Definite Integral to Infinity of Arctangent of p x minus Arctangent of q x over x

Theorem

$\ds \int_0^\infty \frac {\arctan p x - \arctan q x} x \rd x = \frac \pi 2 \ln \frac p q$

where $p$ and $q$ are strictly positive real numbers.

Proof

Note that the integrand is of the form:

$\ds \int_0^\infty \frac {\map f {p x} - \map f {q x} } x \rd x$

where:

$\map f x = \arctan x$

We have, by Derivative of Arctangent Function:

$\map {f'} x = \dfrac 1 {1 + x^2}$

which is continuous on $\R$.

By Limit to Infinity of Arctangent Function:

$\ds \lim_{x \mathop \to \infty} \map f x = \lim_{x \mathop \to \infty} \arctan x = \frac \pi 2$

As $f$ is continuously differentiable and $\ds \lim_{x \mathop \to \infty} \map f x$ exists and is finite, we may apply Frullani's Integral, giving:

\(\ds \int_0^\infty \frac {\arctan p x - \arctan q x} x \rd x\)

\(=\)

\(\ds \paren {\lim_{x \mathop \to \infty} \arctan x - \arctan 0} \ln \frac p q\)

\(\ds \)

\(=\)

\(\ds \frac \pi 2 \ln \frac p q\)

Arctangent of Zero is Zero

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.67$