Integral Integral9

$\begin{aligned} I & = \int_0^\infty \frac 1{\pi^2\csc^2 \frac \pi k}dk \\ & = \frac 1{\pi^2} \int_0^\infty \sin^2 \frac \pi k \ dk & \small \color{#3D99F6} \text{By integration by part} \\ & = \frac {k\sin^2 \frac \pi k}{\pi^2} \bigg|_0^\infty - \color{#3D99F6} \frac 1{\pi^2} \int_0^\infty \frac {-2\pi \sin \frac \pi k \cos \frac \pi k}k d k & \small \color{#3D99F6} \text{Let }x = \frac 1k \implies dx = - \frac {dk}{k^2} \\ & = 0 + \color{#3D99F6} \frac 2\pi \int_0^\infty \frac {\sin x \cos x}x dx \\ & = \frac 2\pi \int_0^\infty \frac {\sin 2x}{2x} dx & \small \color{#3D99F6} \text{Let }u = 2x \implies du = 2\ dx \\ & = \frac 1\pi \color{#3D99F6} \int_0^\infty \frac {\sin u}{u} du \\ & = \frac 1\pi \color{#3D99F6} \text{Si}(\infty) & \small \color{#3D99F6} \text{where }\text{Si}(x) \text{ is the sine integral.} \\ & = \frac 1\pi \times \color{#3D99F6} \frac \pi 2 \\ & = \frac 12 \end{aligned}$

Therefore, $a=\boxed{2}$ .

$\text { We know that } a \ne 0, \text { then } a \text { must be \boxed { 2 } .}$