# Math Help - another complex question

1. ## another complex question

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

$=\int_{C}\frac{1}{z^{6}+1}dz$ 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???

2. 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???

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

3. 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

$\chi$ $\sigma$

4. 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

5. 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.

6. 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.