some math today.

F ind the following integrale and solve the equation for any positive a.

$f\left(x\right)=x^x$

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{x}^x\mathrm{dx}=?$

$x^{x^a}=a$

first note that, this function does not have real roots in the intervals (-1,0) , since that results the
root of a negative number. Therefore function is real in positive × axis.

Let's start with the integral;

we will use following two relations to rewrite the integral

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

$e^x=1+x+\frac{x^2}{2!}+\frac{x3}{3!}+\dots \; <!--\; horizontal\; ellipsis\; -->$

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{x}^x\mathrm{dx}=\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{e}^{x\ln x}\mathrm{dx}=\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\left(1+x\ln x+\frac{{\left(x\ln x\right)}^2}{2!}+\frac{{\left(x\ln x\right)}^3}{3!}+\dots \; <!--\; horizontal\; ellipsis\; -->\right)$

We can use integration by parts for each term , let's first calculate following general form

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{x}^m{\left(\ln x\right)}^n\mathrm{dx}={\left|\frac{x^{m+1}}{m+1}\times \; <!--\; multiplication\; sign\; -->{\left(\ln x\right)}^n\right|}_0^1-\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{x^{m+1}·\; <!--\; middle\; dot\; -->n}{m+1}·\; <!--\; middle\; dot\; -->\frac{{\left(\ln x\right)}^{n-1}}{x}\mathrm{dx}=-\frac{n}{m+1}.\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{x}^m{\left(\ln x\right)}^{n-1}\mathrm{dx}$

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{\left(xln\left(x\right)\right)}^n\mathrm{dx}={\left|\frac{x^{n+1}}{n+1}.{\left(\ln x\right)}^n\right|}_0^1\_n\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\frac{x^n.{\left(\ln x\right)}^{n-1}}{n+1}=-\frac{n}{n+1}\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{x}^n{\left(\ln x\right)}^{n-1}\mathrm{dx}$

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{\left(xln\left(x\right)\right)}^n\mathrm{dx}=-\frac{n}{n+1}·\; <!--\; middle\; dot\; -->\left(-\frac{\left(n-1\right)}{n+1}\right)·\; <!--\; middle\; dot\; -->\left(-\frac{\left(n-2\right)}{n+1}\right)={\left(-1\right)}^n·\; <!--\; middle\; dot\; -->\frac{n!}{{\left(n+1\right)}^{n+1}}$

Therefore,

$\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}{x}^x\mathrm{dx}=1-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}+\dots \; <!--\; horizontal\; ellipsis\; -->$

Now, what is the root of the following equation

where a is any positive number,

$x^{x^a}=a$

first we solve,

$x^a=a\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x=a^{1/a}$

actually this becomes the solution of above ean as well

$x^{x^a}=9\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x^{{\left(a^{1/a}\right)}^9}=a\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x^a=a\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x=a^{1/a}$

It happens to be any number of powers give
the same solution

${\left[\left(x^{x^{x^x}}\right)\right]}^a=a\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x=a^{1/a}$

These are called tower eqns, here is a well known problem:

$x^{xx⋰\; <!--\; up\; right\; diagonal\; ellipsis\; -->}=2\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x=?$

power of × is repeating itself so

$x^2=2\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x=\sqrt{2}$

Note that this function only defined between near 0 and about 4. so following eqn has no solution, if we use same methods we get

$x^{xx⋰\; <!--\; up\; right\; diagonal\; ellipsis\; -->}=4\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x^4=4\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->x=\sqrt{2}$

but this means 4=2 which is incorrect.