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I wonder how I could evaluate $$\lim_{a\rightarrow0}\int_a^\infty\frac{x!}{x^x}dx$$Where $x!$ is defined for all complex numbers with a positive real part. This is inspired by a limit I found on BlackPenRedPen. I tried using the Residue theorem using the quarter circle contour on the complex plane with a singularity at the origin, but I don't know how to use it.enter image description hereThis is the contour I am using. Consider the outer circle as a contour $\Gamma$. Let the smaller semicircle be $\gamma$. Let the radius of the big semicircle be $R$ and the radius of the smaller semicircle be $r$. Then $$\int_0^\infty f(z)dz=\lim_{(r,R)\rightarrow(0,\infty)}\left(\int_\Gamma+\int_\gamma+\int_r^Rf(z)dz+\int_{ri}^{Ri}f(z)dz\right)=2\pi i\sum_k\text{Res}(f, z_k)$$Where $f(z)=\dfrac{z!}{z^z}$, $0^0=1$, and $z_k$ are the poles of $f(z)$. After that I am stuck as I don't think $f(z)$ has any poles. Any ideas?

Kamal Saleh
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    Is that limit supposed to be $\lim_{a\rightarrow 0}$? Also we're using the convention that $x! = \Gamma(x + 1)$ for noninteger $x$, right? – eyeballfrog Feb 14 '23 at 17:10
  • You meant $\lim_{a\to0}$, I think – FShrike Feb 14 '23 at 17:11
  • You have to be really careful of $x^{-x}=\exp(-x\log x)$ as it features branch cuts. It is not holomorphic over the whole plane – FShrike Feb 14 '23 at 17:12
  • @eyeballfrog Yes, thanks! – Kamal Saleh Feb 14 '23 at 17:20
  • @FShrike That was what I meant, thanks! – Kamal Saleh Feb 14 '23 at 17:20
  • @FShrike About your note, does this mean that the integral doesn't exist? Or do we have to take a different path? I am a bit new to Complex Analysis. – Kamal Saleh Feb 14 '23 at 17:22
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    @KamalSaleh The integral exists because $x!/x^x$ is continuous on $[0,\infty)$ and decays very sharply for large $x$. But what it means is, depending on how you define $\log$, you may have to be wary. This may or may not affect residue calculations, because e.g. $x^x$ (principal value) is not holomorphic on $(-\infty,0]$ – FShrike Feb 14 '23 at 17:25
  • Indeed, sometimes we like branch cuts because they allow us to exploit the asymmetries to compute integrals (see "dogbone contours" for examples of this). I was just throwing it out there. To safely do this contour integration you're also going to need to know how to bound $\Gamma$ (which I don't...) – FShrike Feb 14 '23 at 17:26
  • You don't need a small quarter-circle contour centered at $0$. There is a removable discontinuity at that point. – Accelerator Feb 14 '23 at 23:39
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    Interestingly enough, $\int_0^\infty \frac{(a x)^x}{x!}dx$ has a closed form with Lambert W – Тyma Gaidash Feb 15 '23 at 00:23
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    The OEIS entry has information – Тyma Gaidash Feb 15 '23 at 14:22

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Not an answer, just two funny observations: \begin{align} \int_0^\infty\frac{x!}{x^x}\,dx&=\int_0^\infty\frac{dx}{(1+x\log x)^2}; \\\sum_{n=1}^\infty\frac{n!}{n^n}&=\int_0^1\frac{dx}{(1+x\log x)^2}. \end{align} The second one is shown using $(1-z)^{-2}=\sum_{n=1}^\infty nz^{n-1}$ and termwise integration.

For the first one, \begin{align} \int_0^\infty\frac{x!}{x^x}\,dx&=\int_0^\infty\int_0^\infty y^x e^{-y}\,dy\,\frac{dx}{x^x} \\\color{gray}{[\text{exchange}]}\quad&=\int_0^\infty e^{-y}\int_0^\infty(y/x)^x\,dx\,dy \\\color{gray}{[x=yz]}\quad&=\int_0^\infty ye^{-y}\int_0^\infty z^{-yz}\,dz\,dy \\\color{gray}{[\text{exchange}]}\quad&=\int_0^\infty\int_0^\infty ye^{-y(1+z\log z)}\,dy\,dz \\\color{gray}{[\text{do the inner}]}\quad&=\int_0^\infty\frac{dz}{(1+z\log z)^2}. \end{align}

metamorphy
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