MHB Definite integral involving sine and hyperbolic sine

AI Thread Summary
The integral $\int_0^{\infty} \frac{\sin x}{\cos x + \cosh x}\, \mathrm dx$ can be evaluated using exponential forms of the sine, cosine, and hyperbolic cosine functions. The sine function is expressed as $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$, while cosine and hyperbolic cosine are represented as $\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$, respectively. This transformation simplifies the integral into a more manageable form for calculation. The discussion emphasizes the utility of exponential representations in solving integrals involving trigonometric and hyperbolic functions. Ultimately, the integral can be solved effectively by applying these exponential identities.
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Calculate $\displaystyle \int_0^{\infty} \frac{\sin x}{\cos x + \cosh x}\, \mathrm dx.$
 
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If nothing else, you can express these functions as exponentials.

$sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$
$cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$
$cosh(x)= \frac{e^x+ e^{-x}}{2}$
 
Country Boy said:
If nothing else, you can express these functions as exponentials.

$sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$
$cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$
$cosh(x)= \frac{e^x+ e^{-x}}{2}$
Can you derive an infinite series of the integrand using these definitions?
 
Show that $\displaystyle \frac{\sin x }{\cos x + \cosh x} = i \bigg( \frac{1}{1+e^{ix-x}}-\frac{1}{1+e^{-ix-x}}\bigg)$.

Expand the RHS into geometric series to get, for $x \ge 0$:

$\displaystyle \frac{\sin x}{\cos x + \cosh x}=2\sum_{n=1}^{\infty}(-1)^{n-1}\sin(nx)e^{-nx}.$

Source: Yaghoub Sharifi.
 
Using the infinite sum in post #4, the answer follows by switching the order of sum and integral and using the result

$\displaystyle \int_0^{\infty} e^{-ax}\sin{bx} \, \mathrm dx =\frac{b}{a^2+b^2}$

Which can be derived via integration by parts or considering the fact that $\Im \left( e^{-ax+ibx} \right) = e^{-ax}\sin bx.$

See a detailed solution in this blog.
 
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