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The answer can be in the form of two defined constants: $$A = \frac{1}{1^1}+\frac{1}{2^2} + \frac{1}{3^3} + \cdots$$ $$B = \frac{1}{2^2} + \frac{1}{4^4} + \frac{1}{6^6} + \cdots $$

I would highly appreciate it if you included your thought process in the answer itself.


As pointed out by zz20s , the answer to this integral can be seen on the wikipedia page: Sophomore's dream and the page accompanies a proof too

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    $\begingroup$ Sophomore's dream $\endgroup$ – zz20s Jan 9 '18 at 14:47
  • $\begingroup$ $$e^{x\log x}=\sum_{n\geq 0}\frac{(x\log x)^n}{n!}$$ and $$\int_{0}^{1}(x\log x)^n\,dx = \frac{(-1)^n n!}{(n+1)^{n+1}}$$ so the answer is $A-B$. $\endgroup$ – Jack D'Aurizio Jan 9 '18 at 14:58
  • $\begingroup$ It should be $A - 2B$, shouldn't it? $\endgroup$ – Arpit Saxena Jan 9 '18 at 15:06
  • $\begingroup$ Your question is a part of: math.stackexchange.com/questions/2019859/… $\endgroup$ – user90369 Jan 9 '18 at 15:50
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Note: there is no known closed form for $\int x^x \, \mathrm dx$.

But since we have bounds, we can use froggie's approach in this answer.

Recall that: $$\int_0^1 x^x\, \mathrm dx = \sum_{k=0}^\infty \frac{1}{k!}\int_0^1x^k(\log x)^k\, \mathrm dx$$

Make the substitution $u = -\log x$

$$\int_0^1 x^x\, \mathrm dx = \sum_{k=0}^\infty \frac{1}{k!}\int_0^1x^k(\log x)^k\, \mathrm dx = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\int_0^\infty e^{u(k+1)}u^k\, \mathrm du$$$$ = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^{u(k+1)}((k+1)u)^k\, \mathrm du$$

Make the substitution $t = (k+1)u$

$$\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^tt^k\, \mathrm dt = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1),$$ where $\Gamma$ is the usual Gamma function. Since $\Gamma(k+1) = k!$, the final expression is $$ \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$

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I have written a detailed explanation to find the integral of a general expression of the form $f(x) =x^{cx^a} $ from $0$ to $1$ here.

$$\int_{0}^{1} x^{cx^a}\, dx = 1- \frac{c}{(a+1)^2} + \frac{c^2}{(2a+1)^3} - \frac{c}{(3a+1)^4}+\ldots$$

On substituting $c =1$ and $a=1$, we get the desired result.

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