Theorem: Definite Integral to Infinity of Cosine of a x over Hyperbolic Cosine of b x

:$\ds \int_0^\infty \frac {\cos a x} {\cosh b x} \rd x = \frac \pi {2 b} \sech \frac {a \pi} {2 b}$

Proof:
{{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 = Euler's Cosine Identity, {{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 | o = 
      | 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 = 
}}
{{eqn | 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:
:$\ds \pi \map \sech {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 z^2}$
Setting $z = \dfrac a {2 b}$ gives: 
:$\ds \pi \map \sech {\frac {a \pi} {2 b} } = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1} + 4 \paren {\frac a {2 b} }^2}$
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}}