Show that $\int_0^1 \phi^2(x)dx$ does not exist where $\phi(x)=x^{x-1}$

Question

iI am currently studying integral equations from the book "Integral Equations" by Harry Hochstadt. In its second exercise (page $42$) it is asked to (Q.No $2$) show that $\displaystyle \int_0^1 \phi^2(x)dx$ does not exist where $\phi(x)=x^{x-1}$

I was unable to solve the problem, What I done so far is:

I think the integral is improper as at $0$ it has a singularity .

So I considered $\displaystyle \int_0^1 \phi^2(x)dx=\lim_{\epsilon\to 0} \displaystyle \int_{\epsilon}^1x^{2x-2}dx$ and I have to show this diverges.

But I can not proceed further. I think I have to somehow use the fact $x\ge \epsilon$ in the above integral since $x$ is running from $\epsilon$ to $1$.

Can I proceed with my idea from here? It will be very helpful for me if anyone suggest how should I complete the problem.

Thnx in advance.

EDIT:

I tried to plot the graph in Geogebra. It lokks like the following.

So the area under the curve within $0$ and $1$ is likely to be infinity but still I can not prove it.

And the answer posted seems to have some problem. I appologise if I am missing something.Please correct me.

Again thnx in advance. It will be very helpful if I get some suugestion to solve it.

Answer 1

Hint
For $x\le 1$ and $b\ge a$ we have $x^a \le x^b$. Use this estimate with $a = -2$ and $b=2x-2$.