Question:
How do I show that the following integral converges:
$$\mathcal{I}_k:=\int_0^\infty\int_0^\infty\left|\cos\left(s\right)\cdot u^{\frac{1}{2k}-1}\cdot\exp\left(-su\right)\right|\space\text{d}s\space\text{d}u\tag1$$
Where $k$ is a positive integer.
My work:
I started with:
$$\mathcal{I}_k=\int_0^\infty\int_0^\infty\left|\cos\left(s\right)\right|\cdot\left|u^{\frac{1}{2k}-1}\right|\cdot\left|\exp\left(-su\right)\right|\space\text{d}s\space\text{d}u=$$ $$\int_0^\infty u^{\frac{1}{2k}-1}\cdot\left\{\int_0^\infty\left|\cos\left(s\right)\right|\cdot\exp\left(-su\right)\space\text{d}s\right\}\space\text{d}u\tag2$$
$$ \begin{align} &\int_0^\infty\left|\cos s\right| e^{-su}\,ds=\\ &\sum_{k=0}^\infty\left( \int_{2k\pi-\pi/2}^{2k\pi+\pi/2}\cos s\,e^{-su}\,ds- \int_{2k\pi+\pi/2}^{2k\pi+3\pi/2}\cos s\,e^{-su}\,ds \right)-\int_{-\pi/2}^{0}\cos s\,e^{-su}\,ds=\\ &\sum_{k=0}^\infty\left( {1+e^{\pi u}\over1+u^2}e^{-\pi u/2-2k\pi u}+ {1+e^{\pi u}\over1+u^2}e^{-3\pi u/2-2k\pi u} \right)-{e^{\pi u/2}-u\over1+u^2}=\\ &{e^{-\pi u/2}+e^{\pi u/2}\over1+u^2}(1+e^{-\pi u}) \sum_{k=0}^\infty e^{-2k\pi u}-{e^{\pi u/2}-u\over1+u^2}=\\ &{e^{-\pi u/2}+e^{\pi u/2}\over1+u^2}(1+e^{-\pi u}) {1\over 1-e^{-2\pi u}}-{e^{\pi u/2}-u\over1+u^2}=\\ &{e^{\pi u/2}\over1+u^2} {1+e^{-\pi u}\over 1-e^{-\pi u}}-{e^{\pi u/2}-u\over1+u^2}=\\ &{e^{\pi u/2}\over1+u^2} {2e^{-\pi u}\over 1-e^{-\pi u}}+{u\over1+u^2}.\\ \end{align} $$ From here, it should not be difficult to take on. But with the presence of a term $1-e^{-\pi u}$ in the denominator I don't see how the integral could converge.