Definite Integral of \(x^{x}\) from \(x=0\) to \(x=1\)

\(\text{Prove the following integral:}\)

\[\int_{0}^{1}{x^xdx}=\ \frac{1}{{\ 1}^1\ }-\frac{1}{{\ 2}^2\ }+\frac{1}{{\ 3}^3\ }-\frac{1}{{\ 4}^4\ }+\cdots+\frac{{\ \left(-1\right)}^{n-1}\ }{n^n}+\cdots\]

\[\]

\[\]

By expanding \(x^{x}\) as a series:

\[x^x=e^{\ln\left(x^x\right)}=e^{x\ln\left(x\right)}=\sum_{n=0}^{\infty}\frac{{\left(x\ln\left(x\right)\right)}^n}{n!}\]

\[\int_{0}^{1}{x^xdx}=\int_{0}^{1}{\sum_{n=0}^{\infty}\frac{{\left(x\ln\left(x\right)\right)}^n}{n!}dx}=\sum_{n=0}^{\infty}{\frac{1}{n!}\int_{0}^{1}{x^n\left(\ln\left(x\right)\right)^ndx}}\]

\[\]

Make the substitution of \(u=-\ln (x)\), following by \(t=u(n+1)\) and you get that:

\[\]

\[\int_{0}^{1}{x^n\left(\ln\left(x\right)\right)^ndx}=\int_{\infty}^{0}{\left(e^{-u}\right)^n\left(-u\right)^n\left(-e^{-u}\right)du}=\left(-1\right)^n\int_{0}^{\infty}{e^{-u\left(n+1\right)}u^ndu}\]

\[\left(-1\right)^n\int_{0}^{\infty}{e^{-u\left(n+1\right)}u^ndu}=\left(-1\right)^n\int_{0}^{\infty}{e^{-t}\frac{t^n}{\left(n+1\right)^{n+1}}dt}=\frac{\left(-1\right)^n}{\left(n+1\right)^{n+1}}\int_{0}^{\infty}{e^{-t}t^ndt}\]

\[\]

Using the gamma function:

\[\int_{0}^{\infty}{e^{-t}t^ndt}=\Gamma(n-1)=n!\]

\[\]

Therefore:

\[\begin{align*} \int_{0}^{1}{x^xdx}
&= \sum_{n=0}^{\infty}{\frac{1}{n!}\int_{0}^{1}{x^n\left(\ln\left(x\right)\right)^ndx}} \\ &= \sum_{n=0}^{\infty}{\frac{{\left(-1\right)}^n}{n!}\int_{0}^{\infty}{e^{-u\left(n+1\right)}u^ndu}} \\ & = \sum_{n=0}^{\infty}{\frac{\left(-1\right)^n\ }{n!\left(n+1\right)^{n+1}}\int_{0}^{\infty}{e^{-t}t^ndt}} \\ & = \sum_{n=0}^{\infty}\frac{{n!\left(-1\right)}^n\ }{n!\left(n+1\right)^{n+1}} \\ & = \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}\ }{n^n} \\ & = \frac{1}{{\ 1}^1\ }-\frac{1}{{\ 2}^2\ }+\frac{1}{{\ 3}^3\ }-\frac{1}{{\ 4}^4\ }+\cdots+\frac{{\ \left(-1\right)}^{n-1}\ }{n^n}+\cdots \ \ \ (\text{Proven!}) \\ \end{align*}\]

\(\\ \)

\(\\ \)

\(\text{Bonus:}\)

\[x^{-x}=e^{\ln\left(x^{-x}\right)}=e^{{-x}\ln\left(x\right)}=\sum_{n=0}^{\infty}\frac{{(-1)^{n}\left(x\ln\left(x\right)\right)}^n}{n!}\]

\[\int_{0}^{1}{x^{-x}dx}=\int_{0}^{1}{\sum_{n=0}^{\infty}\frac{{(-1)^{n}\left(x\ln\left(x\right)\right)}^n}{n!}dx}=\sum_{n=0}^{\infty}{\frac{(-1)^{n}}{n!}\int_{0}^{1}{x^n\left(\ln\left(x\right)\right)^ndx}}\]

\(\\ \)

\[\begin{align*} \int_{0}^{1}{x^{-x}dx}
&= \sum_{n=0}^{\infty}{\frac{(-1)^{n}}{n!}\int_{0}^{1}{x^n\left(\ln\left(x\right)\right)^ndx}} \\ &= \sum_{n=0}^{\infty}{\frac{{(-1)^{n}\left(-1\right)}^n}{n!}\int_{0}^{\infty}{e^{-u\left(n+1\right)}u^ndu}} \\ & = \sum_{n=0}^{\infty}{\frac{1}{n!\left(n+1\right)^{n+1}}\int_{0}^{\infty}{e^{-t}t^ndt}} \\ & = \sum_{n=0}^{\infty}\frac{n!}{n!\left(n+1\right)^{n+1}} \\ & = \sum_{n=1}^{\infty}\frac{1}{n^n} \\ & = \frac{1}{{\ 1}^1\ }+\frac{1}{{\ 2}^2\ }+\frac{1}{{\ 3}^3\ }+\frac{1}{{\ 4}^4\ }+\cdots+\frac{1}{n^n}+\cdots \end{align*}\]

Note by Jian Hau Chooi
6 days, 13 hours ago

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Nice use of gamma function .

Kelvin Hong - 5 days, 10 hours ago

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With that bonus:

If \[\displaystyle\int_{0}^{1}f(x)dx =\large \displaystyle\sum_{x=1}^{\infty}f(x)\]

then \(f(x)=x^{-x}\).

Blan Morrison - 5 days, 9 hours ago

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Wow! Is this the only solution?

Kelvin Hong - 3 days, 17 hours ago

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Interesting question! For starters, \(\displaystyle\lim_{x\rightarrow \infty}f(x)=0\)

Blan Morrison - 2 days, 21 hours ago

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