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I need to find $\displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\ dx$ where $a > 0$. To do this, I set $f(z) = \displaystyle\frac{\cos z}{z^{2} + a^{2}}$ and integrate along the semi-circle of radius $R$. For the residue at $ia$ I get $\displaystyle\frac{\cos(ia)}{2ia}$. Then letting $R \rightarrow \infty$, the integral over the arc is zero, so I get $\displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\ dx = 2 \pi i \frac{\cos(ia)}{2ia} = \frac{\pi \cos(ia)}{a}$. But this is supposed to be $\displaystyle\frac{\pi e^{-a}}{a}$, so I am doing something wrong.

In a similar problem, I have to evaluate $\displaystyle\int_{-\infty}^{\infty} \frac{x \sin x}{x^{2} + a^{2}}\ dx$ and get $\pi i \sin(ia)$ whereas this is supposed to be $\pi e^{-a}$. I think I am getting some detail wrong in both cases. Can anyone enlighten me?

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    $\begingroup$ The integral over the arc is NOT zero if you are integrating $\cos(x)$ (cosines become hyperbolic cosines for imaginary $x$ which grow exponentially!). It IS, if you replace $\cos x$ with $e^{ix}$ and then take the real part at the very end of the evaluation. $\endgroup$
    – Alex R.
    May 3, 2012 at 20:51
  • $\begingroup$ I was able to derive the result correctly by replacing $\cos(x)$ with $e^{ix}$ as you said, but I do not understand why the integral would not be zero. The length of the curve is $\pi R$. The supremum would be at most $\frac{1}{R^{2} + a^{2}}$. How come this isn't zero? $\endgroup$
    – Pedro
    May 3, 2012 at 21:03
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    $\begingroup$ Again, you are thinking that $\cos(x)$ is between -1 and 1, but this is only true for real $x$. Otherwise you have $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ so try to plug in $x=a+bi$ and you will see that you are gurunteed to have exponential blowup as $|x|\rightarrow \infty$ along any path away from the real axis $\endgroup$
    – Alex R.
    May 3, 2012 at 21:52
  • $\begingroup$ You really need either to use $\cos(x)=\frac{\exp(ix)+\exp(-ix)}{2}$ or $\cos(x)=\Re(\exp(ix))$ so that we work with $\exp(ix)$ over the upper half-plane. $\cos(ix)$ blows up exponentially as the imaginary part of $x$ gets large, so we can't use it in the contour integration. $\endgroup$
    – robjohn
    May 4, 2012 at 2:23

5 Answers 5

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Let $\gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big. $$ \begin{align} \int_{-\infty}^\infty\frac{\cos(x)}{x^2+a^2}\mathrm{d}x\tag{1} &=\Re\left(\int_{-\infty}^\infty\frac{\exp(ix)}{x^2+a^2}\mathrm{d}x\right)\\\tag{2} &=\Re\left(\int_{\gamma}\frac{\exp(ix)}{x^2+a^2}\mathrm{d}x\right)\\\tag{3} &=\Re\left(2\pi i\,\mathrm{Res}\left(\frac{\exp(ix)}{x^2+a^2},ia\right)\right)\\\tag{4} &=\Re\left(2\pi i\,\lim_{z\to ia}\frac{\exp(ix)}{x+ia}\right)\\\tag{5} &=\Re\left(2\pi i\,\frac{\exp(-a)}{2ia}\right)\\[3pt] &=\frac{\pi \exp(-a)}{a}\tag6 \end{align} $$ $(1)$: $\Re(\exp(ix))=\cos(x)$
$(2)$: The integral along the circle back through the upper half-plane vanishes as the circle gets bigger.
$(3)$: There is only one singularity of $\dfrac{\exp(ix)}{x^2+a^2}$ in the upper half-plane, at $x=ia$. The integral along $\gamma$ is the residue of $\dfrac{\exp(ix)}{x^2+a^2}$ at $x=ia$.
$(4)$: The singularity of $\dfrac{\exp(ix)}{x^2+a^2}$ at $x=ia$ is a simple pole. We can compute the residue as $\displaystyle\lim_{x\to ia}(x-ia)\frac{\exp(ix)}{x^2+a^2}=\lim_{x\to ia}\frac{\exp(ix)}{x+ia}$.
$(5)$: plug in $x=ia$.
$(6)$: evaluate


