How to evaluate these?! Part 2

Let $I(a,k)=\int_{0}^{\infty} \dfrac{\cos x}{x+a} ~ e^{-k(x+a)} \text{d}x$

Now partially differentiating the above integral with respect to $k$ , we get $\dfrac{ \partial I}{\partial k}=- \int_{0}^{\infty} \cos x ~ e^{-k(x+a)} \text{d}x=-e^{-ka} \int_{0}^{\infty} \cos x ~ e^{-kx} \text{d}x$

Evaluating the above integral is easy by Integration by parts, which comes out to be $-e^{-ka} \int_{0}^{\infty} \cos x ~ e^{-kx} \text{d}x=- \dfrac{k ~ e^{-ak}}{1+k^2}$

And now, since $I(a,k) \to 0$ as $k \to \infty$ , we deduce that $\int_{0}^{\infty} \dfrac{\cos x}{x+a}\text{d}x= \lim_{k \to 0} I(a,k) = \int_{0}^{\infty} \dfrac{k ~ e^{-ka}}{k^2+1} \text{d}k = \int_{0}^{\infty} \dfrac{u ~ e^{-u}}{u^2+a^2} \text{d}u$

which implies $g(a)=1$ for $a > 0$ .

Hence $g(\pi+e)g(\pi-e)=\boxed{1}$ .