Question: Calculate the value of the integral $ \displaystyle \int_{0}^{\infty} \frac{\cos 3x}{(x^{2}+a^{2})^{2}} dx$ where $a>0$ is an arbitrary positive number.

Thoughts: I don't know how to establish convergence so that a symmetric limit can be used, but if I can do so, then we have that the integral equals $\displaystyle \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos 3x}{(x^{2}+a^{2})^{2}} dx = \displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\cos 3x}{(x^{2}+a^{2})^{2}} dx = Re \left (\displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{e^{i(3x)}}{(x^{2}+a^{2})^{2}} dx \right ) $

which has double poles at $x=ia$ and $x=-ia$, so the integral may be evaluated by calculating residues. Can anyone show me how to solve the problem? All input is appreciated, I am studying for an exam in complex analysis.

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    $\begingroup$ You're more or less there. The last one is what you should use, now all you need is to check that the integrand vanishes fast enough e.g. on a semicircle in the upper half plane. $\endgroup$ – Daniel Fischer Aug 29 '13 at 21:46
  • $\begingroup$ It is important know why you choose the semicircle in the upper half plane. $\endgroup$ – Mhenni Benghorbal Aug 29 '13 at 22:00
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    $\begingroup$ @MhenniBenghorbal I agree, I am not sure why to choose the semicircle in the upper half plane instead of the lower half plane. Can you explain why? $\endgroup$ – Sid Aug 29 '13 at 22:02
  • $\begingroup$ @DanielFischer I tried to show this using an example from my textbook but I got nowhere. Can you show me how this should be done or give me a hint? $\endgroup$ – Sid Aug 29 '13 at 22:14
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    $\begingroup$ Right. And you see that when $y < 0$, i.e. in the lower half-plane, the exponential factor becomes large, while in the upper half-plane it becomes small. $\endgroup$ – Daniel Fischer Aug 29 '13 at 22:32

Let $S=\{-ia, ia\}$, let $ \varphi \colon \Bbb C\setminus S\to \Bbb C, z\mapsto\dfrac{e^{i(3z)}}{(z^2+a^2)^2}$.

Given $n\in \Bbb N$ such that $n> a$, define $\gamma (n):=\gamma _1(n)\lor \gamma _2(n)$ with $\gamma _1(n)\colon [-n,n]\to \Bbb C, t\mapsto t$ and $\gamma _2(n)\colon [0,\pi]\to \Bbb C, \theta \mapsto ne^{i\theta}$, ($\gamma (n)$ is an upper semicircle).

Observe that $S$ is the set of singularities of $\varphi$ and both of them are second order poles.

Therefore $$\operatorname {Res}(\varphi ,ia)=\left.\dfrac{d}{dz}\left(z\mapsto (z-ia)^2\varphi (z)\right)\right\vert_{z=ia}\overset{\text{W.A.}}{=}\left.\dfrac{ie^{i3z}(3ia+3z+2i)}{(z+ia)^3}\right\vert_{z=ia} = \dfrac{e^{-3a}(3a+1)}{4a^3i}.$$

Here is the link for the equality $\text {W.A.}$.

Quick considerations about winding numbers, inside and outside region of $\gamma(n)$, the fact that $n>a$, the fact that $\varphi$ is holomorphic and the residue theorem yield $$\displaystyle \int \limits_{\gamma (n)}\varphi (z)dz=2\pi i\cdot \dfrac{e^{-3a}(3a+1)}{4a^3i}= \dfrac{\pi e~^{-3a}(3a+1)}{2a^3}.$$

On the other hand $\displaystyle \int \limits _{\gamma (n)}\varphi=\int \limits _{\gamma _1(n)}\varphi +\int \limits_{\gamma _2(n)}\varphi \tag {*}$

Note that $$\displaystyle\int \limits _{\gamma _1(n)}\varphi(z)dz=\int \limits _{-n}^n\varphi (t)dt=\int \limits_{-n}^n\dfrac{e^{i(3t)}}{(t^2+a^2)^2}dt=\int \limits_{-n}^n\dfrac{\cos(3t)+i\sin 3t)}{(t^2+a^2)^2}dt=\int \limits _{-n}^n\dfrac{\cos (3t)}{(t^2+a^2)^2}dt.$$ The last equality is due to $t\mapsto \dfrac{\sin (3t)}{(t^2+a^2)^2}$ being an odd function and due to the integral being computed on a symmetric interval.

