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Compute the following integral from $0$ to $\infty$:

$$\int_0^\infty \frac{(\log x)^2}{1+x^2}\, \mathrm dx.$$

Is my function here $\frac{1}{1+x^2}$ and there's a "hole" at $\log 0$ so we plug in $0$ into $\frac{1}{1+x^2}$ and plug into Cauchy's formula or?

A.M.
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homosapien
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  • You can't directly apply Cauchy's formula, since $(0,\infty)$ is not a closed loop. You have tagged this question with "contour integration", so you must surely know about needing a closed contour, and you must have seen examples... what contour do you think you should use to extract this integral? Furthermore, is there a mapping you can apply to this integral to change it into a nicer form? – FShrike Feb 26 '22 at 19:43
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    $$\int_0^\infty\frac{(\log x)^2}{1+x^2},\mathrm{d}x=\int_{-\infty}^\infty\frac{x^2}{2\cosh x},\mathrm{d}x$$ – FShrike Feb 26 '22 at 19:45
  • @FShrike Yeah I couldn't have been able to make that connection to hyperbolic cos. How did you get to that – homosapien Feb 26 '22 at 19:46
  • Use the substitution $u=\log x$. That detail is not important however, I just find integration easier with this substitution. The main thing is that you understand you need to find a contour that includes the integral you want - do you have any ideas? – FShrike Feb 26 '22 at 19:47
  • Sort of stuck on that part @FShrike I know when I graph it, I get the positive $x$ and $y$ values so the curve goes around the positive ordered pairs ? – homosapien Feb 26 '22 at 19:52
  • That's not very specific. When you graph what? I urge you; what is the contour you are using? A simple rectangle will do (for the $\cosh$ form of the integral) – FShrike Feb 26 '22 at 20:02
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    In case you have not been provided with the answer, I will say that: $$\int_0^\infty\frac{(\log x)^2}{1+x^2},\mathrm{d}x=\frac{\pi^3}{8}$$For your reference – FShrike Feb 26 '22 at 20:03
  • @FShrike would the contour then be an upper half circle for the first form of the integral, sorry I am weak at determining contours. I just can't see it for some reason – homosapien Feb 26 '22 at 20:09
  • Let us continue this discussion in chat. – FShrike Feb 26 '22 at 20:12
  • Also, the answers in this and this question might be useful. – Robert Lee Feb 26 '22 at 23:07

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