integration - How to calculate an approximation of $int_0^infty left(x^xoperatorname{ mod }e^xright)frac{dx}{x^x}?$

This morning I was playing with the congruence implemented as Mod[m, n] in Wolfram Language, see the table from this MathWorld. I did these experiments with Wolfram Alpha online calculator:

Plot Mod[x^x, e^x]/x^x, for 0

and also with codes like this

integrate Mod[x^x, e^x]/x^x dx, from x=0 to 3

Question. Is it possible to deduce an approximation of $$\int_0^\infty \left(x^x\operatorname{ mod }e^x\right)\frac{dx}{x^x}?$$
Many thanks.

Answer

$I\approx 3.0547$. Alright well here's one decent way to get an approximation. We have that
$$(x^x\text{ mod } e^x)x^{-x}=\frac{\text{frac}((x/e)^x)}{(x/e)^x}$$

So we have that
$$I=\int_0^{\infty}(x^x\text{ mod } e^x)x^{-x}dx= \int_0^{\infty}\frac{\text{frac}((x/e)^x)}{(x/e)^x}dx=e+\int_e^{\infty}\frac{\text{frac}((x/e)^x)}{(x/e)^x}dx$$
Now let $u=(x/e)^x$. We then have that $du=(x/e)^x\ln(x)dx=u\ln(x)dx$. Thus
$$I=e+\int_1^{\infty}\frac{\text{frac}(u)}{u^2\ln(x)}du$$
Moreover we have that $\ln(x)=W(\ln(u)/e)+1\approx\ln(\ln(u))$, to a fairly accurate degree when $u > 3$. Thus
$$I\approx e+\int_1^{n}\frac{\text{frac}(u)}{u^2(W(\ln(u)/e)+1)}du+\int_n^{\infty}\frac{\text{frac}(u)}{u^2\ln(\ln(u))}du$$
However
$$\int_n^{\infty}\frac{\text{frac}(u)}{u^2\ln(\ln(u))}du\approx \int_n^{\infty}\frac{\text{frac}(u)}{u^2}du=1-\gamma - \int_1^{n}\frac{\text{frac}(u)}{u^2}du$$
(numerically a difference of about $0.002$). So
$$I\approx e+1-\gamma +\int_1^{n}\frac{\text{frac}(u)}{u^2}\cdot\left(\frac{1}{W(\ln(u)/e)+1}-1\right)du$$

Letting $n=100$ we get $I\approx 3.0547$. Of course you may just want to do standard approximations on
$$I=e+\int_1^{\infty}\frac{\text{frac}(u)}{u^2(W(\ln(u)/e)+1)}du$$
since now to integrand doesn't require computing values of extreme magnitudes like $x^x$.