$\int_\mathbb{R} \frac{\cos(x)}{(x^2 + 4)^2}$ using residue calculus

Question

I want to calculate integral $$\int_\mathbb{R} \frac{\cos(x)}{(x^2 + 4)^2}$$

using residue calculus. To do so, I'm going to define $f(z) = \frac{\cos(z)}{(z^2 + 4)^2} = \frac{\cos(z)}{(z-2i)^2(z+2i)^2}$

It has two polars with rank $2$ ($2i$ and $-2i$).

Let's calculate residue:

$$\textrm{res}_{2i}f(z) = \lim_{z \rightarrow 2i} - \frac{2\cos(z) + (z + 2i)\sin(z)}{(z + 2i)^3} = -\frac{2\cos(2i) + 4i\sin(2i)}{(4i)^3}$$

Analogically:

$$\textrm{res}_{-2i}f(z) = -\frac{2\cos(2i) + (-4i)\sin(-2i)}{(-4i)^3}$$

Finally our integral will be equal to:

$$\int_\mathbb{R} \frac{\cos(x)}{(x^2 + 4)^2} = 2\pi i (\textrm{res}_{2i}f + \textrm{res}_{-2i}f) = $$ $$=2 \pi i\frac{-2\cos(2i) - 4i\sin(2i) - 2\cos(-2i) + 4i\sin(-2i)}{(4i^3)}=$$

$$= 2 \pi (\frac{-\cos(2i) + i \sin(2i)}{4^2}) = 2 \pi [\frac{e^{-2}}{4^2}]$$

whereas using wolfram alpha the final results should be formulated as $\frac{3 \pi e^{-2}}{4^2}$ and I want to ask you where do I have the mistake? Is it more calculation error or idea error?

This might help: https://math.stackexchange.com/q/3280689/42969 — Martin R, Feb 13 '22 at 19:58
To evaluate the residues you have to define a closed contour first. It is not clear what contour you use — Svyatoslav, Feb 13 '22 at 20:14

score 1 · Answer 1 · 2022-02-14T02:03:07.937

Define ($R > 2$) $$2\pi i\ \hbox{res}_{2i} f(z) = \int_{C} \dfrac{\cos(z)}{(z^2 + 4)^2} \ dz = \int_{red\ I_1} \dfrac{\cos(z)}{(z^2 + 4)^2} \ dz + \int_{blue\ I_2} \dfrac{\cos(z)}{(z^2 + 4)^2} \ dz $$

Repare that $$\int_{red\ I_1} \dfrac{\cos(z)}{(z^2 + 4)^2} \ dz = \int_{-R}^{R} \dfrac{\cos(x)}{(x^2 + 4)^2} \ dx$$

$$\lim\limits_{R \rightarrow \infty} \int_{-R}^{R} \dfrac{\cos(x)}{(x^2 + 4)^2} \ dx = \int_{-\infty}^{\infty} \dfrac{\cos(x)}{(x^2 + 4)^2} \ dx $$

and

$$\lim\limits_{R \rightarrow \infty} \int_{blue\ I_2} \dfrac{\cos(z)}{(z^2 + 4)^2} \ dz \leq \lim\limits_{R \rightarrow \infty} \pi R M = \lim\limits_{R \rightarrow \infty} O(1/R^3) = 0$$

where $M = max\left\{\left|\dfrac{\cos(z)}{(z^2 + 4)^2}\right|, z = Re^{\theta i} (0 \leq \theta \leq \pi) \right\}$ (I use $|\int_{C} f(z) \ dz| \leq \int_{C} |f(z)| \ dz$)

So

$$\int_{-\infty}^{\infty} \dfrac{\cos(x)}{(x^2 + 4)^2} \ dx = \int_{C} \dfrac{\cos(z)}{(z^2 + 4)^2} \ dz = 2\pi i\ \hbox{res}_{2i} f(z) = \dfrac{3\pi}{4^2e^2} $$

Your $M$ is unbounded as a function of $R$, because $\cos z$ grows exponentially along the imaginary axis. The standard fix is to compute the integral as the real part of $\displaystyle\int_\mathbb{R} \frac{e^{ix}}{(x^2+4)^2} , dx$, and apply Jordan's lemma. — Micah, Feb 14 '22 at 02:55

$\int_\mathbb{R} \frac{\cos(x)}{(x^2 + 4)^2}$ using residue calculus

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