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

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

∫01xxdx= 1 11 −1 22 +1 33 −1 44 +⋯+ (−1)n−1 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=∫01∑n=0∞(xln⁡(x))nn!dx=∑n=0∞1n!∫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(e−u)n(−u)n(−e−u)du=(−1)n∫0∞e−u(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)n∫0∞e−u(n+1)undu=(−1)n∫0∞e−ttn(n+1)n+1dt=(−1)n(n+1)n+1∫0∞e−ttndt\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:

∫0∞e−ttndt=Γ(n−1)=n!\int_{0}^{\infty}{e^{-t}t^ndt}=\Gamma(n-1)=n!

Therefore:

∫01xxdx=∑n=0∞1n!∫01xn(ln⁡(x))ndx=∑n=0∞(−1)nn!∫0∞e−u(n+1)undu=∑n=0∞(−1)n n!(n+1)n+1∫0∞e−ttndt=∑n=0∞n!(−1)n n!(n+1)n+1=∑n=1∞(−1)n−1 nn=1 11 −1 22 +1 33 −1 44 +⋯+ (−1)n−1 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:}

x−x=eln⁡(x−x)=e−xln⁡(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!}

∫01x−xdx=∫01∑n=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}}

\\

∫01x−xdx=∑n=0∞(−1)nn!∫01xn(ln⁡(x))ndx=∑n=0∞(−1)n(−1)nn!∫0∞e−u(n+1)undu=∑n=0∞1n!(n+1)n+1∫0∞e−ttndt=∑n=0∞n!n!(n+1)n+1=∑n=1∞1nn=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}

#Calculus

Note by Jian Hau Chooi
2 years, 10 months ago

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Comments

Nice use of gamma function .

Kelvin Hong - 2 years, 10 months ago

With that bonus:

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

then f(x)=x−xf(x)=x^{-x}.

Blan Morrison - 2 years, 10 months ago

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Wow! Is this the only solution?

Kelvin Hong - 2 years, 10 months ago

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

Blan Morrison - 2 years, 10 months ago
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