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Using substitution ($u=\ln x$), it can be shown that $\int_0^{1/2}\frac{dx}{(x\ln^2x)}$ = $\frac{1}{\ln2}$. However, if we raise the integrand to a number slightly greater than one, this integral diverges, and we're having trouble showing this. We've tried finding a "lower bound" integral, $\int_0^{1/2}f <\int_0^{1/2}\frac{dx}{(x\ln^2x)^{1+\epsilon}}$, such that $\int_0^{1/2}f$ diverges and using that to show that the other integral also diverges. However, we haven't been able to find anything that we can integrate.

The other idea we came to was by looking at the graph of $(x\ln^2x)^{1+\epsilon}$ from 0 to 1/2. This function is bounded above by some constant at $\frac{1}{e^2}$. If we take this constant $C(\epsilon)$ and look at $\int_0^{1/2}\frac{1}{C(\epsilon)\ln^2x}$, this seems like a divergent lower bound integral. Does this seem reasonable?

Thank you!

letslearnmath
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1 Answers1

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The "lower bound function" that you want to use is $x^{-(1+\epsilon/2)}$. It will only work as a lower bound in a sufficiently small neighborhood of zero, but that's all you need. After that it's the same reasoning you've described here.

Ian
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  • I'm sorry, I don't see how this is a lower bound even for a small neighborhood. Would you mind expanding? – letslearnmath Nov 15 '17 at 21:41
  • @wobertson Recall that $\lim_{x \to 0^+} \ln(x) x^\epsilon=0$ for any $\epsilon>0$. – Ian Nov 15 '17 at 21:52

  • < I am so sorry for still not understanding. So I know what you're saying is true; we can write $\lim{x\rightarrow 0}\frac{\ln(x)}{x^{-\epsilon}}$ and use L'Hopital to show that this goes to zero. But I'm having trouble seeing where that comes into play. Are we comparing with $(x\ln^2(x))^{1+\epsilon}$? So this will go to zero because of the $x^{\epsilon}\ln(x)$ but so will $x^{-(1+\epsilon/2)}$ no?

     – letslearnmath Nov 15 '17 at 22:06
  • @wobertson I'm taking your integrand, dividing out $\ln^{2+2\epsilon}(x)$ (a factor that blows up), and multiplying in $x^{-\epsilon/2}$ (a factor that also blows up, but faster than the one I divided out). In the process I decrease the integrand, at least close enough to zero. – Ian Nov 15 '17 at 22:16
  • I feel like I'm seeing the results but still not understanding what's going on. So we have $\frac{1}{xx^{\epsilon}\ln(x)^{2+2\epsilon}}$. And now you're trading in the $x^{\epsilon}\ln(x)^{2+2\epsilon}$ for $x^{\epsilon/2}$, since the former goes to zero faster. Is this correct? – letslearnmath Nov 16 '17 at 02:56
  • Wait yep I think that's it, since $x^{\epsilon/2}$ is bigger, putting it in the denominator would make the integrand smaller, and yet the integral still diverges. I think I've understood it. Thank you!!! We've been working on this one all day, so excited to share – letslearnmath Nov 16 '17 at 03:17