# How to show that $\int_0^1x^{-x}dx = \sum_{n=1}^\infty n^{-n}$? [duplicate]

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How would I go about showing that $$\int_0^1x^{-x}dx = \sum_{n=1}^\infty n^{-n}$$

Right now my numerical analysis class is covering gaussian quadrature but we have also covered interpolation. I'm not sure how to prove this equality without using an estimation for the lefthand integral

## marked as duplicate by Martin R, Yanko, D. Thomine, Community♦Mar 22 at 21:51

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## 1 Answer

Substitute $$y=-\ln x$$ so the integral becomes $$\int_0^1\exp(-x\ln x)dx=\sum_{m\ge 0}\frac{(-1)^m}{m!}\int_0^1 x^m\ln^m xdx\\=\sum_{m\ge 0}\frac{1}{m!}\int_0^\infty y^m\exp[-(m+1)y] dy=\sum_{m\ge 0}\frac{1}{(m+1)^{m+1}}.$$