# Residue theory complex

$$\int_{-\infty}^{\infty}\frac{\cos x}{x^4+5x^2+4}dx$$ Give full justification of your answer, including appropriate bounds for the contributions from all portions of your contour!

I am not sure of how to define the contour. A semi circle with radius R? And then as R goes to infinity the 4 limits go to zero?

What is the correct way to define the curve?

(Original image)

• Yeah, do a semicircle; show the 'round' part of the integral goes to zero; all you're left with when you take the limit is the integral above. Use the residue theorem.
– user98602
Commented Dec 12, 2013 at 18:56
• The limits won't go to zero. After you switch everything into complex variables you'll notice the denominator can be factored into $(z^2+4)(z^2+1) = (z+2i)(z-2i)(z+i)(z-i)$ which are all simple poles. Then Residue theorem would apply to them. Commented Dec 12, 2013 at 19:12

You first need to express cosine as

$$\cos{x} = \frac12 (e^{i x}+e^{-i x})$$

and consider each exponential separately. For the $e^{i x}$, consider

$$\oint_{C_+} dz \frac{e^{i z}}{z^4+5 z^2+4}$$

where $C_+$ is a semicircle in the upper half plane of radius $R$, positively oriented. The contour integral is equal to

$$\int_{-R}^R dx \frac{e^{i x}}{x^4+5 x^2+4}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R \cos{\theta}} e^{-R \sin{\theta}}}{R^4 e^{i 4 \theta}+ 5 R^2 e^{i 2 \theta}+4}$$

You may show that the second integral vanishes in the limit as $R \to \infty$ by showing that its magnitude is bounded by

$$\frac{2}{R^3}\int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^3}\int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{\pi}{R^4}$$

The contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles in the contour $C_+$. You may show that the poles of the integrand are at $z=\pm 2 i$ and $z=\pm i$, so that the poles of interest here are at $z=2 i$ and $z=i$. By the residue theorem, we have

\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i x}}{x^4+5 x^2+4} &= i 2 \pi \left (\frac{e^{-2}}{4 (2 i)^3 + 10 (2 i)} + \frac{e^{-1}}{4 (i)^3+10 (i)} \right )\\ &= \frac{\pi}{6 e} \left (2-\frac1{e}\right )\end{align}

Normally, I would say that we need to consider the other contour in the lower half plane, but because of the symmetry, we need only take the real part of the integral; hence, we are done.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#ff0000}{\int_{-\infty}^{\infty}{\cos\pars{x} \over x^{4} + 5x^{2} + 4} \,\dd x} = {1 \over 3}\int_{-\infty}^{\infty}\cos\pars{x} \pars{{1 \over x^{2} + 1} - {1 \over x^{2} + 4}}\,\dd x \\[3mm]&= {1 \over 3}\int_{-\infty}^{\infty}{\cos\pars{x} \over x^{2} + 1}\,\dd x - {1 \over 6}\int_{-\infty}^{\infty}{\cos\pars{2x} \over x^{2} + 1}\,\dd x = \color{#ff0000}{{1 \over 3}\,{\cal F}\pars{1} - {1 \over 6}\,{\cal F}\pars{2}}\tag{1} \\[3mm]&\mbox{where}\ {\cal F}\pars{\mu} \equiv \Re\int_{-\infty}^{\infty}{\expo{\ic\verts{\mu}x} \over x^{2} + 1}\,\dd x\,, \quad \mu \in {\mathbb R} \end{align}

With the integration contour depicted at this answer end ( the integration over the upper arc vanishes out in the limit $R \to \infty$ ): \begin{align} {\cal F}\pars{\mu} \equiv \Re\int_{-\infty}^{\infty} {\expo{\ic\verts{\mu}x} \over \pars{x - \ic}\pars{x + \ic}}\,\dd x = \Re\bracks{% 2\pi\ic\,{\exp\pars{\ic\verts{\mu}\ic} \over \ic + \ic}} = \pi\expo{-\verts{\mu}} \end{align}

We replace this result in $\pars{1}$: $$\int_{-\infty}^{\infty}{\cos\pars{x} \over x^{4} + 5x^{2} + 4} \,\dd x = {1 \over 3}\,\pars{\pi\expo{-\verts{1}}} - {1 \over 6}\,\pars{\pi\expo{-\verts{2}}}$$ $$\color{#0000ff}{\large\int_{-\infty}^{\infty}{\cos\pars{x} \over x^{4} + 5x^{2} + 4} \,\dd x} =\color{#0000ff}{\large{1 \over 6}\,\pi\expo{-2}\pars{2\expo{} - 1}}$$