This is from a practice question for Putnam

Show that $$\int_{0}^{1}\frac{dx}{x^x} = 1+\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^4}+...$$

It seems to me that this integral is just equals to the discrete sum for all positive integers, but it only integrates from $[0,1]$??

  • $\begingroup$ can you integrate a Taylor expansion of $e^{-x \log x}$ $\endgroup$ – Cato Dec 5 '16 at 15:52
  • 2
    $\begingroup$ Sophomore's Dream $\endgroup$ – Biggs Dec 5 '16 at 16:02
  • $\begingroup$ Please have a look at my hint with the more general series math.stackexchange.com/questions/2019859/… and the comments there. It solves your problem. $\endgroup$ – user90369 Dec 5 '16 at 16:29

$$\frac{1}{x^x} = x^{-x} = e^{-x\ln(x)}$$

Integration range allow you to use Taylor expansion as

$$e^{-x\ln(x)} = \sum_{k = 0}^{+\infty} \frac{(-x\ln(x))^k}{k!}\ \text{d}x$$

Notice that $x\ln(x)$ is well defined in zero and it's zero.

$$\sum_{k = 0}^{+\infty}\frac{1}{k!}\int_0^1 x^k\ln^k(x)\ \text{d}x$$

And hereafter you can have some fun in integrating it by parts $k$ time or simply set a $k$, like $k = 10$ and perform the integration.

Due to the negative exponential, you can actually integrate it over a larger range, but still $[0, M]$


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