Following the same strategy, $$ \begin{align} \int_{-\infty}^\infty\frac{x\sin(x)}{x^2+a^2}\mathrm{d}x &=\Im\left(\int_{-\infty}^\infty\frac{x\exp(ix)}{x^2+a^2}\mathrm{d}x\right)\\ &=\Im\left(\int_{\gamma}\frac{x\exp(ix)}{x^2+a^2}\mathrm{d}x\right)\\ &=\Im\left(2\pi i\,\mathrm{Res}\left(\frac{x\exp(ix)}{x^2+a^2},ia\right)\right)\\ &=\Im\left(2\pi i\,\lim_{z\to ia}\frac{x\exp(ix)}{x+ia}\right)\\ &=\Im\left(2\pi i\,\frac{ia\exp(-a)}{2ia}\right)\\[6pt] &=\pi \exp(-a) \end{align} $$

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  • $\begingroup$ Would it be valid to preform the integral along the lower half plane moving clockwise? $\endgroup$
    – akozi
    Feb 4, 2020 at 1:50
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    $\begingroup$ @akozi: yes. You will have to use $\color{#C00}{-}\Im\left(\int_{-\infty}^\infty\frac{x\exp(\color{#C00}{-}ix)}{x^2+a^2}\mathrm{d}x\right)$ $\endgroup$
    – robjohn
    Feb 4, 2020 at 3:45
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Although the OP is searching for a way forward using contour integration and the residue theorem, I thought it might be instructive to present an approach that uses real analysis only. To that end, we proceed.


Let $g(a)$ be given by the convergent improper integral

$$g(a)=\int_{-\infty}^\infty \frac{\cos(x)}{x^2+a^2}\,dx \tag1$$

Exploiting the even symmetry of the integrand and enforcing the substitution $x\to ax$ reveals

$$g(a)=\frac2{|a|}\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx$$

Now let $f(a)=\frac {|a|}2 g(a)=\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx$.

Since the integral $\int_0^\infty \frac{x\sin(ax)}{x^2+1}\,dx$ is uniformly convergent for $|a|\ge \delta>0$, we may differentiate under the integral in $(3)$ for $|a|>\delta>0$ to obtain

$$\begin{align} f'(a)&=-\int_0^\infty \frac{x\sin(ax)}{x^2+1}\,dx\\\\ &=-\int_0^\infty \frac{(x^2+1-1)\sin(ax)}{x(x^2+1)}\,dx\\\\ &=-\int_0^\infty \frac{\sin(ax)}{x}\,dx+\int_0^\infty \frac{\sin(ax)}{x(x^2+1)}\,dx\\\\ &=-\frac{\pi}{2}\text{sgn}(a)+\int_0^\infty \frac{\sin(ax)}{x(x^2+1)}\,dx\tag4 \end{align}$$

Again, since the integral $\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx$ converges uniformly for all $a$, we may differentiate under the integral in $(4)$ to obtain for $a\ne 0$

$$f''(a)=\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx=f(a)\tag 5$$

Solving the second-order ODE in $(5)$ reveals

$$f(a)=C_1 e^{a}+C_2 e^{-a}$$

Using $f(0)=\pi/2$ and $f'(0^\pm)=\mp \pi/2$, we find that $f(a)=\frac{\pi e^{-|a|}}{2}$. Multiplying by $2/|a|$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{\cos(x)}{x^2+a^2}\,dx=\frac{\pi}{|a|e^{|a|}}}$$

as expected!

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  • $\begingroup$ I just corrected I typo, I hope you don't mind. $\endgroup$
    – Shashi
    Dec 26, 2017 at 10:01
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    $\begingroup$ @shashi Not at all. Much appreciative. $\endgroup$
    – Mark Viola
    Dec 26, 2017 at 16:14
  • $\begingroup$ I am sorry,I saw their end part which is for sine function. $\endgroup$
    – Fawad
    Jan 3, 2019 at 16:06
  • $\begingroup$ (+1) how to generalise this answer for $\int_{0}^{\infty} \frac{cos(bx)}{a^2+x^2} dx$ $\endgroup$
    – M Desmond
    Mar 7, 2019 at 6:04
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    $\begingroup$ @mdesmond Have you tried to proceed as in this post? $\endgroup$
    – Mark Viola
    Mar 7, 2019 at 16:08
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Let us integrate along the contour $\Gamma$, which is the semicircle of radius $R$ described above. Let $C_R$ be the arc of the contour with radius $R$. We may say that

$$\oint_{\Gamma}\frac{\cos z}{z^2+a^2}\, dz=\int_{-R}^{R}\frac{\cos x}{x^2+a^2}\, dx+\int_{C_R}\frac{\cos z}{z^2+a^2}\, dz$$

Note that $\cos x = \operatorname{Re}\,(e^{ix})$. Thus

$$ f(z)=\frac{\operatorname{Re}\,(e^{iz})}{z^2+a^2} = \operatorname{Re}\,\left(\frac{e^{iz}}{(z+ia)(z-ia)}\right) $$

