Arc Length Integral of $x^x$ from 0 to 1 in closed form.

Question

I was recently trying to compute the arc length of $x^x$ from $0$ to $1$ as follows:

$$L=\int_0^1 \sqrt{1+\left(\frac{\text{d}}{\text{d}x}x^x\right)^2} \text{d}x=$$ $$\int_0^1\sqrt{1+x^{2x}(\ln x+1)^2} \text{d}x=$$ Using the infinite series expansion for the binomial theorem with |x|<1, we can rewrite the square root portion as $\sqrt x=x^{1/2}$. We can assume a limit for x=1 in the integral bounds: $$\int_0^1 \sum_{n=0}^∞ \binom{\frac 12}{n} (x^{2x}(\ln x+1)^2)^n \text{d}x=$$ $$\sum_{n=0}^∞ \binom{\frac 12}{n} \int_0^1 e^{2nx\ln(x)}(\ln x+1)^{2n}\text{d}x=$$

It would be maybe easier to expand out the exponential as a series with a different index to have another series with an independent index so that no cauchy products are needed:

$$\sum_{n=0}^∞ \binom{\frac 12}{n} \int_0^1 \sum_{m=0}^∞\frac{2^mn^mx^m\ln^m(x)}{m!}(\ln x+1)^{2n} \text{d}x=$$ $$\sum_{n=0}^∞ \binom{\frac 12}{n} \sum_{m=0}^∞\frac{2^mn^m}{m!} \int_0^1 x^m\ln^m(x)(\ln x+1)^{2n}\text{d}x=$$ Then the exponentiated logarithmic part of this expression, if we only focus on it, can be expanded into another similar summation using a binomials expansion because the index of n is an integer by definition: $$\sum_{n=0}^∞ \binom{\frac 12}{n} \sum_{m=0}^∞\frac{2^mn^m}{m!} \int_0^1 x^m\ln^m(x)\sum_{l=0}^{2n}\binom{2n}{l}\ln^l(x)\text{d}x=$$ $$\sum_{n=0}^∞ \binom{\frac 12}{n} \sum_{m=0}^∞\frac{2^mn^m}{m!} \sum_{l=0}^{2n}\binom{2n}{l} \int_0^1 x^m\ln^{l+m}(x)\text{d}x$$ This integral can be found in terms of the factorial and a “factorial coefficient” part. Using the substitution of u=-ln(x) gets us: $$\int_0^1 x^m\ln^{l+m}(x)\text{d}x=$$ $$(-1)^{l+m}\int_0^ ∞e^{-u(m+1)}u^{l+m}\text{d}u=$$ $$-(-1)^{l+m+1}m^{l+m+1}Γ(l+m+1,-(m+1)\ln(x))|_0^1=$$ $$(-1)^{l+m}m^{l+m+1}(l+m)!$$ The incomplete gamma function is here. This means our final possible answer is: $$L=^? \sum_{n=0}^∞ \binom{\frac 12}{n} \sum_{m=0}^∞\frac{2^mn^m}{m!} \sum_{l=0}^{2n}\binom{2n}{l} (-1)^{l+m}m^{l+m+1}(l+m)!=$$ $$\sum_{n=0}^∞ \binom{\frac 12}{n} \sum_{m=0}^∞\frac{2^mm^{m+1}n^m}{m!} \sum_{l=0}^{2n}\binom{2n}{l} (-1)^{l+m}m^l(l+m)!=^?1.2474...$$

This series diverges maybe because of a bad usage of the summations with the interval of convergence being too small. My main question is if all of these steps are correct as in my previous deleted question. Sorry for the undetailed question. Also, please correct and simplify the problem as much as possible using any widely used function. I.e. do not just do arclength(f(x),a,b).

I tried a similar method which also diverges even though the surface area should not of the same figure but with $(y-x^x)(y-x^{-x})=0$, $0\le x\le 1$ which got me to a similar answer. This one had $$S_y=\int_0^1 x^x \sqrt{1+\left(\frac{\text{d}}{\text{d}x}x^x\right)^2} \text{d}x$$. This one experimentally had 5 summations with different indices by accident when I was trying to calculate the arc length. The area of this figure is simply the difference two sophomore dream integrals.

This is $$A=\sum_{N=1}^ ∞\frac{1+(-1)^N}{N^N}=.507...$$ as n as an index was already defined.

I was also trying to find the perimeter of the lamina and volume of this solid of revolution about both the x and y axes all or which are quite ambitious to figure out and very tricky as I most likely got this wrong except for the area of this figure. Please just figure out the arc length and all will be well:

Here is the desmos demo of this arc length series:

https://www.desmos.com/calculator/gt0hsg40ah

If this is true, please expand the binomial coefficients and simplify the rest. Maybe keep the original form.

