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How would you calculate the following integral? $$\int_0^1 x^xdx$$

I write my answer below. Don't be shy to speak out if you find the mistakes (if there are any) or have a better solution.

Blue
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YANGyu
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  • That is well-known: https://en.wikipedia.org/wiki/Sophomore%27s_dream – Martin R Dec 07 '23 at 07:58
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    Congratulations on re-discovering this. (Do not let the fact that it was already known discourage you.) Keep up the good work. – GEdgar Dec 07 '23 at 17:17

1 Answers1

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It's easy to notice that $$x^x=e^{\ln x^x}=e^{x\ln x}$$ Only to calculate $$\int_0^1 e^{x\ln x}dx$$ According to Taylor-expansion, we have $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ $$\Rightarrow e^{x\ln x}=\sum_{n=0}^\infty\frac{x^n(\ln x)^n}{n!}$$ $$\Rightarrow \int_0^1 e^{x\ln x}dx=\sum_{n=0}^\infty\frac{1}{n!}\int_0^1 x^n(\ln x)^n dx$$ Let $I_n=\int_0^\infty x^n(\ln x)^n dx$, integrate $I_n$ by part $$I_n=\bigg[\frac{x^{n+1}(\ln x)^n}{n+1}\bigg]_0^1-\frac n{n+1}\int_0^1(\ln x)^{n-1}x^ndx=-\frac n{n+1}\int_0^1(\ln x)^{n-1}x^ndx$$ Repeat this process for another $n-1$ times, we will get $$I_n=(-1)^n\frac {n!}{(n+1)^n}\int_0^1x^ndx=(-1)^n\frac {n!}{(n+1)^{n+1}}$$ $$\Rightarrow \int_0^1 e^{x\ln x}dx=\sum_{n=0}^\infty\frac{1}{n!}I_n=\sum_{n=0}^\infty\frac {(-1)^{n}}{(n+1)^{n+1}}$$ $$\Rightarrow \int_0^1 x^xdx=\sum_{n=0}^\infty\frac {(-1)^{n}}{(n+1)^{n+1}}$$

Gary
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YANGyu
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  • That converges fast enough to be calculatable and gives the correct value of about $0.78343051$. – Henry Dec 07 '23 at 08:07