Originally Posted by

**fredrick08** $\displaystyle \int_{-\infty}^{\infty}\frac{1}{x^{6}+1}dx$

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

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

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

let z^4=z^2

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

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

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

please someone help