Question

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

Answer 1

To evaluate the integral, rewrite the integrand as

$x^{-x}=e^{\ln x^{-x}}=e^{-x\ln x}$

Recall that

$e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}\implies x^{-x}=\sum_{n=0}^\infty\frac{(-x\ln x)^n}{n!}$

The leftmost sum is the well-known power series expansion for the function $f(x)=e^x$. In the rightmost sum, we just replace $x$ with $-x\ln x$.

This particular power series has a property called "uniform convergence". Roughly speaking, it's a property that says a sequence of functions $f_n(x)$ converges to some limiting function $f(x)$ in the sense that $f_n(x)$ and $f_{n+1}(x)$ 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

$\displaystyle\int_0^1x^{-x}\,\mathrm dx=\int_0^1\sum_{n=0}^\infty\frac{(-x\ln x)^n}{n!}\,\mathrm dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1(x\ln x)^n\,\mathrm dx$

The integral can be tackled with a substitution,

$x=e^{-u/(n+1)}\implies-(n+1)\ln x=u\implies\mathrm dx=-\dfrac1{n+1}e^{-u/(n+1)}\,\mathrm du$

so that the integral is equivalent to

$\displaystyle\int_0^1(x\ln x)^n\,\mathrm dx=\int_\infty^0\left(e^{-u/(n+1)}\right)^n\left(-\frac u{n+1}\right)^n\left(-\frac1{n+1}e^{-u/(n+1)}\right)\,\mathrm du$

$=\displaystyle\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty e^{-u}u^n\,\mathrm du$

The remaining integral reduces to $n!$, which you can derive for yourself via integration by parts/power reduction.

So we have

$\displaystyle\int_0^1x^{-x}\,\mathrm dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\cdot\frac{(-1)^nn!}{(n+1)^{n+1}}=\sum_{n=0}^\infty\frac1{(n+1)^{n+1}}$

which is the same as

$\displaystyle\sum_{n=1}^\infty\frac1{n^n}=\sum_{n=1}^\infty n^{-n}$

and hence the identity.