Furthermore, $$\int \limits _{\gamma _2(n)}\varphi (z)dz=\int \limits _0^\pi \varphi(ne^{i\theta})\cdot ine^{i\theta}d\theta=\int \limits _0^\pi\dfrac{e^{i\cdot 3ne^{i\theta}}ine^{i\theta}}{(n^2e^{2ni\theta }+a^2)^2}d\theta=n\int \limits _0^\pi i\dfrac{e^{i\cdot 3n(\cos (\theta)+i\sin (\theta))}e^{i\theta}}{(n^2e^{2ni\theta }+a^2)^2}d\theta=\\ =n\int \limits _0^\pi i\dfrac{e^{-3n\sin (\theta)}e^{i(3n\cos (\theta)+\theta)}}{(n^2e^{2ni\theta }+a^2)^2}d\theta,$$

from where one gets $$\left \vert\, \int \limits _{\gamma _2(n)}\varphi (z)dz\right \vert\leq n\int \limits _0^\pi \left \vert i\dfrac{e^{-3n\sin (\theta)}e^{i(3n\cos (\theta)+\theta)}}{(n^2e^{2ni\theta }+a^2)^2}\right \vert d\theta =n\int \limits_0^\pi \left \vert\dfrac{e^{-3n\sin (\theta)}}{(n^2e^{2ni\theta }+a^2)^2}\right \vert d\theta=\\=n\int \limits_0^\pi \dfrac{\left \vert e^{-3n\sin (\theta)}\right \vert}{\left \vert n^2e^{2ni\theta }+a^2\right \vert^2}d\theta \underset{(n>a)}{\leq} n\int \limits _0^\pi \dfrac{e^{-3a\sin (\theta)}}{(n^2-a^2)^2}d\theta=\dfrac{n}{(n^2-a^2)^2}\int \limits _0^\pi e^{-3a\sin (\theta)}d\theta\overset{n\to +\infty}{\longrightarrow} 0$$

Taking the limit in $(*)$ one finally gets $$\dfrac{\pi e~^{-3a}(3a+1)}{2a^3}=\int \limits_{-\infty}^{+\infty} \dfrac{\cos (3t)}{(t^2+a^2)^2}dt.$$

Due to the evenness of $t\to \dfrac{\cos (3t)}{(t^2+a^2)^2}$ it follows that $\displaystyle \int \limits_{0}^{+\infty} \dfrac{\cos (3t)}{(t^2+a^2)^2}dt=\dfrac{\pi e~^{-3a}(3a+1)}{4a^3}$ which agrees with WA.

I regret having started this.

  • $\begingroup$ Does anyone know how can I make the $\vert$ in $\vert _{z=ia}$look bigger? $\endgroup$ – Git Gud Aug 29 '13 at 23:28
  • $\begingroup$ \big\vert, \Big\vert, \bigg\vert, \Bigg\vert produce $\big\vert\;\Big\vert\;\bigg\vert\;\Bigg\vert$, choose your favourite size. $\endgroup$ – Daniel Fischer Aug 29 '13 at 23:41
  • $\begingroup$ @DanielFischer Thanks. $\endgroup$ – Git Gud Aug 29 '13 at 23:43
  • $\begingroup$ Comment downvoter? $\endgroup$ – Git Gud Aug 29 '13 at 23:47
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    $\begingroup$ A better way to match the vertical bar height is to enclose the expression in \left. and \right\vert. The height of the vertical bar matches the height of the enclosed expression. $\endgroup$ – robjohn Aug 30 '13 at 9:13

