{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\cos a x} {\cosh b x} \rd x 	| r = \int_0^\infty \frac {e^{-b x} \paren {e^{i a x} + e^{-i a x} } } {1 - \paren {-e^{-2 b x} } } \rd x 	| c = Cosine Exponential Formulation, {{Defof|Hyperbolic Cosine}} }} {{eqn	| r = \int_0^\infty \paren {e^{\paren {i a - b} x} + e^{-\paren {i a + b} x} } \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n e^{-2 n b x} } \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \int_0^\infty \paren {-1}^n \paren {e^{\paren {i a - \paren {2 n + 1} b} x} + e^{-\paren {i a + \paren {2 n + 1} b} x} } \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\intlimits {-\frac {e^{\paren {i a - \paren {2 n + 1} b} x} } {\paren {2 n + 1} b - i a} } 0 \infty + \intlimits {-\frac {e^{-\paren {i a + \paren {2 n + 1} b} x} } {\paren {2 n + 1} + i a} } 0 \infty} 	| c = Primitive of $e^{a x}$ }}  {{end-eqn}} We have, as $b, n > 0$: {{begin-eqn}} {{eqn	| l = \size {\lim_{x \mathop \to \infty} e^{\paren {i a - \paren {2 n + 1} b} x} } 	| r = \lim_{x \mathop \to \infty} \size {e^{\paren {i a - \paren {2 n + 1} b} x} } 	| c = Modulus of Limit }} {{eqn	| r = \lim_{x \mathop \to \infty} \size {e^{i a x} } \size {e^{-\paren {2 n + 1} b x} } 	| c = Exponential of Sum }} {{eqn	| r = \lim_{x \mathop \to \infty} e^{-\paren {2 n + 1} b x} }} {{eqn	| r = 0 	| c = Exponential Tends to Zero and Infinity }} {{end-eqn}} We similarly have: {{begin-eqn}} {{eqn	| l = \size {\lim_{x \mathop \to \infty} e^{-\paren {i a + \paren {2 n + 1} b} x} } 	| r = \lim_{x \mathop \to \infty} \size {e^{-\paren {i a + \paren {2 n + 1} b} x} } 	| c = Modulus of Limit }} {{eqn	| r = \lim_{x \mathop \to \infty} \size {e^{-i a x} } \size {e^{-\paren {2 n + 1} b x} } 	| c = Exponential of Sum }} {{eqn	| r = \lim_{x \mathop \to \infty} e^{-\paren {2 n + 1} b x} }} {{eqn	| r = 0 	| c = Exponential Tends to Zero and Infinity }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\intlimits {-\frac {e^{\paren {i a - \paren {2 n + 1} b} x} } {\paren {2 n + 1} b - i a} } 0 \infty + \intlimits {-\frac {e^{-\paren {i a + \paren {2 n + 1} b} x} } {\paren {2 n + 1} + i a} } 0 \infty} 	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac 1 {\paren {2 n + 1} b - i a} + \frac 1 {\paren {2 n + 1} b + i a} } 	| c = Exponential of Zero }} {{eqn	| r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac {2 \paren {2 n + 1} b} {\paren {2 n + 1}^2 b^2 + a^2} } 	| c = Difference of Two Squares }} {{eqn	| r = \frac 2 b \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 \paren {\frac a {2 b} }^2} }} {{end-eqn}} By Mittag-Leffler Expansion for Hyperbolic Secant Function: :$\displaystyle \pi \map \sech {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 z^2}$ Setting $z = \dfrac a {2 b}$ gives:  :$\displaystyle \pi \map \sech {\frac {a \pi} {2 b} } = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 \paren {\frac a {2 b} }^2}$ We therefore have:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\cos a x} {\cosh b x} \rd x 	| r = \frac 2 b \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 \paren {\frac a {2 b} }^2} }} {{eqn	| r = \frac {\pi} {2 b} \sech \frac {a \pi} {2 b} }} {{end-eqn}} {{qed}}