Definite Integral of xxx^{x} from x=0x=0 to x=1x=1

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

01xxdx= 1 11 1 22 +1 33 1 44 ++ (1)n1 nn+\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 xxx^{x} as a series:

xx=eln(xx)=exln(x)=n=0(xln(x))nn!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!}

01xxdx=01n=0(xln(x))nn!dx=n=01n!01xn(ln(x))ndx\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)u=-\ln (x), following by t=u(n+1)t=u(n+1) and you get that:

01xn(ln(x))ndx=0(eu)n(u)n(eu)du=(1)n0eu(n+1)undu\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}

(1)n0eu(n+1)undu=(1)n0ettn(n+1)n+1dt=(1)n(n+1)n+10ettndt\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:

0ettndt=Γ(n1)=n!\int_{0}^{\infty}{e^{-t}t^ndt}=\Gamma(n-1)=n!

Therefore:

01xxdx=n=01n!01xn(ln(x))ndx=n=0(1)nn!0eu(n+1)undu=n=0(1)n n!(n+1)n+10ettndt=n=0n!(1)n n!(n+1)n+1=n=1(1)n1 nn=1 11 1 22 +1 33 1 44 ++ (1)n1 nn+   (Proven!)\begin{aligned} \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{aligned}

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Bonus:\text{Bonus:}

xx=eln(xx)=exln(x)=n=0(1)n(xln(x))nn!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!}

01xxdx=01n=0(1)n(xln(x))nn!dx=n=0(1)nn!01xn(ln(x))ndx\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}}

\\

01xxdx=n=0(1)nn!01xn(ln(x))ndx=n=0(1)n(1)nn!0eu(n+1)undu=n=01n!(n+1)n+10ettndt=n=0n!n!(n+1)n+1=n=11nn=1 11 +1 22 +1 33 +1 44 ++1nn+\begin{aligned} \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{aligned}

Note by Jian Hau Chooi
1 year, 2 months ago

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With that bonus:

If 01f(x)dx=x=1f(x)\displaystyle\int_{0}^{1}f(x)dx =\large \displaystyle\sum_{x=1}^{\infty}f(x)

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

Blan Morrison - 1 year, 2 months ago

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Wow! Is this the only solution?

Kelvin Hong - 1 year, 2 months ago

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

Blan Morrison - 1 year, 2 months ago

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Nice use of gamma function .

Kelvin Hong - 1 year, 2 months ago

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