$$ \int_{0}^{\infty} e^{-x \cos a} \cos(x \sin a) \cos (bx) \ dx = \frac{(b^{2}+1) \cos a}{b^{4}+2b^{2} \cos (2a) + 1 }$$
Proof
\begin{align}
\int_{0}^{\infty} e^{-x \cos a} \cos(x \sin a) \cos (bx) \ dx &= \mathrm{Re}\int_{0}^{\infty} e^{-x \cos a} e^{ix \sin a } \cos (bx) \ dx\\ &= \mathrm{Re}\int_{0}^{\infty} e^{-x e^{-ia}} \cos (bx) \ dx \\& = \mathrm{Re}\frac {e^{-ia}}{e^{-2ia}+b^2}\\&= \mathrm{Re}\frac{\cos(a)-i\sin(a)}{\cos(2a)-i\sin(2a)+b^2}\\&= \mathrm{Re}\frac{(\cos(a)-i\sin(a))(\cos(2a)+b^2+i\sin(2a))}{(\cos(2a)+b^2)^2+\sin^2(2a)}\\&= \frac{\cos(2a)\cos(a)+\cos(a)b^2+\sin(2a)\sin(a)}{b^4+2b^2 \cos(2a)+1}\\&=\frac{(1+b^2)\cos(a)}{b^4+2b^2 \cos(2a)+1}
\end{align}
The convergence is justified by the Laplace transform. Since \( |\cos(bx)| \leq 1\) so it is of an exponential order and we can take\( |\cos(bx)| \leq e^{0\, x} \) so the value of \(c=0\). Hence the integral converges to the value for \(\mathrm{Re}(e^{ia})>0\) or \(\cos(a)>0\) which clearly satisfy \(0 \leq a < \frac{\pi}{2}\)
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