<s>[INST] Integrate $\int^\frac{T}{2}_0\frac{dt}{\pi ^2 t^2 (\csc^2 \frac{\pi t}{T}) -T^2}$. 
Evaluate
  $$\int^\frac{T}{2}_0 \frac{dt}{\pi ^2 t^2 (\csc^2 \frac{\pi t}{T})-T^2}.$$

My work. Let $u=\frac{\pi t}{T}$, then $du=\frac{\pi}{T}dt$ and
$$\frac{T}{\pi} \int^\frac{\pi}{2}_0\frac{du}{u^2 T^2 (\csc^2 u)-T^2}=\frac{1}{\pi T} \int^\frac{\pi}{2}_0\frac{du}{u^2 (\csc^2 u) -1}.$$
Please help! I do not know what to do next.
 [/INST] Yes, you are on the right track. For $T>0$,
$$\int^\frac{T}{2}_0 \frac{dt}{\pi ^2 t^2 (\csc^2 \frac{\pi t}{T})-T^2}=\frac{1}{\pi T} \int^\frac{\pi}{2}_0\frac{du}{u^2 \csc^2 u -1}.$$
Now, note that in a right neighbourhood of $0$:
$$0<\frac{1}{u^2 (\csc^2 u) -1}=\frac{\sin^2 u}{u^2-\sin^2 u}=\frac{u^2+o(u^2)}{u^2-(u^2-2\frac{u\cdot u^3}{3!}+o(u^4))}=\frac{u^2+o(u^2)}{\frac{u^4}{3}+o(u^4)}\sim \frac{3}{u^2}$$
so the integral diverges to $+\infty$.
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