I am interested to compute the following integral

$$I=\int_{0}^{+\infty}\frac{\cos x}{a^2+x^2}dx$$

where $a\in\mathbb{R}^+$. Let me explain my first idea. As the integrand is an even function of $x$ then

$$2I=\int_{-\infty}^{+\infty}\frac{\cos x}{a^2+x^2}dx=\lim_{R\to+\infty}\int_{-R}^{R}\frac{\cos x}{a^2+x^2}dx:=\lim_{R\to+\infty}J$$

So, I first focus on computing the $J$ integral by first modifying it as follows

\begin{align*} J&=\int_{-R}^{R}\frac{\cos x}{a^2+x^2}dx=\int_{-R}^{R}\frac{\cos x}{a^2+x^2}dx+i\int_{-R}^{R}\frac{\sin x}{a^2+x^2}dx \\ &= \int_{-R}^{R}\frac{(\cos x+i\sin x)}{a^2+x^2}dx = \int_{-R}^{R}\frac{\exp(ix)}{a^2+x^2}dx \end{align*}

Then, I use the well-known techniques of complex variable theory. First, I replace the real variable $x$ in $J$ with a complex variable $z$ and consider a contour integral over $C=C_1\cup C_2$


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Then, according to the Cauchy's integral theorem and the Residue theorem, I get

\begin{align*} K=J+\int_{C_2}\frac{\exp(iz)}{a^2+z^2}dz &= \int_{C_3}\frac{\exp(iz)}{a^2+z^2}dz=\int_{C_3}\frac{\exp(iz)}{(z+ia)(z-ia)}dz \\ &=2\pi i \frac{\exp(i^2a)}{2ia}=\frac{\pi}{a}\exp(-a) \end{align*}

Next, taking the limit $R\to+\infty$ from the above relation, we obtain


but, we can show that


and then we can obtain the final result


First, please check my steps to see the final result is correct or not. Second, is there any other way to compute $I$?

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    $\begingroup$ a beautiful way: math.stackexchange.com/a/2480371/515527 $\endgroup$ – Zacky Apr 12 '18 at 16:28
  • $\begingroup$ This is the most beautifully explained question I've seen on the math.stackexchange. $(+1)$ if I did not reach my daily voting limit... $\endgroup$ – Mr Pie Apr 12 '18 at 16:28
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    $\begingroup$ and here:math.stackexchange.com/a/272663/515527 $\endgroup$ – Zacky Apr 12 '18 at 16:33
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    $\begingroup$ @zacky: Thanks for the link. How did you search the site to find the links? I wasn't able to find anything! $\endgroup$ – H. R. Apr 12 '18 at 16:34
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    $\begingroup$ I posted a week ago a pretty similar integral: math.stackexchange.com/questions/2723369/… And if you look at the linked questions(on the right) you find some of them, altough I searched alot of time last week to find similar integrals since I also was interested. $\endgroup$ – Zacky Apr 12 '18 at 16:39

A rather exotic approach leading to a known functional equation, which has an exponential solution.

Consider a function ($a>0$):

$$f(a)=a \int_{-\infty}^\infty \frac{\cos x}{a^2+x^2} dx=\int_{-\infty}^\infty \frac{\cos a x}{1+x^2} dx=\pi e^{-a}$$

Let's square it and change the dummy variable:

$$f^2(a)=\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\cos a x \cos a y}{(1+x^2)(1+y^2)} dx dy=$$

$$=\frac{1}{2} \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\cos a (x-y)+ \cos a (x+y)}{(1+x^2)(1+y^2)} dx dy$$

Due to the infinite limits, we can easily make substitutions in the form $x \pm y=t$, which will lead to the following expression under the integral:

$$\frac{\cos a t}{y^2+1} \left(\frac{1}{(y+t)^2+1} +\frac{1}{(y-t)^2+1} \right)$$

We will do partial fraction decomposition to integrate w.r.t. $y$.

$$\frac{1}{((y+t)^2+1)(y^2+1)}=\frac{1}{t (4+t^2)} \left(\frac{2y+3t}{(y+t)^2+1}-\frac{2y-t}{y^2+1} \right)$$

$$\frac{1}{((y-t)^2+1)(y^2+1)}=\frac{1}{t (4+t^2)} \left(\frac{-2y+3t}{(y-t)^2+1}-\frac{-2y-t}{y^2+1} \right)$$

Let's consider separately the 'problematic' integrals, but with finite limits:

$$\int_{-L}^L \frac{2y dy}{(y+t)^2+1}=\int_{-L-t}^{L+t} \frac{2u du}{u^2+1}-2t\int_{-L-t}^{L+t} \frac{du}{u^2+1} $$

The first integral vanishes, the second after taking the limit $L \to \infty$, gives us $-2 \pi t$. In the same way we find the other integral with $(y-t)^2$.

So the two 'problematic' integrals give us:

$$-2 \pi \int_{-\infty}^\infty \frac{\cos at ~dt}{4+t^2}$$

Grouping the other terms we get:


After integration w.r.t. $y$ and adding all the results, we obtain:

$$f^2(a)=2\pi \int_{-\infty}^\infty \frac{\cos at ~dt}{4+t^2}=\pi f(2a)$$

The functional equation:

$$f^2 (a)=\pi f(2a)$$

has a general solution:

$$f(a)=\pi e^{c a}$$

We should have $c<0$, as can be seen by considering the original integral definition and taking the limit $a \to \infty$.

I'm not sure how to prove $c=-1$, but it should be possible.

  • $\begingroup$ (+1) Thanks for the attention and your different approach. :) $\endgroup$ – H. R. Apr 13 '18 at 19:40
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    $\begingroup$ @H.R., thank you. You could also check out this answer math.stackexchange.com/a/1841104/269624. That's the method I planned to use initially, but this user already has a great solution this way. Which is why I made up another one $\endgroup$ – Yuriy S Apr 13 '18 at 20:30

For any $a>0$, $$ I(a)=\int_{0}^{+\infty}\frac{\cos(x)}{x^2+a^2}\,dx = \frac{1}{a}\int_{0}^{+\infty}\frac{\cos(ax)}{1+x^2}\,dx = \frac{J(a)}{a} $$ and the Laplace transform of $J(a)$ is given by $$ \int_{0}^{+\infty}J(a) e^{-sa}\,da = \int_{0}^{+\infty}\int_{0}^{+\infty}\frac{\cos(ax)e^{-sa}}{1+x^2}\,dx\,da $$ or, by invoking Fubini's theorem and integration by parts: $$ \int_{0}^{+\infty}\frac{s}{(1+x^2)(s^2+x^2)}\,dx =\frac{\pi}{2(1+s)}$$ by partial fraction decomposition. $\mathcal{L}^{-1}$ then gives $J(a)=\frac{\pi}{2}e^{-a}$ and $I(a)=\frac{\pi}{2a}e^{-a}$ as wanted.


Another way is to prove $$\int_\mathbb{R}\dfrac{a}{\pi}\dfrac{e^{ikx}}{a^2+x^2}=\exp -a|k|$$for $a>0$, by noting we're just trying to compute the characteristic function of a Cauchy distribution. The inversion theorem implies we need only check this characteristic function gives the right pdf. To prove $$\int_\mathbb{R}\exp (-ikx-a|k|)dk=\dfrac{2a}{a^2+x^2},$$write the left-hand side as the sum of integrals either side of $k=0$. The left-hand side is then $$\dfrac{1}{a+ix}+\dfrac{1}{a-ix},$$as required.


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