Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x by Cosecant of p x
Jump to navigation
Jump to search
Theorem
- $\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \arctan \frac b p - \arctan \frac a p$
where:
- $a$ and $b$ are non-negative real numbers
- $p$ is a non-zero real number.
Proof
Fix $p$ and set:
- $\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x$
for all $\alpha \ge 0$.
Then:
- $\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \map I a - \map I b$
We have:
\(\ds \map {I'} \alpha\) | \(=\) | \(\ds \frac \d {\d \alpha} \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {e^{-\alpha x} } {x \csc p x} } \rd x\) | Definite Integral of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds -\int_0^\infty e^{-\alpha x} \sin p x \rd x\) | Derivative of $e^{a x}$, Definition of Cosecant | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac p {\alpha^2 + p^2}\) | Definite Integral to Infinity of $e^{-a x} \sin b x$ |
so:
\(\ds \map I \alpha\) | \(=\) | \(\ds -p \int \frac 1 {\alpha^2 + p^2} \rd \alpha\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\arctan \frac \alpha p + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ |
for all $\alpha \ge 0$ and constant $C \in \R$.
We therefore have:
\(\ds \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x\) | \(=\) | \(\ds \map I a - \map I b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-\arctan \frac a p + C} - \paren {-\arctan \frac b p + C}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \arctan \frac b p - \arctan \frac a p\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.88$