I want to find the sum of the following series

$$\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}$$

Using theorems on integration and differentiation of series. I can set $t=\ln x+1$ so that I get $$\sum_{n \geq 1} \frac{t^n}{n^n}$$

But then I don't see how to proceed, since that $n^n$ is difficult to see as the result of a differentiation or integration. How can I see it?

  • 2
    $\begingroup$ If you calculate $\displaystyle \int\limits_0^1 (x^a)^{x^b}dx$ then you will get a series which includes your series. E.g. $\displaystyle \int\limits_0^1 x^x dx=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$ . $\endgroup$ – user90369 Nov 18 '16 at 13:25
  • $\begingroup$ @user90369 why don't you make this to a full anwer? $\endgroup$ – user159517 Nov 18 '16 at 13:29
  • $\begingroup$ @user159517 : Then I have to calculate it for which I have no time. But I hope someone will find a suitable link. :-) $\endgroup$ – user90369 Nov 18 '16 at 13:31

Only a hint:

For $\,a\geq 0\,$ it's $$\sum\limits_{k=0}^\infty\frac{x^k}{(ak+1)^{k+1}}=\int\limits_0^1 t^{-xt^a}dt $$

and the base for this result is calculating $\enspace\int\limits_0^1 t^b(\ln t)^cdt\enspace$.

Therefore with $a:=1$ and $x$ substituted by $\ln x+1$ we get $$\sum_{n \geq 1} \frac{(\ln x +1)^n}{n^n}=(\ln x+1)\int\limits_0^1 t^{-t(\ln x +1)}dt \enspace.$$

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  • $\begingroup$ It's very interesting to know which is the function that corresponds to the series $\sum_{n \geq 1} \frac{z^n}{n^n}$ ! $\endgroup$ – G Cab Nov 20 '16 at 17:03
  • $\begingroup$ could you please detail or give reference for the passage from $\enspace\int\limits_0^1 t^b(\ln t)^cdt\enspace$ to the rest ? thanks indeed. $\endgroup$ – G Cab Nov 22 '16 at 18:45
  • $\begingroup$ What you need here is $\int\limits_0^1 t^{ak}(\ln t)^k dt$ with $k\in\mathbb{N}_0$ . I haven't used a reference but you can calculate yourself : $\int\limits_0^1 t^{ak} (\ln t)^k dt = \frac{t^{ak+1}}{ak+1} (\ln t)^k|_0^1 - \frac{k}{ak+1}\int\limits_0^1 t^{ak} (\ln t)^{k-1} dt$ for $k\in\mathbb{N}$ and $ak+1\neq 0$ . $\endgroup$ – user90369 Nov 23 '16 at 11:22
  • $\begingroup$ $a>0$, of course ... $\endgroup$ – user90369 Jul 14 '17 at 11:10

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