Because $f(z)$ can be written ias $g(z)e^{iaz}$ and suffices all of the conditions of Jordan's lemma, we see that the integral along the arc tends to zero when $R\to \infty$. Thus

$$ \lim_{R\to\infty}\oint_{\Gamma}\frac{\cos z}{z^2+a^2}\, dz=\lim_{R\to\infty} \int_{-R}^{R}\frac{\cos x}{x^2+a^2}\, dx+0 $$

To solve the LHS, we find the residues. The residue, similar to the one you found is $$ b=\frac{e^{i^2a}}{ia+ia}=\frac{e^{-a}}{2ia} $$

Thus

$$ \displaystyle\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}\, dx= \lim_{R\to\infty}\oint_{\Gamma}\frac{\cos z}{z^2+a^2}\, dz=\operatorname{Re}\,\left(\displaystyle\int_{-\infty}^{\infty} \frac{e^{iz}}{z^{2} + a^{2}}\, dz\right) =\operatorname{Re}\,(2\pi ib)=\operatorname{Re}\,(2\pi i\frac{e^{-a}}{2ia}) = \frac{\pi e^{-a}}{a} $$


Similarly, with $\displaystyle\int_{-\infty}^{\infty} \frac{x \sin x}{x^{2} + a^{2}}\ dx$ we change $\sin x$ to $\operatorname{Im}\,(e^{ix})$ and make the function complex-valued.

$$g(z)=\frac{z\sin z}{z^2+a^2}=\operatorname{Im}\,\left(\frac{ze^{iz}}{(z+ia)(z-ia)}\right)$$

Jordan's lemma again can be used to show that the integral around the arc tends to zero as $R\to\infty$. We then proceed to find the residues. The residue of pole $z_1=ia$ is

$$b=\frac{iae^{-a}}{2ia}=\frac{e^{-a}}{2}$$

$z_1$ is the unique pole in the contour, so upon multiplying its residue, $b$, with $2\pi i$ we find: $$ \displaystyle\int_{-\infty}^{\infty} \frac{x \sin x}{x^{2} + a^{2}}\, dx= \lim_{R\to\infty}\oint_{\Gamma} \frac{x \sin x}{x^{2} + a^{2}}\, dz= \operatorname{Im}\,\left(\displaystyle\int_{-\infty}^{\infty} \frac{ze^{iz}}{(z+ia)(z-ia)}\, dz\right)= \operatorname{Im}\,(2\pi ib)= \operatorname{Im}\,(2\pi i \frac{e^{-a}}{2})= \operatorname{Im}\,(\pi e^{-a}i)= \pi e^{-a} $$

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  • $\begingroup$ It would be good to describe the contour along which you are integrating. $\endgroup$
    – robjohn
    May 4, 2012 at 0:34
  • $\begingroup$ @robjohn I am using the one described above: "integrate along the semi-circle of radius R"... $\endgroup$
    – Argon
    May 4, 2012 at 1:09
  • $\begingroup$ @Argon Maybe also mention one needs to use Jordan's lemma to estimate the contribution from the arc for the $\int \frac{x\sin x}{x^2+a^2} dx $ integral, just the usual Length * Max estimate fails there. $\endgroup$ May 4, 2012 at 1:19
  • $\begingroup$ @RagibZaman Very true, I will include this. $\endgroup$
    – Argon
    May 4, 2012 at 1:49
  • $\begingroup$ @robjohn the contour he uses is clearly stated the first sentence, re-read the solution. Very nice solution Argon, thanks a lot! $\endgroup$ Dec 17, 2013 at 5:59
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The integrals $\int_{0}^{\infty}\frac{\cos (x)}{x^{2}+a^{2}} \, dx$ and $\int_{0}^{\infty} \frac{x \sin (x)}{x^{2}+a^{2}} \, dx$ are specific cases of the integral $$I(s,a) = \int_{0}^{\infty} x^{s-1} \, \frac{\sin\left(\frac{\pi s}{2}-x \right)}{x^{2}+a^{2}} \, dx = \frac{\pi}{2} \, a^{s-2} e^{-a}, \quad 0 < s< 3, \ a>0. \tag{1}$$

(This integral appears as an exercise on page 154 of the textbook An Introduction to the Theory of Functions of a Complex Variable by E.T. Copson. Copson attributes it to Cauchy.)

Using the branch of $z^{s-1}$ that corresponds to the branch of the logarithm where $- \frac{\pi}{2} <\arg(z) \le \frac{3 \pi}{2}$, we can show $(1)$ by integrating the function $$f(z) =z^{s-1} \, \frac{e^{- i \pi s/2}e^{iz}}{z^{2}+a^{2}} $$ counterclockwise around an indented contour that consists of the real axis and the large semicircle above it. (The indentation is needed to avoid the branch point at the origin.)

