MHB Integral Challenge #1: Prove 0 ≤ a < π/2

[polygamma]

Show that for $ \displaystyle 0 \le a < \frac{\pi}{2}$,

$$ \int_{0}^{\infty} e^{-x \cos a} \cos(x \sin a) \cos (bx) \ dx = \frac{(b^{2}+1) \cos a}{b^{4}+2b^{2} \cos (2a) + 1 }$$When I post integral challenge problems in the future, I'll just number them.

 

[alyafey22]

Show that for $ \displaystyle 0 \le a < \frac{\pi}{2}$,

$$ \int_{0}^{\infty} e^{-x \cos a} \cos(x \sin a) \cos (bx) \ dx = \frac{(b^{2}+1) \cos a}{b^{4}+2b^{2} \cos (2a) + 1 }$$When I post integral challenge problems in the future, I'll just number them.

Spoiler

\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 $$0 \leq a < \frac{\pi}{2}$$