Poles of $\frac{1}{1+x^4}$

Question

The integral I'd like to solve with contour integration is $\int^{\infty }_{0}\dfrac {dx}{x^{4}+1}$ and I believe the simplest way to do it is using the residue theorem. I know the integrand has four simple poles at ${\dfrac {\pi}{4}}$,${\dfrac {3\pi}{4}}$, ${\dfrac {-\pi}{4}}$,${\dfrac {-3\pi}{4}}$ but I am stumped on how to calculate the residues at those poles. Since my contour is in the y>0 space I only need the poles at $\dfrac{\pi}{4}$ and $\dfrac{3\pi}{4}$

I know that the integrand $f(z)=\dfrac{p(z)}{q(z)}$ where $q(z_{0})=0$, $q'(z_{0})\neq0$, $p(z_{0})\neq0$ so the Residue of $f(z)$ at $z=z_{0}$ is $\dfrac {p(z_{0})}{q'(z_{0})}=\dfrac {1}{4z^3}$. How do I calculate the integral using the residue theorem?

The derivative of $1$ is zero, $\frac{1}{4z^3 + 1}$ should be $\frac{1}{4z^3}$. — dustin, Dec 09 '14 at 22:36
Fix it. Don't delete a question, it's better to just leave it than to delete it. — MPW, Dec 09 '14 at 22:41

score 1 · Accepted Answer · answered Dec 09 '14 at 22:43

1

To simplify your calculation (besides the error already pointed out): $$ \frac{1}{4z_j^3} = \frac{z_j}{4z_j^4} = -\frac{z_j}{4} $$ since $z_j^4 = -1$.

answered Dec 09 '14 at 22:43

mrf

43,639

score 1 · Answer 2 · answered Dec 09 '14 at 23:07

By the residue theorem, $\int^{\infty }_{0}\dfrac {dx}{x^{4}+1}= \dfrac {1}{2}(2i\pi\sum_{i}{Res[f(z_{i})]})=i\pi (\dfrac {-i\sqrt{2}}{4})=\dfrac{\sqrt{2}\pi}{4}$

The sum of the residues is simply $\dfrac{-1}{4}(e^\dfrac{i\pi}{4}+e^\dfrac{3i\pi}{4})=\dfrac{-\sqrt{2}i}{4}$

Poles of $\frac{1}{1+x^4}$

2 Answers2