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- Thread starter dwsmith
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- Jan 31, 2012

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No I think log is supposed to be viewed as the real log in this integral.

If you have Lang's book, I can tell you what page it is on.

- Jan 31, 2012

- 253

---------- Post added at 11:42 ---------- Previous post was at 10:29 ----------

Define the cut along the negative imaginary axis (including the origin). Then the contour is exactly the same as the contour for $\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx $.

let $\displaystyle f(z) = \frac{(\log z)^{2}}{1+z^{2}}$ and recall that $\displaystyle \log z = \ln z + i \arg z$

$\displaystyle \int_{-R}^{-r} f(x) \ dx + \int_{C_{r}} f(z) \ dz + \int_{r}^{R} f(x) \ dx + \int_{C_{R}} f(z) \ dz = 2 \pi i \ \text{Res} [f,i] $

$\displaystyle = 2 \pi i \lim_{z \to i} \ (z-i) \frac{(\log z)^2}{1+z^{2}} = 2 \pi i \frac{(\log i)^{2}}{2i} = \pi \left( \ln |i| + i \arg{i} \right)^{2} = \pi \left(\frac{i \pi}{2} \right)^{2} = \frac{-\pi^{3}}{4}$

so we have $\displaystyle \int_{-R}^{-r} \frac{(\ln |x| + i \pi)^{2}}{1+x^{2}} \ dx + \int_{C_{r}} f(z) \ dz + \int_{r}^{R} \frac{(\ln|x| + i0)^{2}}{1+x^{2}} \ dx + \int_{C_{R}} f(z) \ dz = \frac{-\pi^{3}}{4}$

If you let $r$ to go to zero and $R$ go to infinity, the second the fourth integrals will go to zero. You can show that using the ML inequality.

$\displaystyle \text{PV} \left( \int_{-\infty}^{0} \frac{(\ln |x|)^{2} + 2 \pi i \ln |x| - \pi^{2}} {1+x^{2}} + \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx \right)$

$ \displaystyle =\text{PV} \left( \int_{0}^{\infty} \frac{(\ln x)^{2} + 2 \pi i \ln x - \pi^{2}} {1+x^{2}} + \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx \right) = \frac{-\pi^{3}}{4} $

now equate the real parts on both sides

$ \displaystyle 2 \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx - \pi^{2} \int_{0}^{\infty} \frac{1}{1+x^{2}} \ dx = \frac{-\pi^{3}}{4} $ (I dropped the PV label since both integrals converge.)

so $\displaystyle 2 \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx - \frac{-\pi^{3}}{2} = \frac{-\pi^{3}}{4} \implies \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx = \frac{\pi^{3}}{8}$

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So the contour is the semi-circle in the upper half plane?

---------- Post added at 11:42 ---------- Previous post was at 10:29 ----------

Define the cut along the negative imaginary axis (including the origin). Then the contour is exactly the same as the contour for $\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx $.

let $\displaystyle f(z) = \frac{(\log z)^{2}}{1+z^{2}}$ and recall that $\displaystyle \log z = \ln z + i \arg z$

$\displaystyle \int_{-R}^{-r} f(x) \ dx + \int_{C_{r}} f(z) \ dz + \int_{r}^{R} f(x) \ dx + \int_{C_{R}} f(z) \ dz = 2 \pi i \ \text{Res} [f,i] $

$\displaystyle = 2 \pi i \lim_{z \to i} \ (z-i) \frac{(\log z)^2}{1+z^{2}} = 2 \pi i \frac{(\log i)^{2}}{2i} = \pi \left( \ln |i| + i \arg{i} \right)^{2} = \pi \left(\frac{i \pi}{2} \right)^{2} = \frac{-\pi^{3}}{4}$

so we have $\displaystyle \int_{-R}^{-r} \frac{(\ln |x| + i \pi)^{2}}{1+x^{2}} \ dx + \int_{C_{r}} f(z) \ dz + \int_{r}^{R} \frac{(\ln|x| + i0)^{2}}{1+x^{2}} \ dx + \int_{C_{R}} f(z) \ dz = \frac{-\pi^{3}}{4}$

If you let $r$ to go to zero and $R$ go to infinity, the second the fourth integrals will go to zero. You can show that using the ML inequality.

$\displaystyle \text{PV} \left( \int_{-\infty}^{0} \frac{(\ln |x|)^{2} + 2 \pi i \ln |x| - \pi^{2}} {1+x^{2}} + \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx \right)$

$ \displaystyle =\text{PV} \left( \int_{0}^{\infty} \frac{(\ln x)^{2} + 2 \pi i \ln x - \pi^{2}} {1+x^{2}} + \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx \right) = \frac{-\pi^{3}}{4} $

now equate the real parts on both sides

$ \displaystyle 2 \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx - \pi^{2} \int_{0}^{\infty} \frac{1}{1+x^{2}} \ dx = \frac{-\pi^{3}}{4} $ (I dropped the PV label since both integrals converge.)

so $\displaystyle 2 \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx - \frac{-\pi^{3}}{2} = \frac{-\pi^{3}}{4} \implies \int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} \ dx = \frac{\pi^{3}}{8}$

- Jan 31, 2012

- 253