Complex Integral: Struggling to Integrate [(lnx)^2](1+x^2)

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SUMMARY

The discussion focuses on the integration of the complex function \(\frac{(\ln x)^2}{1+x^2}\) from zero to infinity. A branch cut along the negative y-axis is established to handle the logarithmic term, leading to an equivalent integral from minus infinity to zero. The solution involves substituting \(\ln(x)\) with \(\ln|x| + \pi i\) on the negative real axis, resulting in three integrals, where one matches the original integral, another can be solved using basic calculus, and the third evaluates to zero due to the properties of imaginary parts.

PREREQUISITES
  • Complex analysis, specifically contour integration techniques.
  • Understanding of branch cuts in complex functions.
  • Familiarity with logarithmic properties in complex analysis.
  • Basic calculus for evaluating integrals.
NEXT STEPS
  • Study complex contour integration methods in detail.
  • Learn about branch cuts and their implications in complex functions.
  • Explore the properties of logarithms in complex analysis.
  • Practice evaluating integrals involving complex functions and logarithmic terms.
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Mathematicians, students of complex analysis, and anyone interested in advanced integration techniques involving logarithmic functions.

sachi
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We have to integrate [(lnx)^2)/(1+(x^2)) from zero to infinity.
I have set up the correct complex integral with a branch cut along the negative y-axis, but I end up with an integral of [(lnx)^2](1+x^2) from minus infinity to zero. I'm not sure how to deal with this.
 
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sachi said:
We have to integrate [(lnx)^2)/(1+(x^2)) from zero to infinity.
I have set up the correct complex integral with a branch cut along the negative y-axis, but I end up with an integral of [(lnx)^2](1+x^2) from minus infinity to zero. I'm not sure how to deal with this.

On the negative real axis replace ln(x) by ln|x|+Pi*i. Expand the square and you'll have 3 integrals, one is the same as the thing you're after, one can be evaluated using first year calculus stuff, and one is necessarily zero (compare imaginary parts)
 

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