Question

INTEGRATE:
\[\Large \int\limits_{-\infty}^{\infty}\frac{ 1 }{ x^4+1 }dx\]

Answer 1

which technique i should use
Residue? or something else

Answer 2

Residue would be the simplest way, yes.

Answer 3

yeah but it is consuming much time :(

Answer 4

Factor the denominator (so you'll have to find the fourth roots of -1), then compute. The integral will be equivalent to
\[2\pi i\left(\large\sum_{\text{Im}z>0}\text{Res}f(z)+\frac{1}{2}\sum_{\text{Im}z=0}\text{Res}f(z)\right)\]

Answer 5

i'm considering
\[\int\limits_{C}^{}f(z)dz\]
so
\[f(z)=\frac{ 1 }{ z^4+1 }\]
which gives me
\[\Large z^4=-1\rightarrow z=[\cos(2n+1)\pi+i \sin(2n+1)\pi]^{1/4}\]
right?

Answer 6

\[\Large z=[\cos(2n+1)(\pi/4)+isin(2n+1)(\pi/4)]\]

Answer 7

so for four poles i have to replace
n=0,1,2,3 right?

Answer 8

Yes that's right. The beauty of the contour you're using (I'm assuming a semicircle in the upper half-plane?) is that you need only consider the residues of the poles in the upper half-plane.