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Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis

I need help to evaluate the integral of

$$\int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx$$ where $a$ is a real number. I know the answer is $e^{-|a|}$.

I think contour integration can do the job by evaluating $e^{iaz}/(z^2+1)$ but I don't remember the details from complex analysis in college.

My question is "Is there a way to compute this integration WITHOUT using complex analysis method?" For example, by change of variables with trigonometric function, integration by part, or something like that.


marked as duplicate by Aryabhata, user17762, Micah, Alexander Gruber, Hagen von Eitzen Jan 8 '13 at 7:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Let $$I(a) = \int_{-\infty}^{\infty} \dfrac{\cos(ax)}{\pi(1+x^2)}dx$$ $$I''(a) = -\int_{-\infty}^{\infty} \dfrac{x^2\cos(ax)}{\pi(1+x^2)}dx$$ Hence, $$I''(a)-I(a) = - \dfrac1{\pi}\int_{-\infty}^{\infty} \cos(ax) dx = f(a)$$ The Cauchy principal value of $f(a) = 0$. Hence, we get that $$I(a) = c_1e^a + c_2e^{-a}$$

Note that $$\vert I(a) \vert \leq \int_{-\infty}^{\infty} \dfrac{dx}{\pi(1+x^2)} = I(0) = \dfrac2{\pi} \arctan(2N \pi) < 1$$ If $a>0$, this implies $c_1 = 0$. Further, $I(0) = 1$ gives us $I=e^{-a}$. Now since $I(-a) = I(a)$ for $a>0$, we get that $$I(a) = e^{-\vert a \vert}$$ for all $a \in \mathbb{R}$.

  • $\begingroup$ Why should $\int_{-2N\pi}^{2N \pi} \cos(a x) dx$ be zero? Take $a = 0$ for example. Note that the period of $\cos(a x)$ is $2 \pi /a$ if $a \neq 0$. $\endgroup$ – WimC Jan 8 '13 at 5:29
  • $\begingroup$ @WimC Yes. Valid point. However, I believe it can be fixed. Consider $a$ to be non-zero rational, then we can find $N$ such that $\int_{-2 N \pi}^{2 N \pi} \cos(ax) dx = 0$ and then by continuity argument, we should be able to get that $I''(a) - I(a) = 0$. $\endgroup$ – user17762 Jan 8 '13 at 5:56
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    $\begingroup$ @Marvis, I think you should mention that $\int_{\infty}^{\infty} dx \: cos(a x) = 2 \pi \delta(x)$, so that $I(a)$ is related to the Green function of the differential operator you posted. $\endgroup$ – Ron Gordon Jan 8 '13 at 6:27
  • $\begingroup$ @RonGordon I notice that but answers are already closed. $\endgroup$ – Felix Marin Oct 26 '13 at 9:50
  • $\begingroup$ @RonGordon I discuss the solution with the Dirac delta function here ( math.stackexchange.com/questions/540129/… ) since answer are closed for this question. $\endgroup$ – Felix Marin Oct 26 '13 at 12:09

Let $\displaystyle f(a) = \int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}\;{dx}$. Consider the Laplace transform of $f(a)$.

$$\begin{aligned}\mathcal{L}(f(a)) & = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}e^{-as}\;{da}\;{dx} \\&= \int_{0}^{\infty}\frac{s}{(1+x^2)(s^2+x^2)}\;{dx} \\& = \frac{\pi}{2(s+1)}.\end{aligned} $$

Thus $ \displaystyle f(a) =\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-|a|} $ and your integral is $\displaystyle \frac{2}{\pi}f(a) = e^{-|a|}.$

  • $\begingroup$ @MhenniBenghorbal thank you! :] $\endgroup$ – NeverBeenHere Jan 8 '13 at 8:21
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    $\begingroup$ Nice answer (+1) $\endgroup$ – user 1357113 Jan 10 '13 at 22:01

