Compute residues for $\int_{-\infty}^{\infty} \frac{x \sin x}{1+x^4}dx$

Question

I am trying to compute

$$\int_{-\infty}^{\infty} \frac{x \sin x}{1+x^4}dx$$

This question has been asked here, but the asker and answerer skip over the calculation of the residues.

Define the curve $\gamma$ to be the semicircle in the top half plane with radius $R$ and straight edge on the real axis from $-R$ to $R$.

Then by the residue theorem and Jordan's lemma, taking $R \to \infty$, I know that

$$\int_{-\infty}^{\infty} \frac{z e^{iz}}{1+z^4}dz = \oint_{\gamma} \frac{ze^{iz}}{1+z^4} dz = 2 \pi i \sum_{\alpha \text{ roots of } 1+z^4, \Im(\alpha)>0} \text{Res}\left(\frac{z e^{iz}}{1+z^4}, \alpha \right)$$

The relevant roots are $\alpha=e^{\frac{\pi i}{4}}, e^{\frac{3 \pi i}{4}}$.

How can I go about calculating these residues? I know that it's the coefficient of $z^{-1}$ the Laurent series at $\alpha$, or also $\frac{1}{2 \pi i} \oint_{\partial D(\alpha, r)} \frac{z e^{iz}}{1+z^4} dz$. I'm struggling to calculate either of these. What do I do?

Answer 1

$z^4+1=(z-w_1)\cdots(z-w_4)$, where $w_j=e^{2\pi ij/4}$, for $j=1, 2,3,4$. You are interested in $w_1$ and $w_3$. Since they are poles of order $1$ you calculate the residues as $$\lim_{z\to w_1}(z-w_1) \frac{ze^{iz}}{1+z^4}, $$ and same for $w_3$.

Answer 2

For $n$-th order pole at $z=a$, the general residue formula is:

$$Res(f, a)=\frac{1}{(n-1)!}\lim_{z\to a}\frac{d^{n-1}}{dz^{n-1}}((z-a)^nf(z))$$

For simple pole, $n=1$, we get

$$Res(f, a)=\lim_{z\to a}~(z-a)f(z)$$

Answer 3

$\newcommand{\res}{\operatorname{Res}}$For simple poles it is easier to use the following approach. For higher order poles, the use of Taylor and Laurent expansions dramatically simplifies the procedure. The limit formula is correct but generally unwieldy.

Task: evaluate $\res(f;\zeta)$ where $f$ is meromorphic with a simple pole at $\zeta$.

Identify the "problematic" part of $f$ and express $f=h/g$ where $h$ is analytic in a neighbourhood of $\zeta$ and $g$ is analytic in a neighbourhood of, and has a simple zero at, $\zeta$. Then quite simply: $$\res(f;\zeta)=h(\zeta)/g'(\zeta)$$

Example:

Evaluate: $$\res\left(z\mapsto\frac{ze^{iz}}{1+z^4};e^{\pi i/4}\right)$$

I just bring out the $ze^{iz}$ term as it is analytic at the given point, and $1+z^4$ has a simple zero at $e^{\pi i/4}$ and derivative $4z^3$ so: $$\res=\frac{(e^{\pi i/4})e^{ie^{\pi i/4}}}{4e^{3\pi i/4}}=\frac{1}{4i}\exp(-2^{-1/2})\exp(i2^{-1/2})$$The other residue is easily evaluated with the same method to be: $$-\frac{1}{4i}\exp(-2^{-1/2})\exp(-i2^{-1/2})$$In sum we get: $$\frac{1}{2e^{1/\sqrt{2}}}\sin\frac{1}{\sqrt{2}}$$So that: $$\int_{\Bbb R}\frac{x\cdot\sin x}{1+x^4}\,\mathrm{d}x=\frac{\pi}{e^{1/\sqrt{2}}}\sin\frac{1}{\sqrt{2}}$$