If you are interested in another method:

$$\begin{aligned}f(t)=\int_0^{\infty} \frac{\cos 3xt}{(x^2+a^2)^2}\,dx\Rightarrow \mathcal{L}\{f(t)\} &=\int_0^{\infty}e^{-st}\int_0^{\infty}\frac{\cos 3xt}{(x^2+a^2)^2}\,dx\,dt\\&=\int_0^{\infty}\frac{1}{(x^2+a^2)^2}\int_0^{\infty} e^{-st}\cos 3xt\,dt\,dx\\&=\int_0^{\infty}\frac{s\,dx}{(x^2+a^2)^2(9x^2+s^2)}\\&=\frac{3\pi}{2a^2(3a+s)^2}+\frac{\pi s}{4a^3(3a+s)^2}\end{aligned}$$

Which follows from a quick partial fraction decomposition.


$$\mathcal{L}^{-1}\left\{\frac{3\pi}{2a^2(3a+s)^2}\right\}+\mathcal{L}^{-1}\left\{\frac{\pi s}{4a^3(3a+s)^2}\right\}=\frac{3\pi t}{2a^2e^{3at}}+\frac{\pi(1-3at)}{4a^3e^{3at}}$$

By setting, $t=1$, we get:

$$\int_0^{\infty}\frac{\cos 3x\,dx}{(x^2+a^2)^2}=\frac{\pi(3a+1) }{4a^3e^{3a}}$$


Since you're studying for an exam, I'll try to be a bit more vague. The function you're considering is an even function. So


Thus, it suffices to compute the symmetric limit. Using this contour, the residue theorem shows that for $R>0$ sufficiently large,

$\int_{-R}^{R}\frac{e^{3ix}}{(x^2+a^2)^2}\ dx+\int_{\gamma_R}\frac{e^{3iz}}{(z^2+a^2)^2}\ dz=2\pi i\cdot \text{res}_{ia}\left(\frac{e^{3iz}}{(z^2+a^2)^2}\right)$

Since $ia$ is the only pole in the semicircular region (where $\gamma_R$ is the semicircular portion of the contour [as indicated in the picture]). Jordan's Lemma shows that the second integral vanishes as $R\to \infty$, so it suffices to compute the residue (don't forget that $ia$ is a double pole). After taking the real part of both sides, you'll have the value of the integral you desire.

Note: Jordan's Lemma is sort of an unnecessary tool for this specific problem, but I chose to refer you to it since it comes up a lot when doing contour integration, and knowing it saves a lot of time and reduces redundant arguments/calculations.

  • $\begingroup$ Thanks, this is what I wanted to do. But don't we need to show that the integral converges to show that we can calculate the symmetric limit? Also, is the reason that we choose to study the upper semicircle instead of the lower semicircle that $a$ is positive (and therefore $ia$ is in the upper half-plane)? To add one more point, I think we need to show that integral along the upper semicircle tends to 0 as R tends to $\infty$, like David Fischer said. I'm trying to show that as we speak. $\endgroup$ – Sid Aug 29 '13 at 22:26
  • $\begingroup$ I'll answer your questions in order. Typically, when doing contour integration, $\int_{-\infty}^{\infty}f(x)\ dx$ means $\lim_{R\to \infty}\int_{-R}^{R} f(x)\ dx$. In the second equation in my answer, the right hand side is constant and the second integral is going to zero - consequently, the first integral converges to the value on the right hand side (work it out! :)). The upper circle is preferred since the real part of $e^{iz}$ becomes small there. Also, showing that the integral tends to zero along the upper semicircle is precisely what Jordan's Lemma gives you. $\endgroup$ – Adam Azzam Aug 29 '13 at 22:38

Answering my comment. Note that, when you parametrize the upper have of the circle $z=Re^{I\theta},\, 0\leq \theta\leq \pi$and dealing with the integral

$$ \int_{C_R}\frac{e^{iz}}{z^2+a^2}dz $$

the integrand becomes

$$ \Bigg|\frac{e^{iz}}{z^2+a^2}\Bigg| \leq \frac{\Big|e^{i R e^{R\theta}}\Big|}{R^2-a^2} = \frac{e^{-R \sin(\theta)}}{R^2-a^2}. $$

Now, you can see that $R\sin(\theta ) \geq 0 $ for $0\leq \theta\leq \pi$ which insures that

$$ \lim_{R\to \infty} \frac{e^{-R \sin(\theta)}}{R^2-a^2} = 0 .$$


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