Write the integral as

$$\frac{1}{\pi} \int_{-\infty}^{\infty} dx \: \frac{\exp (i a x)}{ (1+x^2)}$$

which is still real because the imaginary part vanishes over the symmetric interval. Now consider the following complex integral

$$ \int_{C} dz \: \frac{\exp (i a z)}{ (1+z^2)}$$

where $C$ is the following contour for $a > 0$:


The value of this integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand inside $C$. These poles are at $z = \pm i$, and the residues of the integrand at these poles are

$$\mathrm{Res}_{z=\pm i} \frac{\exp (i a z)}{ (1+z^2)} = \pm \frac{\exp{(\mp a)}}{i 2}$$

For the contour $C$, only the residue at $z=i$ is inside, so the value of the integral is

$$ \int_{C} dz \: \frac{\exp (i a z)}{ (1+z^2)} = i 2 \pi \frac{\exp{(-a)}}{i 2} = \pi \exp{(-a)}$$

This integral may also be expressed in terms of the integral over the two individual components of contour $C$:

$$ \int_{C} dz \: \frac{\exp (i a z)}{ (1+z^2)} = \int_{-R}^{R} dx \: \frac{\exp (i a x)}{ (1+x^2)} + i R \int_{0}^{\pi} d \phi \: \exp{(i \phi)} \frac{\exp (i a R \exp{(i \phi)})}{ (1+R^2 \exp{(i 2 \phi)})} $$

where $R$ is the extent of $C$ along the $\Re{z}$ axis. Note that the second integral on the right-hand side results from a substitution $z = R \exp{(i \phi)}$ and corresponds to the integral along the semicircle. We take the limit as $R \rightarrow \infty$. Note that the first integral becomes the integral we seek, and we want to show that the second integral vanishes in this limit. In fact, it turns out that

$$ \left | i R \int_{0}^{\pi} d \phi \: \exp{(i \phi)} \frac{\exp (i a R \exp{(i \phi)})}{ (1+R^2 \exp{(i 2 \phi)})} \right | \approx \frac{1}{R} \int_{0}^{\pi} d \phi \: \exp{(-a R \cos{\phi})}, \: \: (R \rightarrow \infty) $$

which only converges when $a>0$; for this case, the integral vanishes as $R \rightarrow \infty$, and we can say:

$$\frac{1}{\pi} \int_{-\infty}^{\infty} dx \: \frac{\exp (i a x)}{ (1+x^2)} = \exp{(-a)} , \: \: a>0$$

When $a<0$, we flip the contour about the $\Re{z}$ axis and use the pole at $z=-i$ for the residue. In this case, we find that

$$\frac{1}{\pi} \int_{-\infty}^{\infty} dx \: \frac{\exp (i a x)}{ (1+x^2)} = \exp{(a)} , \: \: a<0$$

Combining these results and returning to the original integral expression, we get the result you sought:

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

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    $\begingroup$ OP requested without complex analysis. $\endgroup$ – WimC Jan 8 '13 at 5:43
  • $\begingroup$ Yes, but I saw the part about not remembering the details of complex analysis and thought it could help anyway. $\endgroup$ – Ron Gordon Jan 8 '13 at 5:45
  • $\begingroup$ @rlgordonma Your clear explanation is very useful to help me remember what I have learned in complex analysis. Thank you. $\endgroup$ – Patrick Li Jan 8 '13 at 14:43
  • $\begingroup$ @Patrick: You're welcome. Actually, your question gave me a chance to exercise my Mathematica graphics a bit! $\endgroup$ – Ron Gordon Jan 8 '13 at 15:08
  • $\begingroup$ Dear Ron - Maybe you recall I expressed appreciation for your contribution. Not that this is quid pro quo, but maybe you would not mind my asking a general question, as I am currently self-studying a course in CA with substantially more intention than my prior endeavors. <<Other than cultivating the art and perseverance that it takes to solve these integrals, do they have any practical applications. By that I mean to other realms of math itself.>> I have seen in Edwards "RZF" there are a few integrals (a motivation). But other than coursework, I haven't seen much. Thanks Best regards, $\endgroup$ – user12802 May 16 '15 at 21:14

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