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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? $\endgroup$ – IAmNoOne Nov 4 '14 at 3:04
  • $\begingroup$ 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. $\endgroup$ – Colin Nov 4 '14 at 3:09
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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 :) $\endgroup$ – Colin Nov 4 '14 at 3:09
  • $\begingroup$ @Colin, well technically complex integration can simplify that last integral. $\endgroup$ – IAmNoOne Nov 4 '14 at 3:10
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    $\begingroup$ If the OP wants to avoid complex numbers, though, it would be doable using integration by parts. $\endgroup$ – GFauxPas Nov 4 '14 at 3:11
  • $\begingroup$ 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! $\endgroup$ – Colin Nov 4 '14 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. $\endgroup$ – Colin Nov 4 '14 at 3:52
  • $\begingroup$ You should do whatever you find easier! $\endgroup$ – GFauxPas Nov 4 '14 at 3:55

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