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

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

With a view to expressing the primitive in the form:

$\ds \int f g' \rd t = f g - \int f' g \rd t$

let:

\(\ds f\) \(=\) \(\ds \sin p x \sin q x\)
\(\ds \leadsto \ \ \) \(\ds f'\) \(=\) \(\ds p \cos p x \sin q x + q \sin p x \cos q x\) Product Rule for Derivatives and Derivative of $\sin a x$
\(\ds g'\) \(=\) \(\ds \frac 1 {x^2}\)
\(\ds \leadsto \ \ \) \(\ds g\) \(=\) \(\ds -\frac 1 x\) Chain Rule for Derivatives


So:

\(\ds \int_0^\infty \frac {\sin p x \sin q x} {x^2} \rd x\) \(=\) \(\ds \intlimits {-\frac {\sin p x \sin q x} x} 0 \infty + \int_0^\infty \frac {p \cos p x \sin q x + q \sin p x \cos q x} x \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds 0 + \int_0^\infty \frac {p \cos p x \sin q x + q \sin p x \cos q x} x \rd x\) evaluating limits using L'Hôpital's Rule and Product Rule for Limits of Functions
\(\ds \) \(=\) \(\ds p \cdot \int_0^\infty \frac {\cos p x \sin q x} x \rd x + q \cdot \int_0^\infty \frac {\sin p x \cos q x} x \rd x\) Linear Combination of Definite Integrals


Case $0 < p = q$

\(\ds \) \(\) \(\ds p \cdot \int_0^\infty \frac {\cos p x \sin q x} x \rd x + q \cdot \int_0^\infty \frac {\sin p x \cos q x} x \rd x\)
\(\ds \) \(=\) \(\ds 2 p \cdot \int_0^\infty \frac {\cos p x \sin p x} x \rd x\) Substitute $p$ for $q$
\(\ds \) \(=\) \(\ds 2 p \cdot \frac \pi 4\) Integral to Infinity of $\dfrac {\sin p x \cos q x} x$
\(\ds \) \(=\) \(\ds \frac {\pi p} 2 = \frac {\pi q} 2\)

$\Box$


Case $0 < p \le q$:

Suppose $0 < p < q$.

Then:

\(\ds \) \(\) \(\ds p \cdot \int_0^\infty \frac {\cos p x \sin q x} x \rd x + q \cdot \int_0^\infty \frac {\sin p x \cos q x} x \rd x\)
\(\ds \) \(=\) \(\ds p \cdot \frac \pi 2 + q \cdot 0\) Integral to Infinity of $\dfrac {\sin p x \cos q x} x$
\(\ds \) \(=\) \(\ds \frac {\pi p} 2\) simplifying

Adjoin to the case where $p = q$.

$\Box$


Case $p \ge q > 0$:

Suppose $p > q > 0$.

Then:

\(\ds \) \(\) \(\ds p \cdot \int_0^\infty \frac {\cos p x \sin q x} x \rd x + q \cdot \int_0^\infty \frac {\sin p x \cos q x} x \rd x\)
\(\ds \) \(=\) \(\ds p \cdot 0 + q \cdot \frac \pi 2\) Integral to Infinity of $\dfrac {\sin p x \cos q x} x$
\(\ds \) \(=\) \(\ds \frac {\pi q} 2\) simplifying

Adjoin to the case where $p = q$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Integral_to_Infinity_of_Sine_p_x_Sine_q_x_over_x_Squared&oldid=690655"