To evaluate the integral, rewrite the integrand as
![x^{-x}=e^{\ln x^{-x}}=e^{-x\ln x}](https://tex.z-dn.net/?f=x%5E%7B-x%7D%3De%5E%7B%5Cln%20x%5E%7B-x%7D%7D%3De%5E%7B-x%5Cln%20x%7D)
Recall that
![e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}\implies x^{-x}=\sum_{n=0}^\infty\frac{(-x\ln x)^n}{n!}](https://tex.z-dn.net/?f=e%5Ex%3D%5Cdisplaystyle%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7Bx%5En%7D%7Bn%21%7D%5Cimplies%20x%5E%7B-x%7D%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7B%28-x%5Cln%20x%29%5En%7D%7Bn%21%7D)
The leftmost sum is the well-known power series expansion for the function
. In the rightmost sum, we just replace
with
.
This particular power series has a property called "uniform convergence". Roughly speaking, it's a property that says a sequence of functions
converges to some limiting function
in the sense that
and
get arbitrarily close to one another. If you have an idea of what "convergence" alone means, then you can think of "uniform convergence" as a more powerful form of convergence.
Long story short, this property allows us to interchange the order of summation/integration to write
![\displaystyle\int_0^1x^{-x}\,\mathrm dx=\int_0^1\sum_{n=0}^\infty\frac{(-x\ln x)^n}{n!}\,\mathrm dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1(x\ln x)^n\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1x%5E%7B-x%7D%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5E1%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7B%28-x%5Cln%20x%29%5En%7D%7Bn%21%7D%5C%2C%5Cmathrm%20dx%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7B%28-1%29%5En%7D%7Bn%21%7D%5Cint_0%5E1%28x%5Cln%20x%29%5En%5C%2C%5Cmathrm%20dx)
The integral can be tackled with a substitution,
![x=e^{-u/(n+1)}\implies-(n+1)\ln x=u\implies\mathrm dx=-\dfrac1{n+1}e^{-u/(n+1)}\,\mathrm du](https://tex.z-dn.net/?f=x%3De%5E%7B-u%2F%28n%2B1%29%7D%5Cimplies-%28n%2B1%29%5Cln%20x%3Du%5Cimplies%5Cmathrm%20dx%3D-%5Cdfrac1%7Bn%2B1%7De%5E%7B-u%2F%28n%2B1%29%7D%5C%2C%5Cmathrm%20du)
so that the integral is equivalent to
![\displaystyle\int_0^1(x\ln x)^n\,\mathrm dx=\int_\infty^0\left(e^{-u/(n+1)}\right)^n\left(-\frac u{n+1}\right)^n\left(-\frac1{n+1}e^{-u/(n+1)}\right)\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1%28x%5Cln%20x%29%5En%5C%2C%5Cmathrm%20dx%3D%5Cint_%5Cinfty%5E0%5Cleft%28e%5E%7B-u%2F%28n%2B1%29%7D%5Cright%29%5En%5Cleft%28-%5Cfrac%20u%7Bn%2B1%7D%5Cright%29%5En%5Cleft%28-%5Cfrac1%7Bn%2B1%7De%5E%7B-u%2F%28n%2B1%29%7D%5Cright%29%5C%2C%5Cmathrm%20du)
![=\displaystyle\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty e^{-u}u^n\,\mathrm du](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cfrac%7B%28-1%29%5En%7D%7B%28n%2B1%29%5E%7Bn%2B1%7D%7D%5Cint_0%5E%5Cinfty%20e%5E%7B-u%7Du%5En%5C%2C%5Cmathrm%20du)
The remaining integral reduces to
, which you can derive for yourself via integration by parts/power reduction.
So we have
![\displaystyle\int_0^1x^{-x}\,\mathrm dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\cdot\frac{(-1)^nn!}{(n+1)^{n+1}}=\sum_{n=0}^\infty\frac1{(n+1)^{n+1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1x%5E%7B-x%7D%5C%2C%5Cmathrm%20dx%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7B%28-1%29%5En%7D%7Bn%21%7D%5Ccdot%5Cfrac%7B%28-1%29%5Enn%21%7D%7B%28n%2B1%29%5E%7Bn%2B1%7D%7D%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac1%7B%28n%2B1%29%5E%7Bn%2B1%7D%7D)
which is the same as
![\displaystyle\sum_{n=1}^\infty\frac1{n^n}=\sum_{n=1}^\infty n^{-n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5En%7D%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20n%5E%7B-n%7D)
and hence the identity.
Answer:
x-4
Step-by-step explanation:
(x^2-16)/(x+4) = x-4
Markup (or price spread) is the difference between the selling price of a good or service and cost.
Nope
Step-by-step explanation:
Volume of rectangular prism :
= Base Area × h = l × w × h
Volume of triangular prism :
= Base Area × h
Cuz the base area of triangular prism, the volume is :
(½ × l × w) × h
= ½ lwh
So, the volume of a triangular prism is half of a rectangular prism