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This is another problem that I got stuck on during my self-study of Complex Variables.

$$\int_{0}^{\infty} \frac{\cos(ax)}{(1+x^2)^2}\mathrm dx \quad\text{ with }a>0$$

This is equivalent to $$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(ax)}{(1+x^2)^2}\mathrm dx$$

I know the following:

(1) There is a pole at $x=i$

(2) We work with the half-circle with radius $R$.

(3) I think I used the identity Re$(e^{ai\theta})$

After that I get a bit murky.

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You can interchange integration and real part extraction: $\int \mathrm{Re}=\mathrm{Re}\int$. So do you know how to apply the residue theorem to $$\int_\gamma \frac{e^{iaz}}{(1+z^2)^2}dz?$$ – anon Oct 12 '11 at 7:11
Not really...I'm trying to work on getting down how to apply the residue theorem. – emka Oct 12 '11 at 7:46
up vote 8 down vote accepted

Let $$\begin{equation*} f(z)=\frac{e^{iaz}}{\left( 1+z^{2}\right) ^{2}}=\frac{e^{iaz}}{ (z-i)^{2}(z+i)^{2}}. \end{equation*}\tag{1} $$ The residue of $f(z)$ at $z=i$ is $$ \begin{eqnarray*} \underset{z=i}{\mathrm{res}}f(z) &=&\frac{1}{(2-1)!}\lim_{z\rightarrow i}\frac{d}{dz}\left((z-i)^2f(z)\right)\\&=&\lim_{z\rightarrow i}\frac{d}{dz}\left( \frac{e^{iaz}}{(z+i)^{2}}\right)=\lim_{z\rightarrow i}\frac{ iae^{iaz}(z+i)^{2}-e^{iaz}2\left( z+i\right) }{(z+i)^{4}} \\ &=&-\frac{1}{4}i\left( a+1\right) e^{-a}. \end{eqnarray*}\tag{2} $$

enter image description here

Let $C_{R}$ denote the boundary of the upper half of the disk $ \left\vert z\right\vert =R$, described counterclockwise (see picture). By the residue theorem $$ \begin{eqnarray*} \int_{-R}^{R}\frac{e^{iax}}{\left( 1+x^{2}\right) ^{2}}dx+\int_{C_{R}}\frac{ e^{iaz}}{\left( 1+z^{2}\right) ^{2}}dz &=&2\pi i \underset{z=i}{\ \mathrm{ res}} f(z)e^{iaz} \\ &=&\frac{1}{2}\pi \left( a+1\right) e^{-a}. \end{eqnarray*}\tag{3} $$ Then $$ \begin{eqnarray*} \text{Re}\int_{-R}^{R}\frac{e^{iax}}{\left( 1+x^{2}\right) ^{2}}dx+\text{Re} \int_{C_{R}}\frac{e^{iaz}}{\left( 1+z^{2}\right) ^{2}}dz &=&\text{Re}\frac{1}{2}\pi \left( a+1\right) e^{-a}, \\ \\ \int_{-R}^{R}\frac{\cos ax}{\left( 1+x^{2}\right) ^{2}}dx+\text{Re} \int_{C_{R}}\frac{e^{iaz}}{\left( 1+z^{2}\right) ^{2}}dz &=&\frac{1}{2}\pi \left( a+1\right) e^{-a}. \end{eqnarray*}\tag{4} $$

When $\left\vert z\right\vert =R$, we have

$$\left\vert \frac{1}{(1+z^{2})^{2}}\right\vert =\frac{1}{\left\vert z+i\right\vert ^{2}\left\vert z-i\right\vert ^{2}}\leq \frac{1}{\left\vert \left\vert z\right\vert -|i\right\vert ^{2}\left\vert \left\vert z\right\vert -\left\vert i\right\vert \right\vert ^{2}}=\frac{1}{(R-1) ^{4}}=:M_R\tag{5},$$

which means that $M_R>0$ and $$ \begin{equation*} \lim_{R\rightarrow \infty }M_R= \lim_{R\rightarrow \infty }\frac{1}{(R-1) ^{4}}=0. \end{equation*}\tag{6} $$

Then we can apply the Jordan's lemma for every positive constant $a$ and conclude that

$$ \begin{equation*} \lim_{R\rightarrow \infty }\int_{C_{R}}\frac{e^{iaz}}{\left( 1+z^{2}\right) ^{2}}dz=0. \end{equation*}\tag{7} $$ Consequently,
$$ \begin{equation*} \int_{-\infty }^{\infty }\frac{\cos ax}{\left( 1+x^{2}\right) ^{2}}dx=\frac{1 }{2}\pi \left( a+1\right) e^{-a} \end{equation*}\tag{8} $$ and $$ \begin{equation*} \int_{0}^{\infty }\frac{\cos ax}{\left( 1+x^{2}\right) ^{2}}dx=\frac{1}{4} \pi \left( a+1\right) e^{-a}. \end{equation*}\tag{9} $$

Note: Exercise 3 on page 265 of Complex Variables and Applications by James Brown and Ruell Churchill generalizes this integral to

$$ \begin{equation*} \int_{0}^{\infty }\frac{\cos ax}{( b^2+x^{2}) ^{2}}dx=\frac{1}{4b^3} \pi \left( ab+1\right) e^{-ab}\qquad (a>0,b>0). \end{equation*} $$

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This is amazing...thanks for your help. – emka Oct 12 '11 at 19:05
@EMKA. You are welcome. Glad to help. – Américo Tavares Oct 12 '11 at 19:25

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