Integrals like $\int_{-\infty}^\infty\frac{\cos x}{x^2 + 1}dx$ appear in Fourier analysis and in physics. The half-disk method handles them too, with one change: replace $\cos(ax)$ by the exponential $e^{iaz}$, which stays small in the upper half-plane.
Key idea
For $a > 0$ and a rational $R(x)$ with no real poles that tends to $0$ at $\pm\infty$:
$$\int_{-\infty}^{\infty}R(x)e^{iax}\,dx = 2\pi i\sum_{\operatorname{Im} z_j > 0}\operatorname{Res}\left(R(z)e^{iaz}, z_j\right),$$
and the cosine and sine integrals are its real and imaginary parts (for real $R$). For example, for $k, b > 0$,
$$\int_{-\infty}^{\infty}\frac{\cos(kx)}{x^2 + b^2}\,dx = \frac{\pi e^{-kb}}{b}.$$
Why it works
On the upper half-plane, $z = x + iy$ with $y \ge 0$, so $|e^{iaz}| = |e^{iax}e^{-ay}| = e^{-ay} \le 1$: the exponential does not grow, and when $\deg Q \ge \deg P + 2$ the ML bound kills the arc as before. (When $R$ decays only like $\frac{1}{x}$, a sharper estimate called Jordan's lemma still kills it.) For the example: the upper pole is $bi$, with residue $\frac{e^{ik(bi)}}{2bi} = \frac{e^{-kb}}{2bi}$, and $2\pi i$ times that is $\frac{\pi e^{-kb}}{b}$, a real number, which is the cosine integral (the sine part is $0$ by symmetry).
Worked examples
Worked example: A cosine integral
Evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{\cos(2x)}{x^2 + 1}\,dx$.
- Use $f(z) = \frac{e^{2iz}}{z^2 + 1}$; the upper pole is $i$.
- $\operatorname{Res}(f, i) = \frac{e^{2i \cdot i}}{2i} = \frac{e^{-2}}{2i}$.
- $2\pi i\cdot\frac{e^{-2}}{2i} = \pi e^{-2}$. Real part: $\pi e^{-2} \approx 0.425$.
Answer: $\pi e^{-2}$
Worked example: A sine integral
Evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{x\sin x}{x^2 + 9}\,dx$.
- Use $f(z) = \frac{ze^{iz}}{z^2 + 9}$; the upper pole is $3i$. Here the top has degree only one less, so Jordan's lemma is needed for the arc.
- $\operatorname{Res}(f, 3i) = \frac{3i\,e^{-3}}{6i} = \frac{e^{-3}}{2}$, so $\int f = 2\pi i\cdot\frac{e^{-3}}{2} = \pi i\,e^{-3}$.
- The sine integral is the imaginary part: $\pi e^{-3}$.
Answer: $\pi e^{-3}$
- Replace $\cos(ax)$ or $\sin(ax)$ by $e^{iaz}$ (with $a > 0$).
- Find the poles of $R(z)$ in the upper half-plane and the residues of $R(z)e^{iaz}$ there.
- Multiply the sum by $2\pi i$; take the real part for cosine, the imaginary part for sine.
Worked example: A residue to check
Find the residue of $\dfrac{e^{3iz}}{z^2 + 4}$ at $z = 2i$, and use it to evaluate $\displaystyle\int_{-\infty}^{\infty}\frac{\cos 3x}{x^2 + 4}\,dx$.
- Simple pole: $\frac{e^{3i(2i)}}{2(2i)} = \frac{e^{-6}}{4i}$.
- $2\pi i\cdot\frac{e^{-6}}{4i} = \frac{\pi e^{-6}}{2}$, which matches $\frac{\pi e^{-kb}}{b}$ with $k = 3$, $b = 2$.
Answer: $\frac{e^{-6}}{4i}$; $\frac{\pi e^{-6}}{2}$
Common mistake
Replacing $\cos x$ by $\cos z$. On the real line they agree, but in the upper half-plane $\cos z = \frac{e^{iz} + e^{-iz}}{2}$ contains $e^{-iz}$, whose size is $e^{y}$: it explodes on the arc, and the arc integral does not vanish. Integrate $R(z)e^{iaz}$ (with $a > 0$, so the exponential decays upward) and take the real or imaginary part at the end. If $a < 0$, close in the lower half-plane instead.