another complex question

**fredrick08** · May 27th 2010, 11:30 PM

$\int_{-\infty}^{\infty}\frac{1}{x^{6}+1}dx$

$=\int_{C}\frac{1}{z^{6}+1}dz$ i think.... a bit confused because there will be singularities in UHP and LHP...

$=\int_{C}\frac{1}{(z^{2}+1)(z^{4}-z^{2}+1)}dz$

$=\int_{C}\frac{1}{(z-i)(z+i)(z^{4}-z^{2}+1)}dz$

let z^4=z^2

$z^{2}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{3}\\i}{2}$

$\\\\z=\{-i,i,\frac{1}{2}(\sqrt{3}+i),\frac{1}{2}(\sqrt{3}-i),\frac{1}{2}(-\sqrt{3}+i),\frac{1}{2}(-\sqrt{3}-i)\}$

now from here... how the hell do i do the residues.. the singularities are all over the place???

please someone help

**Drexel28** · May 27th 2010, 11:36 PM

Originally Posted by fredrick08

$\int_{-\infty}^{\infty}\frac{1}{x^{6}+1}dx$

$=\int_{C}\frac{1}{z^{6}+1}dz$ i think.... a bit confused because there will be singularities in UHP and LHP...

$=\int_{C}\frac{1}{(z^{2}+1)(z^{4}-z^{2}+1)}dz$

$=\int_{C}\frac{1}{(z-i)(z+i)(z^{4}-z^{2}+1)}dz$

let z^4=z^2

$z^{2}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{3}\\i}{2}$

$\\\\z=\{-i,i,\frac{1}{2}(\sqrt{3}+i),\frac{1}{2}(\sqrt{3}-i),\frac{1}{2}(-\sqrt{3}+i),\frac{1}{2}(-\sqrt{3}-i)\}$

now from here... how the hell do i do the residues.. the singularities are all over the place???

please someone help

What do you mean how do you find the residues?? How else does one usually find them?

**chisigma** · May 27th 2010, 11:54 PM

The poles of the function $f(z)= \frac{1}{1+z^{6}}$ are at $z= e^{\pm i \pi \frac{1+2k}{6}}$ , $k=0,1,2$ . Only the residues of the poles with positive imaginary part give contribution of the integral that is...

$\int_{-\infty}^{+\infty} \frac{dx}{1+x^{6}} = 2 \pi i \sum_{k=0}^{2} r_{k}$ (1)

... where...

$r_{0} = \lim_{z \rightarrow e^{i \frac{\pi}{6}}} \frac{z-e^{i \frac{\pi}{6}}}{1+z^{6}}$

$r_{1} = \lim_{z \rightarrow e^{i \frac{\pi}{2}}} \frac{z-e^{i \frac{\pi}{2}}}{1+z^{6}}$

$r_{2} = \lim_{z \rightarrow e^{i \frac{5 \pi}{6}}} \frac{z-e^{i \frac{5 \pi}{6}}}{1+z^{6}}$ (2)

Kind regards

$\chi$ $\sigma$

**fredrick08** · May 28th 2010, 03:41 PM

Originally Posted by chisigma

The poles of the function $f(z)= \frac{1}{1+z^{6}}$ are at $z= e^{\pm i \pi \frac{1+2k}{6}}$ , $k=0,1,2$ . Only the residues of the poles with positive imaginary part give contribution of the integral that is...

$\int_{-\infty}^{+\infty} \frac{dx}{1+x^{6}} = 2 \pi i \sum_{k=0}^{2} r_{k}$ (1)

... where...

$r_{0} = \lim_{z \rightarrow e^{i \frac{\pi}{6}}} \frac{z-e^{i \frac{\pi}{6}}}{1+z^{6}}$

$r_{1} = \lim_{z \rightarrow e^{i \frac{\pi}{2}}} \frac{z-e^{i \frac{\pi}{2}}}{1+z^{6}}$

$r_{2} = \lim_{z \rightarrow e^{i \frac{5 \pi}{6}}} \frac{z-e^{i \frac{5 \pi}{6}}}{1+z^{6}}$ (2)

Kind regards

$\chi$ $\sigma$

srry im having a lot of trouble with these limits?? dont they all goto zero.. as z approaches to what equal in the top line????

srry dumb question.... Thankyou very much for ur help

**HallsofIvy** · May 29th 2010, 01:54 AM

No, the limits are not all 0. If $z= e^{i\pi/6}$ the numerator is 0 but $z^6= e^{i\pi}= -1$ so the denominator, $z^6+ 1$ is also 0. Use L'Hopital's rule to find the limits.

**igopogo** · May 30th 2010, 07:27 PM

I think there's a simpler way. Consider

$f(z)=\frac{p(z)}{q(z)}$ where

$p(z) = 1$ and

$q(z)=z^6+1$

Then $p(z_i) \neq 0$ , $q(z_i)=0$ and $q'(z_i) \neq 0$ for each of your singularities.

Then the residue at $z_i$ is $\frac{p(z_i)}{q'(z_i)}=\frac{1}{6{z_i}^5}$

There are usually singularities in the UHP and LHP with these problems, but the LHP singularities are never inside the contour that you construct.

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