}-\frac{1}{(2r)^{2r}}. $$ Inserting this in (*), we get the simplified expression $$ V=4\pi\sum_{r=1}^\infty\left(\frac{1}{(2r)^{2r}}-\frac{1}{(2r+1)^{2r}}\right), $$ which converges very rapidly: $$ \begin{array}{cll} r&\frac{1}{(2r)^{2r}}-\frac{1}{(2r+1)^{2r}}&\mbox{partial sum up to $r$}\\ 1&0.138888888888889&0.138888888888889\\ 2&0.00230625&0.141195138888889\\ 3&1.29336107552305\cdot10^{-5}&0.141208072499644\\ 4&3.63740716499719\cdot10^{-8}&0.141208108873716\\ 5&6.14456710570468\cdot10^{-11}&0.141208108935161\\ 6&6.92346723666235\cdot10^{-14}&0.141208108935231\\ 7&5.57378713899953\cdot10^{-17}&0.141208108935231 \end{array} $$ giving $V=4\pi\cdot0.1412081089352\ldots=1.77447343063\ldots$.