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According to Sophomore's Dream:

$$\int_{0}^{1} x^x dx = -\sum_{n=1}^{\infty} (-n)^{-n}$$

Now, I want to find $\int_{0}^{2} x^x dx$. Because they are quite similar, is there any way I could use Sophomore's Dream to help me find an answer for the integral in question? Or do I have to use other methods? Either way, what would I do?

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    $\begingroup$ you can also go back to the proof and apply the change of variable $x = 2 e^{y}$ in the obtained $\int_0^2 x^n (\log x)^n dx$ terms $\endgroup$
    – reuns
    Commented Apr 16, 2016 at 21:17
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    $\begingroup$ According to fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function $$\int_0^2 x^x dx=Sphd(1\:;\:2)$$ and from numerical calculus $\simeq 2.83388$ Infinite series are given in the above paper. $\endgroup$
    – JJacquelin
    Commented Apr 17, 2016 at 4:30

1 Answer 1

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Use the series $$x^x=\sum_{n=0}^\infty \frac {a_n}{b_n}\,(x-1)^n$$ The coefficients $a_n$ and $b_n$ correspond respectively to sequences $A082525$ and $A082526$ in $OEIS$.

$$\int_0^2 x^x\,dx=2\sum_{n=0}^\infty \frac {a_{2n}}{(2n+1)\,b_{2n}}$$ Computing partial sums and converting the result to decimals

$$\left( \begin{array}{cc} p & \sum_{n=0}^{2p} \\ 1 & 2.66667 \\ 2 & 2.80000 \\ 3 & 2.82143 \\ 4 & 2.82663 \\ 5 & 2.82899 \\ 6 & 2.83037 \\ 7 & 2.83125 \\ 8 & 2.83185 \\ 9 & 2.83226 \\ 10 & 2.83256 \\ 20 & 2.83355 \\ 30 & 2.83373 \\ 40 & 2.83380 \\ 50 & 2.83383 \\ 60 & 2.83384 \\ 70 & 2.83385 \\ 80 & 2.83386 \\ \end{array} \right)$$

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