Computing $\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx$ using residue calculus

Solution 1:

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

Solution 2:

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

$$g(a)=\frac2a\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx$$

Now let $f(a)=\frac a2 g(a)=\int_0^\infty \frac{\cos(ax)}{x^2+1}\,dx$

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

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

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

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

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

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

Using $f(0)=\pi/2$ and $f'(0)=-\pi/2$, we find that $C_1=0$ and $C_2=\frac{\pi}{2}$ and hence $f(a)=\frac{\pi e^{-a}}{2}$. Multiplying by $2/a$ yields the coveted result

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

as expected!

Solution 3:

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

Solution 4:

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 both 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 $s$.