How to calculate the integral of $x^{-x}$?

Asked Apr 22 '20 at 13:07

Active Apr 22 '20 at 13:17

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So, I know the integral of $\frac{1}{x^x}$ has not an elementary anti-derivative..

The question is to prove that:
The definite Integral of $x^{-x}$ from $0$ to $1$ equals $$\frac{1}{1^1} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} +...$$

This problem is an exercise at a book I have bought in the chapter of “definite integrals, improper integrals and integration of series”

Since the requested solution is given in the form of a series, I guess sb has to work by transforming $x^{-x}$ as a series (that’s how the writer solved a similar example) but this requires finding a series which converges at this function..

Any type of help would be appreciated thanks in advance!

edited Apr 22 '20 at 13:17

jameselmore

5,207

asked Apr 22 '20 at 13:07

Grigoris

1

This called the “Sophomore’s dream” integral. Worth looking that up. – Physical Mathematics Apr 22 '20 at 13:11
Haha yeah. Just changed that. I though the sophomore dream had a different sign on the exponent but looked it up. – Physical Mathematics Apr 22 '20 at 13:12
1

Possible duplicate of this. – Mason Apr 22 '20 at 13:21
Haha noway. I googled it and I saw freshman’s and sophomore’s dream. Too good to be true haha. I love maths even more after that. – Grigoris Apr 22 '20 at 13:25
"I thought the sophomore dream had a different sign on the exponent" It can refer to either or the pair, whereas the Freshman's dream is very different indeed. – J.G. Apr 22 '20 at 13:48

How to calculate the integral of $x^{-x}$?

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