I'm looking to evaluate $$\int_0^1 x^{1/x} dx$$ So far, I have that $$ \int_0^1 x^\frac{1}{x} dx \Rightarrow \int_0^1 e^{\frac{\ln x}{x}} dx \Rightarrow \int_0^1 \sum_{n \geq 0} \frac{\ln^nx}{x^nn!} dx $$ However, I know that $ \int_0^1 \frac{\ln x}{x} dx $ diverges, so I can't see a justification for switching the integral and sum. Furthermore, substituting $x = e^\frac{-u}{n+1}$ yields that the above is equivalent to (assuming my calculations were right) $$ \sum_{n\geq0} \frac{(-1)^n}{(1-n)^{1+n}} $$ which is nonsensical. This thought process can be used to solve the "sophomore's dream" integral, which is $\int_0^1 x^{-x} dx = \sum_{n\geq1} n^{-n}$, to which I have two questions:

  1. Why can't the same process be used? What changes between the two expressions? I suspect is has something to do with switching the integral and sum, but being a power series for $e^x$ I believe it should be justifiable.
  2. If the above is true, what is another approach to take for this integral?

For some context, I have barely cracked into analysis, so if the answer is rudimentary, this is why. However, its very clear that the original integral will have a finite value, and being a cousin of the sophomore's dream, I suspect it could have a solution of the same form.

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    $\begingroup$ A175999. $\endgroup$ – g.kov Dec 28 '19 at 2:56
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    $\begingroup$ What value does the integrand take at $x=0,$ for example? Is it continuous at this endpoint? Is this a proper integral? $\endgroup$ – Allawonder Dec 28 '19 at 8:58
  • $\begingroup$ My key opens my lock; why wouldn't it open any other lock? That's just the way things work. $\endgroup$ – Ivan Neretin Dec 28 '19 at 12:11
  • $\begingroup$ @Allawonder It oscillates between $-\infty$ and $\infty$ depending on n, but to that I would ask why $e^{\frac{lnx}{x}}$ can't be represented as a power series in the first place because I know for a fact its integral converges. $\endgroup$ – Suchetan Dontha Dec 28 '19 at 14:30
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    $\begingroup$ In this case, the exponent in $e^{f(x)}$ is unbounded, so every partial sum of the power series is a very bad approximation to the integrand on part of the interval. So bad that the approximation isn't even integrable. The same thing happens if you use the exponential series in $\Gamma(a) = \int_0^{\infty} t^{a-1}e^{-t}\,dt$. $\endgroup$ – Daniel Fischer Dec 28 '19 at 15:43

The difference between $x^x$ and $x^{1/x}$ is not obvious when you restrict your view to the real line. However, the singularity of $x^{1/x}$ at 0 is much more ill-behaved than that of $x^x$. If you approach 0 from along the negative axis the difference is clear: For $t\in \mathbb R^+$:\begin{eqnarray} (-t)^{-t} &=& e^{-t\log(-t)} = e^{-t\log t - i\pi t} = t^{-t}\left(\cos(\pi t) - i \sin(\pi t)\right)\\ (-t)^{\frac{1}{-t}} &=& e^{\frac{\log(-t)}{-t}} = e^{-\frac{\log t}{t} - i\frac{\pi}{t}} = t^{-\frac1t}\left(\cos\left(\frac\pi t\right) - i \sin\left(\frac\pi t\right)\right) \end{eqnarray} The first is well behaved as $t$ approaches 0, whereas the second diverges to infinity in magnitude while oscillating through all possible arguments. Similarly, approaching $x=0$ from any direction other than exactly the positive $x$ directly, $x^{1/x}$ behaves very badly while $x^x$ approaches 1.

This means that any trick to evaluating the integral needs to tread carefully around the point $x=0$. The method of expanding as a series does not succeed; as you observe, none of the terms of the series $\sum_{n=0}^\infty \frac1{n!} \left(\frac{\log x}{x}\right)^n$ is integrable on $(0,1)$. It's somewhat analogous to try evaluating $\int_0^\infty e^{-x}dx$ by expanding $e^{-x}$ as a series.

