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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).2012-12-19
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    and $\lim_{r\to\infty} \frac{\log(r)}r=0$ of course.2012-12-19
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    thanks. but how is that helpful for epsilon approaches zero?2012-12-19
  • 0
    See example $V$ [here](http://en.wikipedia.org/wiki/Methods_of_contour_integration).2012-12-19

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