Why is $x^{1.36^x}$ such a good approximate to $\int_{0}^{x}t^t dt$?

Question

So, once again I was experimenting on Desmos and found that $\int_{0}^{x}t^t dt$ can be approximated pretty well by the function $x^{1.36^x}$.

It roughly becomes more accurate as $x$ approaches to infinity, but I wonder whether there is an exact of that number in the top e.g.

What number $n$ makes the approximation of the $\int_{0}^xt^tdt$ by the function $x^{n^x}$ almost perfect as $x$ approaches infinity?

Or a better way of phrasing is:

What choice of $n$ makes the ratio between $x^{n^x}$ and $\int_{0}^xt^tdt$ approach $1$ as $x$ approaches infinity?

I would like an answer as to why the approximate is such a good one and why it incorporates $x$ to the power of an exponential function.

Answer 1

WEll, $1.36^2$ is close to 2, and $1.36^3$ is close to 3, and $1.36^4$ is close to 4. Thus in general $1.36^x$ isn't that far away from $x$ for $x$ in $[1,5]$ Furthermore, $1.36^x < x$ for some values of $x$ in $[1,5]$ and $1.36^x > x$ for other such $x$. Thus letting $y(x)= 1.36^x$, one would expect that $\int_{1}^a f(x,y(x))dx$ and $\int_1^a f(x,x)dx$ for a continuous function $f$ that is increasing in both variables s.t. $f(u,v)=u^v$.

In fact, it is not that hard to check that say $\frac{2^2}{2^{1.36^2}} = 2^{2-1.36^2}$ which satisfies $1<2^{2-1.36^2} < 1.1$.

Now for large $x$ though, note that $1.36^x >> x$, thus $\int_0^a x^{1.36^x} dx >> \int_0^a x^x$ for say $a > 20$.

Answer 2

Expanding on my comment: $$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$ so for $z=x\ln(x)$ we get: $$x^x=\sum_{n=0}^\infty\frac{\left[x\ln(x)\right]^n}{n!}=1+x\ln(x)+\frac{x^2}2\ln^2(x)+...$$ Now integrate termwise