In terms of some sophomore dream like identity, I have in the past unsuccessfully tried a lot of things to attack this and similar integrals (for example, see my question from 2 years ago), and I don't expect anything as exciting as the sophomore's dream. However, there's one nice pseudo-identity that's too nice to pass up:$$ \int_0^1 x^{1/x} dx = \sum_{n=0}^\infty (-1)^n (n+2)^n $$ The sum clearly diverges, but its Borel sum is your integral. As proof: $$ \sum_{n=0}^\infty (-1)^n (n+2)^n \frac{z^n}{n!} = \frac{1}{2z}\frac{d}{dz} W(z)^2 $$ Where W is the Lambert W-function. Then we have the Borel sum:\begin{eqnarray} \sum_{n=0}^\infty (-1)^n (n+2)^n &\stackrel{B}{=}& \int_0^\infty e^{-z} \left(\frac{1}{2z}\frac{d}{dz} W(z)^2\right)dz \\&=& \int_0^\infty e^{-z} \frac{W(z)}{z} W'(z)dz = \int_0^\infty e^{-z} e^{-W(z)} W'(z) dz\\ &=& \int_0^\infty \exp\left(\frac{\log \left(e^{-W(z)}\right)}{e^{-W(z)}}\right) e^{-W(z)}W'(z)dz\\ &=& \int_0^1 x^{\frac1x} dx \end{eqnarray}

That's probably the nicest identity you'll find for this integral.

One thing you might think to try is expanding the series at $x=1$, but this doesn't actually help: Recall $-\sum_{n=1}^\infty H_n (1-x)^n = \frac{\log x}{x}$, where $H_n$ is the $n$th harmonic number. Using this and Bell polynomials, we find $$ x^{1/x} = e^{\frac{\log x}{x}} = \sum_{n=0}^\infty B_n(-1! H_1, -2! H_2,\dots, -n! H_n)\frac{(1-x)^n}{n!} $$ so theoretically $\int_0^1 x^{1/x} dx = \sum_{n=0}^\infty \frac{B_n(-1! H_1,\dots, -n! H_n)}{(n+1)!}$, but this sum does not converge. (Of course, we trivially have that it is Abel summable to $\int_0^1 x^{1/x} dx$). The sequence $a_n = \frac{B_n(-1! H_1,\dots, -n! H_n)}{n!}$ is very difficult to analyze. Since every antiderivative of $x^{1/x}$ behaves in a similarly bad way near $x=0$, $a_n/n^p$ is not summable for any $p$. Numerical calculations suggest $a_n$ diverges around as fast $e^{(\log n)^2}$. I don't expect there's a different summability method that would work better here than Abel summation.

One last trick which isn't as exciting as the sophomore's dream, but might interest you, we can use the Abel-Plana formula to find (using the substitution suggested by Allawonder in the comments to convert the integral to $\int_0^\infty (z+1)^{-(z+3)}dz$):$$ \int_0^1 x^{1/x} dx = \sum_{n=1}^\infty n^{-n-2} -\frac12 - 2\int_0^\infty \frac{e^{t\arctan t}\sin\left(\frac t2\log(1+t^2) + 3\arctan t\right)}{\left(e^{2\pi t} - 1\right)(1+t^2)^{\frac32}}dt $$

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    $\begingroup$ Outstanding solution! I wish I had seen your old post a week ago as it would have saved me several hours from tackling this problem. On a side note, I never knew about Borel summations, and upon further research you've opened my eyes to perturbation theory, which even looking at momentarily gets me excited to learn more about! Another side note, this problem arose when I tried to tackle $\int_0^1 x^{x^{x...}} dx$, as I found that it should be equivalent to $1 - \int_0^1 x^{\frac{1}{x}} dx$, which I have a feeling you would be able to better approach. $\endgroup$ – Suchetan Dontha Jan 10 at 4:04
  • $\begingroup$ Thanks! There's a small catch with $\int_0^1 x^{x^{x^\cdots}} dx$, which is that $x^{x^{x^\cdots}}$ doesn't converge for $x<e^{-e}$. I'm guessing you worked out that identity for the analytic continuation of it on the region of convergence (which has closed form $W(-\log x)/(-\log x)$)? Numerically it checks out but I'm not sure how you solved it. $\endgroup$ – Dark Malthorp Jan 10 at 4:27
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    $\begingroup$ I completely forgot about its convergence! I was going off of it graphically. When I plotted x^1/x I just noticed that it was the reflection of x^x^x... on y = x, which is no surprise as they're inverses. With that I thought that solving the integral of x^1/x would be simpler. Quite honestly I don't really have the background to do more than that. $\endgroup$ – Suchetan Dontha Jan 10 at 4:45
  • $\begingroup$ Ah of course! Geometry is my weakness. That does give me an idea though, I may amend my answer if I find something. $\endgroup$ – Dark Malthorp Jan 10 at 4:47
  • $\begingroup$ I've done some more research, and for $x \in (0, e^{-e})$ you can analytically continue $x^{x^{x...}}$ by defining it as the inverse of $y^{1/y}$ on $y \in (0, e)$. I think that is why it checks out numerically, and I believe its integral from 0 to 1 should be the Borel sum of $\sum_0^{\infty} (-1)^n n^{n-2}$. $\endgroup$ – Suchetan Dontha Jan 10 at 23:25

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