This formula appears in Gradshteyn and Ryzhik as formula 2.674.4. However it's too simple to have been included in Victor Moll's set of proofs. I've attacked it using integration by parts using both $u = {e^{ax}}$ and $u = \cos bx\cosh cx$. Both methods end up with circular integration by parts. Writing ${I_{cch}} \equiv \int{\cos bx\cosh cx}$ etc, we find: $$\begin{array}{l} {I_{cch}} \to {I_{sch}} + {I_{csh}}\\ {I_{sch}} \to {I_{cch}} + {I_{ssh}}\\ {I_{csh}} \to {I_{ssh}} + {I_{cch}}\\ {I_{ssh}} \to {I_{csh}} + {I_{sch}} \end{array}$$ Now this can be solved using matrix algebra, but it's getting very messy. I've tried as well writing the trigonometric and hyperbolic functions as exponents, but it's also a bit messy. Just wondering if anyone has a simpler approach?
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$\begingroup$ Just a thought, but maybe try complex integrable? $\endgroup$– IAmNoOneCommented Nov 4, 2014 at 3:04
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$\begingroup$ I started down the road of writing it as just the real component of the complex integral in which cos and cosh are written in terms of e. Quite messy algebra. $\endgroup$– ColinCommented Nov 4, 2014 at 3:09
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2 Answers
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Write $\displaystyle \cosh cx = \frac{1}{2} \left( e^{cx} + e^{-cx} \right)$ and then use or rederive standard results for $\displaystyle \int e^{\alpha x} \cos(\beta x) \ dx$.
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$\begingroup$ So no need for complex, and write only one of the functions in terms of e? I like it :) $\endgroup$– ColinCommented Nov 4, 2014 at 3:09
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$\begingroup$ @Colin, well technically complex integration can simplify that last integral. $\endgroup$– IAmNoOneCommented Nov 4, 2014 at 3:10
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1$\begingroup$ If the OP wants to avoid complex numbers, though, it would be doable using integration by parts. $\endgroup$– GFauxPasCommented Nov 4, 2014 at 3:11
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$\begingroup$ And that standard result is G&R formula 2.663(3), and application of it indeed gives their result 2.674(4). Nice one Simon :) Much simpler than the other approaches I tried - cheers! $\endgroup$– ColinCommented Nov 4, 2014 at 3:51
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If you want to invoke complex numbers, you can use Euler's formula:
$e^{ibx} = \cos bx + i \sin bx$
$\implies e^{ax}e^{ibx} = e^{ax}\cos bx + i e^{ax} \sin bx$
$\displaystyle \implies e^{ax}e^{ibx}\frac{(e^{cx}+e^{-cx})}2 = e^{ax}\cos bx \cosh cx + i e^{ax}\sin bx \cosh cx$
Take the integral of the LHS and then take the real part.
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$\begingroup$ Thanks GFauxPas, I had tried that approach but the algebra got messy. Simon's suggestion to just change one of the functions to exponential form I think works best. $\endgroup$– ColinCommented Nov 4, 2014 at 3:52
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$\begingroup$ You should do whatever you find easier! $\endgroup$– GFauxPasCommented Nov 4, 2014 at 3:55