Near the origin, $f(z)$ behaves like a constant times $z^{s-1}$. As long as $s >0$, the contribution from the semicircular indentation about the branch point at the origin will vanish as its radius goes to zero.

So letting the radius of the large semicircle go to $\infty$ and applying Jordan's lemma, we get, for $0 < s< 3$, $$\begin{align} \int_{-\infty}^{0} (|x|e^{i \pi})^{s-1} \, \frac{e^{- i \pi s/2} e^{ix}}{x^{2}+a^{2}} \, dx + \int_{0}^{\infty} x^{s-1} \frac{e^{- i \pi s/2}e^{ix}}{x^{2}+a^{2}} \, dx &= 2 \pi i \operatorname{Res} [f(z), e^{i \pi /2}a] \\ &= 2 \pi i \left( (e^{i \pi /2}a)^{s-1} \frac{e^{- i \pi s/2} e^{-a}}{2ia}\right)\\ & = \frac{\pi}{i} \, a^{s-2} e^{-a}. \end{align}$$

But the left side of the equation can be written as $$-\int_{0}^{\infty} u^{s-1} \, \frac{e^{i \pi s/2}e^{-iu}}{u^{2}+a^{2}} \, du+ \int_{0}^{\infty} x^{s-1} \frac{e^{- i \pi s/2}e^{ix}}{x^{2}+a^{2}} \, dx = 2i \int_{0}^{\infty} x^{s-1} \, \frac{\sin\left(x- \frac{\pi s}{2} \right)}{x^{2}+a^{2}} \, dx.$$ The result then follows.

As a side note, notice that the value of $I(s,1)$ does not depend on $s$.

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  • $\begingroup$ Note the link is not working. $\endgroup$
    – MathArt
    Apr 2 at 8:17
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Following method makes no use of contour integration and Leibniz integral rule, $\mathcal{P}\{\}$ is Cauchy principal value.

Assume $a>0$,

$\int_{-\infty}^\infty\frac{\cos(x)}{x^2+a^2}\text{d}x=\frac{1}{2a}\Re\Bigg\{\int_{-\infty}^\infty\left(\frac{e^{xi}}{a+xi}+\frac{e^{xi}}{a-xi}\right)\text{d}x\Bigg\}$

$$\operatorname{Ei}(z):=-\mathcal{P}\Bigg\{\int_1^\infty\frac{e^{zt}}{t}\text{d}t\Bigg\}$$

\begin{align*}\operatorname{Ei}(z)-\operatorname{Ei}(\overline{z})=-2i\mathcal{P}\Bigg\{\int_1^{\infty}\frac{e^{\Re(z)t}\sin(\Im(z)t)}{t} \text{d}t\Bigg\}\\=-2i\mathcal{P}\Bigg\{\int_{\Im(z)}^\infty\frac{e^{\frac{\Re(z)}{\Im(z)}t}\sin(t)}{t}\text{d}t\Bigg\}\end{align*}

Let $z\mapsto \overline{z}$,

$$\operatorname{Ei}(z)-\operatorname{Ei}(\overline{z})=2i\mathcal{P}\Bigg\{\int_{-\Im(z)}^\infty\frac{e^{-\frac{\Re(z)}{\Im(z)}t}\sin(t)}{t}\text{d}t\Bigg\}$$

We have for $\Re(z)>0$,

$$\lim_{\Im(z)\to\infty}\operatorname{Ei}(z)-\operatorname{Ei}(\overline{z})=2\pi i$$

Clearly for $\Re(z)<0$,

$$\lim_{\Im(z)\to\infty}\operatorname{Ei}(z)-\operatorname{Ei}(\overline{z})=0$$

Using above limits and the fact that $\operatorname{Ei}'(z)=\frac{e^z}{z}$,

\begin{align*}\int_{-\infty}^\infty\frac{e^{xi}}{a+xi}\text{d}x=-ie^{-a}\int_{a-\infty i}^{a+\infty i}\frac{e^t }{t} \text{d}t=-ie^{-a}\lim_{z\to\infty}\left(\operatorname{Ei}(a+zi)-\operatorname{Ei}(a-zi)\right)\\=2\pi e^{-a}\end{align*} \begin{align*}\int_{-\infty}^\infty\frac{e^{xi}}{a-xi}\text{d}x=ie^{a}\int_{-a-\infty i}^{-a+\infty i}\frac{e^{t}}{t}\text{d}t=ie^a\lim_{z\to\infty}\left(\operatorname{Ei}(-a+zi)-\operatorname{Ei}(-a-zi)\right)\\=0\end{align*}

$$\implies\int_{-\infty}^\infty\frac{\cos(x)}{x^2+a^2}\text{d}x=\frac{\pi}{ae^a}$$

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