i cant find this integral $\int_0^1 x^{-x} \mathrm{dx}$

Question

$$\int_0^1 x^{-x} \mathrm{dx} $$

Google "Sophomore's dream" for more about it. – Daniel Fischer Sep 24 '13 at 17:47 — Daniel Fischer, Sep 24 '13 at 17:47
here it is: $\int_0^1 x^{-x} \mathrm{d} x$. – azimut Sep 24 '13 at 19:04 — azimut, Sep 24 '13 at 19:04

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

1

Hint:

$$\int_0^1{x^{-x}dx} = \int_0^1{e^{\log x^{-x}}dx} = \int_0^1{\sum_{k=1}^{\infty}\frac{(-x)^k\log^k x}{k!}}dx= \sum_{k=0}^\infty \frac{1}{k!}\int_0^1 (-1)^kx^k(\log x)^k\,dx $$ for more detail see here how to integrate $\int\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx$

edited Apr 13 '17 at 12:21

Community

1

answered Sep 24 '13 at 17:57

M.H

11,498
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score 0 · Answer 2 · answered Sep 24 '13 at 17:46

0

The value is approximately 1.29129, as shown in this link:

http://www.wolframalpha.com/input/?i=definite+integral+of+x^%28-x%29+from+0+to+1

answered Sep 24 '13 at 17:46

score 0 · Answer 3 · answered Sep 24 '13 at 18:19

0

It's equal to 1 + 1/4 + 1/27 + 1/64 + .... +1/n^n ...

answered Sep 24 '13 at 18:19

Syd Henderson

2,061

Predictably, I proved the similar result for x^x when I was a sophomore. – Syd Henderson Sep 24 '13 at 18:26

i cant find this integral $\int_0^1 x^{-x} \mathrm{dx}$

3 Answers3