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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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    See example $V$ [here](http://en.wikipedia.org/wiki/Methods_of_contour_integration).2012-12-19

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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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    @lkhs: You are welcome !2012-12-20