MHB Integration--Cauchy Principal Value, Residue, line integration, poles

[Dustinsfl]

$$
\int_0^{\infty}\frac{(\log x)^2}{1 + x^2}dx = \frac{\pi^3}{8}
$$

I have no idea what to do with this integral. I can't see it is even and do 1/2 the integral from -infinity to infinity since log -x doesn't make sense.

 

[polygamma]

$\log x$ is a multivalued function, so there needs to be a branch cut somewhere. Have you used the keyhole contour before?

 

[Dustinsfl]

$\log x$ is a multivalued function, so there needs to be a branch cut somewhere. Have you used the keyhole contour before?

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.

 

[polygamma]

Let me correct myself. You're going to let $\displaystyle f(z) = \frac{\log^{2} z}{1+z^{2}}$ and $\log z$ is a multivalued function.

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

 

[Dustinsfl]

Let me correct myself. You're going to let $\displaystyle f(z) = \frac{\log^{2} z}{1+z^{2}}$ and $\log z$ is a multivalued function.

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

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

 

[polygamma]

Yes, but with a small semi-circle about the origin to avoid the logarithmic singularity there. A keyhole contour was not needed.