Closed form for $\int_0^1\left(\frac1x-1\right)^x\mathrm dx$?

Question

Is there a closed form for $I=\int_0^1\left(\frac1x-1\right)^x\mathrm dx\approx 0.838104577482$ ?

That is, can $I$ be expressed in terms of known functions (elementary or otherwise) or established constants? I know this is not is not a strict definition of "closed form"; I'm just trying to see what can be said about this integral.

Wolfram does not give a closed form.

Here is the graph of $y=\left(\frac1x-1\right)^x$.

(It is close to the cubic curve $y=4x^3-8x^2+3x+1$ for $0<x<1$.)

Context

I'm interested in Pascal's triangle, in particular, numbers that represent the entire triangle.

I found that if we take the $n$th root of each number in Pascal's triangle, where $n$ is each number's row number, then the mean value of the resulting numbers for the entire triangle is $2\int_0^1\left(\frac1x-1\right)^x\mathrm dx\approx 1.67620915496$. I'm wondering if this value has a closed form.

Related fun fact: If we take the log of the $n$th root of each number in Pascal's triangle, where $n$ is each number's row number, then the mean value of the resulting numbers for the entire triangle is $\dfrac12$ and the variance is $\dfrac{21-2\pi^2}{36}$.

Related integrals

$\int_0^1 \frac{1}{x^x}\mathrm dx\approx 1.2913$ (which equals $\sum\limits_{n=1}^\infty \frac{1}{n^n}$) does not have a closed form.

$\int_0^\infty \frac{1}{x^x}\mathrm dx\approx 1.9955$ does not have a closed form (but it can be shown to be less than $2$).

Answer 1

$$A=\left(\frac{1}{x}-1\right)^x \qquad \implies \qquad \log(A)=x \log\left(\frac{1}{x}-1\right)$$ Expand using Taylor $$\log(A)= -x \log (x)-x^2-\frac{x^3}{2}-\frac{x^4}{3}-\frac{x^5}{4}-\frac{x^6}{5 }-\frac{x^7}{6}+O\left(x^8\right)$$ Exponentiate and Taylor again $$A=e^{\log(A)}=1+\sum_{n=1}^\infty (-1)^n\,\frac{ P_n(L)}{n!}\, x^n \qquad L=\log(x)$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(L) \\ 1 & L \\ 2 & L^2-2 \\ 3 & L^3-6 L+3 \\ 4 & L^4-12 L^2+12 L+4 \\ 5 & L^5-20 L^3+30 L^2+20 L-30 \\ 6 & L^6-30 L^4+60 L^3+60 L^2-180 L+66 \\ 7 & L^7-42 L^5+105 L^4+140 L^3-630 L^2+462 L \\ 8 & L^8-56 L^6+168 L^5+280 L^4-1680 L^3+1848 L^2-496 \\ \end{array} \right)$$ We shall use $$I_{m,n}=\int_0^1 x^m\, \log ^n(x)\,dx= (-1)^n \frac {n!} {(m+1)^{n+1} }$$

Using the terms given in the table, we have $$\frac{14020440620128474409998907}{16727798278915463577600000}$$ which is $\color{red}{0.8381}52$

Answer 2

$\DeclareMathOperator\W{W}$ After a bit of integral algebra and an inverse substitution, one uses the product logarithm $\W(x)$:

$$\int_0^1\left(\frac1x-1\right)^xdx=\int_0^1\frac{dx}{\W(-x\ln(x))+1}$$

Also, Lagrange reversion, after simplification, gives:

$$\frac1{\W(x)+1}=1+\sum_{n=1}^\infty\frac{(-nx)^n}{n!},|x|<\frac1e$$

Therefore:

$$\int_0^1\frac{dx}{\W(-x\ln(x))+1}= \int_0^1 1+\sum_{n=1}^\infty\frac{(nx\ln(x))^n}{n!}dx,x\ne\frac1e$$

The single point where the series diverges can be ignored. The integral then is:

$$\boxed{\int_0^1\left(\frac1x-1\right)^xdx=1+\sum_{n=1}^\infty\frac{(-n)^n}{(n+1)^{n+1}}}$$

To put it in terms of a “sophomore dream” type result, the OP’s mean is:

$$\boxed{\int_0^1\frac{(1-x)^{x-1}}{x^x}dx=2+2\sum_{n=2}^\infty\frac{(1-n)^{n-1}}{n^n}}$$

shown here: