Integral to Infinity of Sine p x Cosine q x over x

From ProofWiki
Jump to navigation Jump to search

Theorem

$\ds \int_0^\infty \frac {\sin p x \cos q x} x \rd x = \begin {cases} \dfrac \pi 2 & : p > q > 0 \\

0 & : 0 < p < q \\ \dfrac \pi 4 & : p = q > 0 \end {cases}$


Proof

\(\ds \int_0^\infty \frac {\sin p x \cos q x} x \rd x\) \(=\) \(\ds \int_0^\infty \frac 1 2 \cdot \frac {\sin \paren {\paren {p + q} x} + \sin \paren {\paren {p - q} x} } x \rd x\) Werner Formula for Sine by Cosine
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^\infty \frac {\sin \paren {\paren {p + q} x} } x \rd x + \frac 1 2 \int_0^\infty \frac {\sin \paren {\paren {p - q} x} } x \rd x\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to \infty} \frac 1 2 \int_0^t \frac {\sin \paren {\paren {p + q} x} } x \rd x + \lim_{t \mathop \to \infty} \frac 1 2 \int_0^t \frac {\sin \paren {\paren {p - q} x} } x \rd x\) Definition of Improper Integral on Closed Interval Unbounded Above
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to \infty} \frac 1 2 \int_0^{t \paren {p + q} } \frac {\sin u} u \rd u + \lim_{t \mathop \to \infty} \frac 1 2 \int_0^t \frac {\sin \paren {\paren {p - q} x} } x \rd x\) Integration by Substitution: $u = \paren {p + q} x$
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to \infty} \frac 1 2 \int_0^{t \paren {p + q} } \frac {\sin u} u \rd u + \lim_{t \mathop \to \infty} \frac 1 2 \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v\) Integration by Substitution: $v = \paren {p - q} x$
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^\infty \frac {\sin u} u \rd u + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v\) Definition of Improper Integral on Closed Interval Unbounded Above: by hypothesis $p + q$ is (strictly) positive
\(\ds \) \(=\) \(\ds \frac \pi 4 + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v\) Dirichlet Integral


Case $p > q > 0$:

\(\ds \) \(\) \(\ds \frac \pi 4 + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v\)
\(\ds \) \(=\) \(\ds \frac \pi 4 + \frac 1 2 \int_0^\infty \frac {\sin v} v \rd v\) Definition of Improper Integral on Closed Interval Unbounded Above: by hypothesis $p - q$ is (strictly) positive
\(\ds \) \(=\) \(\ds \frac \pi 4 + \frac 1 2 \cdot \frac \pi 2\) Dirichlet Integral
\(\ds \) \(=\) \(\ds \frac \pi 2\) simplifying

$\Box$


Case $0 < p < q$:

\(\ds \) \(\) \(\ds \frac \pi 4 + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v\)
\(\ds \) \(=\) \(\ds \frac \pi 4 + \frac 1 2 \int_0^{-\infty} \frac {\sin v} v \rd v\) Definition of Improper Integral on Closed Interval Unbounded Below: by hypothesis $p - q$ is (strictly) negative
\(\ds \) \(=\) \(\ds \frac \pi 4 - \frac 1 2 \int_0^\infty \frac {\sin w} w \rd w\) Integration by Substitution: $w = -v$
\(\ds \) \(=\) \(\ds \frac \pi 4 - \frac 1 2 \cdot \frac \pi 2\) Dirichlet Integral
\(\ds \) \(=\) \(\ds 0\) simplifying

$\Box$


Case $p = q > 0$:

\(\ds \) \(\) \(\ds \frac \pi 4 + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^{t \paren {p - q} } \frac {\sin v} v \rd v\)
\(\ds \) \(=\) \(\ds \frac \pi 4 + \frac 1 2 \lim_{t \mathop \to \infty} \int_0^0 \frac {\sin v} v \rd v\) $p - q = 0$ by hypothesis
\(\ds \) \(=\) \(\ds \frac \pi 4 + \frac 1 2 \cdot 0\) Definite Integral on Zero Interval, independently of $t$
\(\ds \) \(=\) \(\ds \frac \pi 4\) simplifying

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Integral_to_Infinity_of_Sine_p_x_Cosine_q_x_over_x&oldid=680466"