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I'd like to integrate

$$I(a)=\int \frac{\sin(x)}{\cosh(ax)+\cos(x)}dx$$

by changing $sin$ etc... into their exponential representation.

Using $e^{ix} = \cos(x) + i \sin(x)$ and $e^{ax} = \cosh(ax)+\sinh(ax)$ we have

$$\frac{1}{i} \int \frac{e^{ix}-e^{-ix}}{e^{ax}+e^{-ax}+e^{ix}+e^{-ix}}dx =I(a)$$

If one substitutes $e^x = y$ we get $y^i$, can we do this? Can one get this method to work? A very similar integral has an answer:

http://integralsandseries.prophpbb.com/topic397.html?sid=77128c640169d5c07a9b32a5e7c35bc2

mike
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bolbteppa
  • 4,389

1 Answers1

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Consider the integral \begin{align} I(a) = \int \frac{\sin(x) }{ \cos(x) + \cos(ax) } \ dx. \end{align} This integral can be readily evaluated as is given by \begin{align} I(a) = \frac{1}{a^{2}-1} \ \left[ (a+1) \ln\left( \cos\left(\frac{(1-a)x}{2}\right)\right) - (a-1) \ln\left( \cos\left(\frac{(1+a)x}{2}\right)\right) \right]. \end{align} Now letting $a \rightarrow i a$ leads to \begin{align} I(ia) &= \frac{-1}{a^{2}+1} \ \left[ (1+ia) \ln\left( \cos\left(\frac{(1-ia)x}{2}\right)\right) + (1-ia) \ln\left( \cos\left(\frac{(1+ia)x}{2}\right)\right) \right] \\ &= \frac{-1}{a^{2}+1} \ \left[ \ln\left(\cos\left(\frac{(1-ia)x}{2}\right) \cos\left(\frac{(1+ia)x}{2}\right) \right) + ia \ \ln\left(\frac{\cos\left(\frac{(1-ia)x}{2}\right)}{\cos\left(\frac{(1+ia)x}{2}\right)}\right) \ \right] \\ I(ia) &= \frac{-1}{a^{2}+1} \ \left[ \ln\left(\frac{\cos(x) + \cosh(ax)}{2}\right) + ia \ \ln\left( \frac{1+i \tan(x/2) \tanh(ax/2)}{1-i \tan(x/2) \tanh(ax/2)}\right) \right]. \end{align} The second term can be reduced as follows. It is known that \begin{align} x+iy = \sqrt{x^{2}+y^{2}} \ e^{i \tan^{-1}(y/x)} \end{align} which helps lead to \begin{align} \ln(x+iy) = \frac{1}{2} \ \ln(x^{2}+y^{2}) + i \tan^{-1}(y/x). \end{align} From this result it is seen that \begin{align} \ln\left( \frac{1+i \tan(x/2) \tanh(ax/2)}{1-i \tan(x/2) \tanh(ax/2)}\right) = 2 i \ \tan^{-1}\left(\tan\left(\frac{x}{2}\right) \ \tanh^{-1}\left( \frac{ax}{2} \right) \right) \end{align} and now $I(ia)$ becomes \begin{align} I(ia) &= \frac{2a}{a^{2}+1} \ \tan^{-1}\left(\tan\left(\frac{x}{2}\right) \ \tanh^{-1}\left( \frac{ax}{2} \right) \right) - \frac{1}{a^{2}+1} \ \ln\left(\frac{\cos(x) + \cosh(ax)}{2}\right). \end{align} Hence the desired integral value is \begin{align} \int \frac{\sin(x) }{ \cos(x) + \cosh(ax) } \ dx = \frac{2a}{a^{2}+1} \ \tan^{-1}\left(\tan\left(\frac{x}{2}\right) \ \tanh^{-1}\left( \frac{ax}{2} \right) \right) - \frac{1}{a^{2}+1} \ \ln\left(\frac{\cos(x) + \cosh(ax)}{2}\right). \end{align}

Leucippus
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  • "The integral can be readily evaluated as given..." I think this first step is the crucial one; perhaps you could give some clue as to how the integral resolves to this value? – abiessu Jun 02 '14 at 13:13
  • Interesting solution, though the hope was to explicitly stay with the e^ix, e^-ix along with the e^ax and e^-ax stuff, if you have any idea how to get that to work it'd be fantastic to see. – bolbteppa Jun 02 '14 at 14:12