Integration of $\int {{e^{ax}}\cos bx\cosh cx\,dx}$

Question

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?

Answer 1

Write $\displaystyle \cosh cx = \frac{1}{2} \left( e^{cx} + e^{-cx} \right)$ and then use or rederive standard results for $\displaystyle \int e^{\alpha x} \cos(\beta x) \ dx$.

Answer 2

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.