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I came across the problem of finding the sum

$\sum_{k=1}^{\infty}\frac{1}{(2k)^k}$ and I proved that it is equal to the integral $$\int_{0}^{\infty}e^{-2x+xe^{-2x}}dx$$ but I'am unable to solve this integral even though I can find numerical approximations using Wolfram.

I welcome any solutions(Exact Value of the sum),comments,suggestions and possible generalisations to this problem.

nmasanta
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Sundara Narasimhan
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    Maybe it's time to define a new special function: $$\operatorname{Sun}(z) = \sum_{n=1}^\infty \frac{z^n} {n^n} $$In particular your sum is just $\operatorname{Sun} \left(\frac12\right) $ :) – Zacky Jul 13 '19 at 11:38
  • Yes ,unless we know something nice (or find something nice) about this function,this is not heading anywhere. – Sundara Narasimhan Jul 13 '19 at 12:21
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    @Zacky No need to define a new function $\operatorname{Sun}(z)$ as, funny enough, it is actually related to the Sophomore dream function $\operatorname{Sphd}(\alpha;z)$ :) – omegadot Jul 13 '19 at 12:34

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Interestingly, your sum is related to the so-called Sophomore dream function $\operatorname{Sphd} (\alpha;x)$ which is defined as $$\operatorname{Sphd}(\alpha;x) = \int_0^x t^{\alpha t} \, dt.$$ Of course, whether you accept this function as a member of the "standard" canon of special functions or not is still debatable.

In any event, to see this, enforcing a substitution of $x \mapsto -\frac{1}{2} \ln x$ in your integral yields \begin{align} \int_0^\infty e^{-2x + xe^{-2x}} \, dx &= \frac{1}{2} \int_0^1 e^{-\frac{x}{2} \ln x} \, dx\\ &= \frac{1}{2} \int_0^1 x^{-x/2} \, dx\\ &= \frac{1}{2} \operatorname{Sphd} \left (-\frac{1}{2};1 \right ). \end{align} Thus $$\sum_{n = 1}^\infty \frac{1}{2^n n^n} = \frac{1}{2} \operatorname{Sphd} \left (-\frac{1}{2};1 \right ).$$

Comments

  1. In general $$\sum_{n = 1}^\infty \frac{x^n}{n^n} = x \int_0^1 x t^{-xt} \, dt = x \operatorname{Sphd} (-x;1).$$

  2. For a discussion on what "counts" as a known function, see here.

  3. Finally, as they say, a function is only as important as it is useful.

omegadot
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