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When evaluating real integrals involving log, I am having trouble with the step that involves finding a bound on circular segments. Let me explain what I mean:

If, for example, we have $$ \int_0^\infty \frac{(\log(x))^2}{1+x^2} \, \mathrm{d}x $$ We consider the complex integral $$ \int\frac{(\log(z))^2}{1+z^2} \, \mathrm{d}z $$ along a path on which the function is analytic. In this case, our path, gamma, would be made of four segments:

  1. from $\epsilon$ to $R$ along the positive real axis,
  2. from $R$ to $-R$ along a circle in the upper half plane
  3. from $-R$ to $-\epsilon$ on the negative real axis
  4. from $-\epsilon$ to $\epsilon$ along a circle in the upper half plane

(in this way we can consider the branch of log excluding the negative imaginary axis)

I understand that you then proceed to show that integrals 2 and 4 reduce to zero as $R$ approaches infinity and $\epsilon$ approaches zero. This is where I have trouble. Most resources simply say, "show f is bounded".

What is the typical procedure for finding a bound for this type of function involving log? (Or even not involving log.)

I'm sorry for the messy latex and I would be very appreciative of any guidance you can provide.

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Use that $$|\log z| \le \big|\ln|z|\big| + 2\pi$$ (for the natural branch; replace $2\pi$ by other suitable constants for other branches). –  mrf Dec 19 '12 at 6:35
and $\lim_{r\to\infty} \frac{\log(r)}r=0$ of course. –  Raymond Manzoni Dec 19 '12 at 13:58
thanks. but how is that helpful for epsilon approaches zero? –  lkhs Dec 19 '12 at 15:16
See example $V$ here. –  Mhenni Benghorbal Dec 19 '12 at 20:13

1 Answer 1

Let's look at the part $4.$ with $\ z:=\epsilon\, e^{i\theta}$ : $$I_\epsilon=\int_{c_\epsilon} \frac{(\log(z))^2}{1+z^2}\,dz=\int_{-\pi}^0 \frac{\bigl(\log\bigl(\epsilon\, e^{i\theta}\bigr)\bigr)^2}{1+\epsilon^2\, e^{2i\theta}}\,\epsilon\, e^{i\theta}d\theta$$ $$I_\epsilon=\int_{-\pi}^0 \frac{(\log(\epsilon)+i\theta)^2}{1+\epsilon^2\, e^{2i\theta}}\,\epsilon\, e^{i\theta}d\theta$$ and we get the majoration : $$|I_\epsilon|\le\int_0^{\pi} \frac{(|\log(\epsilon)|+\pi)^2}{1-\epsilon^2}\,\epsilon\, d\theta$$ i.e. : $$|I_\epsilon|\le \pi\, \epsilon\,\frac{(|\log(\epsilon)|+\pi)^2}{1-\epsilon^2}$$

The most 'determinant' parameter is the $\ \epsilon\ $ at the front.
Use $\ \displaystyle\lim_{\epsilon\to 0^+}\ (\log(\epsilon))^n\,\epsilon=0\ $ for $\ n>0\ $ to conclude.

If you have more questions please ask them here.

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Is this the correct idea?: We develop the bound on the 3rd line by considering the absolute values of the numerator, denominator, and epsilon term. The numerator can be expanded, then bounded using the triangle inequality, which shows us that the numerator is largest for theta = pi. The magnitude of the epsilon term is epsilon. Do we then try to find a lower bound for the denominator? Since the second term in the denominator may have a non-zero argument then the magnitude of the denominator is somewhere between $(1-\epsilon^2)$ and $(1+\epsilon^2)$. So we take it to be $(1-\epsilon^2)$? –  lkhs Dec 19 '12 at 23:15
@lkhs: yes your analysis concerning the numerator and denominator are both right (we want a majoration of the numerator and a minoration of the denominator). Note that the minoration for the denominator is valid only for $\epsilon<1$ (I should have specified this) ; we could have supposed too that $\epsilon<1/2$ for example and replaced the denominator by $3/4$. For this specific problem we could remove the integral by replacing $\theta$ by its largest value (this is not always possible). –  Raymond Manzoni Dec 19 '12 at 23:58
Thanks very much! –  lkhs Dec 20 '12 at 0:36
@lkhs: You are welcome ! –  Raymond Manzoni Dec 20 '12 at 0:38

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