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.


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

enter image description here

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.

  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – AlexR Feb 10 '15 at 9:45
  • $\begingroup$ Thank you for the suggestion.I will surely keep in mind. $\endgroup$ – usermath Feb 10 '15 at 9:46

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

  • $\begingroup$ Thank you for your help. So in this way $\int_0^1\phi^2(x)dx>\int_0^1\frac{1}{x^2}dx$ which divergs. Am I right? $\endgroup$ – usermath Feb 10 '15 at 9:49
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    $\begingroup$ Wait a minute..If $x=\frac12,b=2,a=1$ then $\frac14$ is not greater than $\frac12$. Is this violating the inequality?Or I am making something silly mistake? $\endgroup$ – usermath Feb 10 '15 at 9:54

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