Who's up to the challenge?1

$\displaystyle\sinh{x}=\frac{e^x-e^{-x}}{2}$ so $\displaystyle e^{-x}\sinh{x} = \frac{1}{2}- \frac{e^{-2x}}{2}$

$\displaystyle\int_{0}^{\infty}( \frac{1}{2}- \frac{e^{-2x}}{2}) \cos^2{x}dx$

The area under $\frac{\cos^2{x}}{2}$ clearly diverges, but now the trick is to find that the area of $\frac{\cos^2{x}}{2e^{2x}}$ converges.

$\frac{\cos^2{x}}{2e^{2x}}$ is less than $\frac{1}{2e^{2x}}$ for all $x>0$ so by the comparison test (and logic!) $\displaystyle\int_{0}^{\infty}\frac{\cos^2{x}}{2e^{2x}}$ converges. Clearly then, a divergent integral minus a convergent integral diverges.

The value is clearly positive but -a/b is negative. At least we can predict the integral is not in the form -a/b, so we can answer 1000.

$\displaystyle \int_0^\infty e^{-x} \sinh x \cos^2 x dx \space = \space \int_0^\infty e^{-x} \dfrac{(e^x-e^{-x})}{2} \cos^2 x \, dx \\ = \displaystyle \int_0^\infty e^{-x} \dfrac{(e^x-e^{-x})}{2} \dfrac{\cos{2 x}+1}{2} dx = \dfrac{1}{4} \int_0^\infty (1-e^{-2x}) (\cos{2x}+1) dx$

Upon expanding this expression, we see have terms like the constant, $1$ , and $\cos{2x}$ , which do not converge when their integrals are evaluated at infinity. Therefore, the integral diverges.