Evaluate the integral \[ \int_\RR {dx \over 1 + x^4} .\]
What are the poles of \({1\over 1 + z^4}\) ?
Solution
Let \(f(z) = 1/(1 + z^4)\). For \(R>1\), let \(\Gamma_R\) be the closed contour consisting of the segment \([-R, R]\) and the arc \(C_R\): \(z = Re^{it}\), \(t \in [0, \pi]\).
1. The poles of \(f\) are the simple poles \(z_k = e^{i\pi(1 + 2k)/4}\), \(k = 0, 1, 2, 3\).
\(1 + z^4 = 0\) if and only if \(z^4 = -1 = e^{i\pi}\), so the zeros of \(1+z^4\) are \(z_0 = e^{i\pi/4}\), \(z_1 = e^{3i\pi/4}\), \(z_2 = e^{5i\pi/4}\), \(z_3 = e^{7i\pi/4}\). Each is simple, since the derivative \(4z^3\) does not vanish at any of them.
2. The poles inside \(\Gamma_R\) are \(z_0 = e^{i\pi/4}\) and \(z_1 = e^{3i\pi/4}\).
By step 1, \(z_0\) and \(z_1\) have positive imaginary part and modulus \(1<R\); \(z_2\) and \(z_3\) have negative imaginary part.
3. \(\Res_{z_0} f + \Res_{z_1} f = -\dfrac{i\sqrt2}{4}\).
At a simple zero \(z_k\) of \(1+z^4\), \(\Res_{z_k} f = 1/(4 z_k^3)\). Since \(z_0^3 = e^{3i\pi/4}\) and \(z_1^3 = e^{9i\pi/4} = e^{i\pi/4}\), \[\Res_{z_0} f + \Res_{z_1} f = \frac14\left(e^{-3i\pi/4} + e^{-i\pi/4}\right) = \frac14\left(-\frac{\sqrt2}{2} - i\frac{\sqrt2}{2} + \frac{\sqrt2}{2} - i\frac{\sqrt2}{2}\right) = -\frac{i\sqrt2}{4}.\]
4. \(\displaystyle\int_{C_R} f(z)\,dz \to 0\) as \(R \to \infty\).
On \(C_R\), \(\abs{1 + z^4} \ge \abs{z}^4 - 1 = R^4 - 1\), so \(\abs{f(z)} \le 1/(R^4 - 1)\). The arc has length \(\pi R\), so the integral is bounded by \(\pi R /(R^4 - 1) \to 0\).
5. \(\displaystyle\int_\RR \frac{dx}{1+x^4} = \boxed{\frac{\pi}{\sqrt2}}\).
3. Letting \(R\to\infty\) and using step 4, the integral over \([-R,R]\) converges to the same value.
6. Q.E.D.
Step 5 evaluates the integral, and step 1 lists the poles.