Thanks, and please give me feedback!

Answer 1

One trivial solution to this problem that was just noticed was using the definition of the integral with one variable and the other estimation techniques likely give the same summation such as the trapezoid rule: $$\int_a^b f(x) dx=\sum_{k=1}^n f(x_k=a+iΔx)Δx, Δx=\frac {b-a}{n}$$

This means $Δx=\frac {1-0}{n}$ and $$P_1= \lim_{n\to ∞ }\frac 1n \sum_{k=1}^n \sqrt{1+\bigg(\frac kn\bigg)^{\frac{2k}{n}}(ln(x)+1)^2}=.8142...$$ and the perimeter, being unable to combine the two square root terms, is $$P=\lim_{n\to ∞ } \bigg(\sqrt{1+\bigg(\frac kn\bigg)^{\frac{2k}{n}}(ln(x)+1)^2}+ \sqrt{1+\bigg(\frac kn\bigg)^{-\frac{2k}{n}}(ln(x)+1)^2}\bigg) =3.0847...$$ Now for the surfaces of revolution: demo of what the surfaces look like! Just ignore the “sides” and “little errors” of the y surface of revolution as that is a consequence of the bounds because Math3d does not have the productlog which would be used to graph it with the inverse function using the x surface of revolution formula there. The artifacts from the graph include the incomplete “spindle” and overlapping of the two surfaces which is just from approximation and technology errors.

The surface area about the x-axis is the sum of the two surface areas as the two “sides” of the graph do not intersect. The sum of square roots into one square root was tried, but did not work in the sum which is curious,$\sqrt a \pm\sqrt b=\sqrt{a+b\pm \sqrt{ab}}$

$$S_x= 2π\lim_{n\to ∞ }\frac 1n \sum_{k=1}^n\bigg(\frac kn\bigg)^{\frac{k}{n}} \bigg(\sqrt{1+\bigg(\frac kn\bigg)^{\frac{2k}{n}}(ln(x)+1)^2}+ \sqrt{1+\bigg(\frac kn\bigg)^{-\frac{2k}{n}}(ln(x)+1)^2}\bigg)\bigg)=17.46376...$$

The surface area for f(x)=y around the y axis would have the Δx multiplied out by the factored $x_k$:

$$S_y= 2π \lim_{n\to ∞ } \frac 1{n^2} \sum_{k=1}^nk \biggl(\sqrt{1+\bigg(\frac kn\bigg)^{\frac{2k}{n}}(ln(x)+1)^2}+ \sqrt{1+\bigg(\frac kn\bigg)^{-\frac{2k}{n}}(ln(x)+1)^2}\biggr) =7.6398...$$

The volumes of revolution are surprisingly solvable using summation notation, and probably other special functions and the definition or the reimann integral defenition, but that would be “boring”, and are fun to see the steps behind them. The interchanging of the integral and summation proof is left as an exercise and probably could be done with Fubuni’s theorem or a convergence theorem. One crucial step in these solutions is using the power series for $e^x$ and the binomial theorem of $(1\pm x)^n$.

For the volume about the x-axis and no other axis, for simplicity, of the red lamina, it would be easier to use the shell technique. The area of a shell, or a ring, is $πR^2$ where R is defined in the integral as the difference of the two functional radii with each squared reminiscent of the Pythagorean theorem. The shape is the sum of the volumes of the cylinders as the width approach zero so that we get the integral, expand into a power series of the natural base exponential function and factor to get over the defined x in our problem because we are using the ring method over this axis: $$V_x=π\int_0^1 r_1^2-r_2^2 dx= π\int_0^1 x^{-2x}-x^{2x}dx=π\int_0^1 \sum_{n=0}^ ∞\frac{2^nx^nln^n(x)((-1)^n-1)}{n!}dx$$

The next step in finding the exact form of this volume would be to isolate the integrand into the integral and factor out the constants to then integrate similarly like in the last step of the question and simplify. $$V_x=\sum_{n=0}^ ∞\frac{2^n((-1)^n-1)}{n!}\int_0^1 x^nln^n(x)dx=\frac π2 \sum_{n=1}^ ∞ \frac{2^n((-1)^n+1)}{n^n}=3.34229990…$$

Finally, the volume about the line y=1 may be more easily done using the cylinder method will give $V_y=2πrh$ which will be integrated over the x values because when using this rule, one integrates over the opposite “side” than the axis in 2d euclidean space. If we used the other rule for this volume or the cylinder rule to find V_x, we would have to get ugly values of the incomplete gamma function as $$.6902...=\sqrt[-e]e\le y \le \sqrt[e]e=1.4446…$$

