MATH SOLVE

4 months ago

Q:
# So, my buddy is new to doing Calculus and needs help understanding this equation, it would be very appeciated for some help

Accepted Solution

A:

To evaluate the integral, rewrite the integrand as[tex]x^{-x}=e^{\ln x^{-x}}=e^{-x\ln x}[/tex]Recall that[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}\implies x^{-x}=\sum_{n=0}^\infty\frac{(-x\ln x)^n}{n!}[/tex]The leftmost sum is the well-known power series expansion for the function [tex]f(x)=e^x[/tex]. In the rightmost sum, we just replace [tex]x[/tex] with [tex]-x\ln x[/tex].This particular power series has a property called "uniform convergence". Roughly speaking, it's a property that says a sequence of functions [tex]f_n(x)[/tex] converges to some limiting function [tex]f(x)[/tex] in the sense that [tex]f_n(x)[/tex] and [tex]f_{n+1}(x)[/tex] 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[tex]\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[/tex]The integral can be tackled with a substitution,[tex]x=e^{-u/(n+1)}\implies-(n+1)\ln x=u\implies\mathrm dx=-\dfrac1{n+1}e^{-u/(n+1)}\,\mathrm du[/tex]so that the integral is equivalent to[tex]\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[/tex][tex]=\displaystyle\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty e^{-u}u^n\,\mathrm du[/tex]The remaining integral reduces to [tex]n![/tex], which you can derive for yourself via integration by parts/power reduction.So we have[tex]\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}}[/tex]which is the same as[tex]\displaystyle\sum_{n=1}^\infty\frac1{n^n}=\sum_{n=1}^\infty n^{-n}[/tex]and hence the identity.