Improper Trigonometric Integrals

Integrals of a rational function multiplied by $\cos(\alpha x)$ or $\sin(\alpha x)$ use

e^{i\alpha x} = \cos(\alpha x) + i\sin(\alpha x)

and compare real and imaginary parts.

Existence

Let $R(z) = \frac{P(z)}{Q(z)}$ where $P$ and $Q$ are polynomials with real coefficients of degrees $m$ and $n$ . Suppose

$Q(x) \neq 0$ for all $x \in \mathbb{R}$
$n \ge m + 1$

Then for each $\alpha > 0$ , the Cauchy principal values

\text{P.V.} \int_{-\infty}^{\infty} R(x)\cos(\alpha x)\,\text{d}x \quad \text{and} \quad \text{P.V.} \int_{-\infty}^{\infty} R(x)\sin(\alpha x)\,\text{d}x

exist.

Evaluation

Under the same conditions with $\alpha > 0$ , let

F(z) = R(z) e^{i\alpha z}

If $z_1, z_2, \ldots, z_k$ are the poles of $F$ in the upper half-plane, then

\text{P.V.} \int_{-\infty}^{\infty} R(x) e^{i\alpha x}\,\text{d}x = 2\pi i \sum_{j=1}^{k} \text{Res}(F, z_j)

Consequently,

\text{P.V.} \int_{-\infty}^{\infty} R(x)\cos(\alpha x)\,\text{d}x = -2\pi \sum_{j=1}^{k} \text{Im}\left(\text{Res}(F, z_j)\right)

\text{P.V.} \int_{-\infty}^{\infty} R(x)\sin(\alpha x)\,\text{d}x = 2\pi \sum_{j=1}^{k} \text{Re}\left(\text{Res}(F, z_j)\right)

Examples:

$\text{P.V.} \int_{-\infty}^{\infty} \frac{x\sin x}{x^2+4}\,\text{d}x = \frac{\pi}{e^2}$
$\text{P.V.} \int_{-\infty}^{\infty} \frac{\cos x}{x^4+4}\,\text{d}x = \frac{\pi(\cos 1 + \sin 1)}{4e}$

Worked Examples

Cosine Integral

Evaluate $\text{P.V.} \int_{-\infty}^{\infty} \frac{\cos x}{x^2+1}\,\text{d}x$ .

Solution Reveal Hide

Here $m = 0$ , $n = 2 \ge m + 1$ , $\alpha = 1$ , and $F(z) = \frac{e^{iz}}{z^2+1}$ . The only pole in the upper half-plane is $z = i$ .

\text{Res}(F, i) = \frac{e^{i \cdot i}}{2i} = \frac{1}{2ie}

Then $\text{Res}(F, i) = -\frac{i}{2e}$ , so $\text{Im}\left(\text{Res}(F, i)\right) = -\frac{1}{2e}$ and $\text{Re}\left(\text{Res}(F, i)\right) = 0$ .

\text{P.V.} \int_{-\infty}^{\infty} \frac{\cos x}{x^2+1}\,\text{d}x = -2\pi \cdot \left(-\frac{1}{2e}\right) = \frac{\pi}{e}

\text{P.V.} \int_{-\infty}^{\infty} \frac{\sin x}{x^2+1}\,\text{d}x = 2\pi \cdot 0 = 0

Sine Integral

Show that $\text{P.V.} \int_{-\infty}^{\infty} \frac{x \sin x}{x^2+4}\,\text{d}x = \frac{\pi}{e^2}$ .

Solution Reveal Hide

Here $m = 1$ , $n = 2 \ge m + 1$ , $\alpha = 1$ , and $F(z) = \frac{z e^{iz}}{z^2+4}$ . The only pole in the upper half-plane is $z = 2i$ .

\text{Res}(F, 2i) = \frac{2i \cdot e^{-2}}{4i} = \frac{1}{2e^2}

\text{P.V.} \int_{-\infty}^{\infty} \frac{x \sin x}{x^2+4}\,\text{d}x = 2\pi \,\text{Re}\left(\frac{1}{2e^2}\right) = \frac{\pi}{e^2}

Two Poles

Show that $\text{P.V.} \int_{-\infty}^{\infty} \frac{\cos x}{x^4+4}\,\text{d}x = \frac{\pi(\cos 1 + \sin 1)}{4e}$ .

Solution Reveal Hide

Here $F(z) = \frac{e^{iz}}{z^4+4}$ . The roots of $z^4 = -4$ in the upper half-plane are $z = 1 + i$ and $z = -1 + i$ .

For a simple pole $z_k$ , using $z_k^4 = -4$ ,

\text{Res}(F, z_k) = \frac{e^{iz_k}}{4z_k^3} = -\frac{z_k e^{iz_k}}{16}

For $z = 1 + i$ , $e^{iz} = e^{-1} e^{i}$ .

\text{Res}(F, 1+i) = -\frac{(1+i)e^{i}}{16e}

For $z = -1 + i$ , $e^{iz} = e^{-1} e^{-i}$ .

\text{Res}(F, -1+i) = \frac{(1-i)e^{-i}}{16e}

Expand the products.

(1+i)e^{i} = (\cos 1 - \sin 1) + i(\cos 1 + \sin 1)

(1-i)e^{-i} = (\cos 1 - \sin 1) - i(\cos 1 + \sin 1)

The sum of the residues is

\frac{-2i(\cos 1 + \sin 1)}{16e} = -\frac{i(\cos 1 + \sin 1)}{8e}

\text{P.V.} \int_{-\infty}^{\infty} \frac{\cos x}{x^4+4}\,\text{d}x = -2\pi \left(-\frac{\cos 1 + \sin 1}{8e}\right) = \frac{\pi(\cos 1 + \sin 1)}{4e}