The defined surface would is a modified cylinder which has radius of 1-x as the most the radius can to to is the max of the x value in the lamina with the radius being the “remaining value”, i.e. x+r=1,and height of $x^{-x}-x^x$ just like when finding the area of the lamina we subtracted the “inner” curve from the “outer” curve to find the rectangle height. This all means our volume is: $$V_y=2π\int_0^1 rh\mathrm dx=2π\int_0^1 (x^x-x^{-x})(1-x)dx=2π\int_0^1 x^{-x}-x^x-x^{1-x}+x^{1+x}dx$$

Then we do the same as the last volume problem and we already know the integral of $x^{-x}-x^x$ from 0 to 1 as that is simply the conjugate of the area in the question. This summation will not be enough to integrate because we will have a new integration problem with the $x^{1\pm x}$ terms. This is why a second summation will be needed. Obviously, just like in the last example, we can combine all summations into one at the end with the same index and only another one for the inner sum. This is simply because they are in the same problem and will converge and integrating similarly to the question is: $$\int_0^1 x^{1+x}-x^{1-x}dx=\int_0^1 \sum_{n=0}^ ∞ \bigg(\frac{(1+x)^n ln^n(x)}{n!}-\frac{(1-x)^n ln^n(x)}{n!}\bigg)dx= \sum_{n=0}^ ∞ \frac 1{n!} \int_0^1 (1+x)^n ln^n(x)-(1-x)^n ln^n(x)dx= \sum_{n=0}^ ∞ \frac 1{n!} \int_0^1 ln^n(x)\sum_{k=0}^n\binom{n}{k}x^k-ln^n(x)\sum_{k=0}^n(-1)^nx^kdx=\sum_{n=0}^∞\sum_{k=0}^n \binom{n}{k} \frac {(-1)^{n+1}}{(k+1)^{n+1}}+ \sum_{n=0}^∞\sum_{k=0}^n \binom{n}{k} \frac {(-1)^{n+k+1}}{(k+1)^{n+1}}=-.225439...$$.

Negating this expression and multiplying by 2π will give the volume about the y-axis as a bonus value: $$V_y^*=2π\int_0^1 x(x^{-x}-x^x)dx= -2π\bigg(\sum_{n=0}^∞\sum_{k=0}^n \binom{n}{k} \frac {(-1)^{n+1}}{(k+1)^{n+1}}+ \sum_{n=0}^∞\sum_{k=0}^n \binom{n}{k} \frac {(-1)^{n+k+1}}{(k+1)^{n+1}}\bigg)=2π\sum_{n=0}^{\infty}\sum_{k=0}^n\binom nk \frac{(-1)^{k+n}+(-1)^n}{k^n} =2π\sum_{n=0}^{\infty}\sum_{k=0}^n\binom nk \frac{(-1)^n+1}{(-k)^n}=1.416476…$$

Thanks to @Uwe for finding a method for simplifying this result:$$V_y^*=\sum_{n=1}^{\infty}(2n+1)^{-2n}$$

All in all, the answer can be partially simplified if a substitution is made and a desmos shortcut which was accidentally discovered and is linked below is carried out: $$V_y=2π\biggl(\sum_{n=0}^∞\sum_{k=0}^{n+1} \binom{n}{k} \frac {(-1)^{n+1}}{(k+1)^{n+1}}+\sum_{n=0}^∞\sum_{k=0}^{n+1} \binom{n}{k} \frac {(-1)^{n+k+1}}{(k+1)^{n+1}}+\sum_{n=1}^∞ \frac{(-1)^n+1}{n^n}\biggl)= 2π\sum_{n=1}^∞\biggl(\frac{(-1)^n+1}{n^n}+\sum_{k=0}^{n-1} \binom{n-1}{k-1} \frac {(-1)^{n+1}((-1)^k+1)}{k^n}\biggl)=-2π\sum_{n=1}^∞\sum_{k=1}^{n-1} \binom{n-1}{k-1}\frac{(-1)^k+1}{(-k)^n}=$$ $$-2π\sum_{n=1}^∞\sum_{k=1}^{n-1} \binom{n-1}{k-1}\frac{(-1)^{k+n}+(-1)^n}{k^n}=1.774473…$$ Proof of $V_y$

There is a simplified from @Uwe as well: $$V_y= 4\pi\sum_{r=1}^\infty\left((2r)^{-2r}-(2r+1)^{-2r}\right) $$