$\lim_{n→∞}∫_0^{\pi/2}\frac{\sin nx}{\sin x}dx=\frac\pi2$

hbghlyj

$\frac1{\sin x}\notin L^1(0,\frac\pi2)$，故Riemann–Lebesgue lemma不适用。

SageMath

ListPlot[{Pi/2, 2, Pi/2, 4/3, Pi/2, 26/15, Pi/2, 152/105, Pi/2, 526/315, Pi/2, 5156/3465, Pi/2}]

hbghlyj

$$\left\{\begin{aligned}I_{2n-1}&=\frac\pi2
\\I_{2 n}&=\sum_{k=1}^{n}\frac{2(-1)^{k-1}}{2 k-1}\end{aligned}\right.$$
artofproblemsolving.com/community/c7h534270p3061505

math.stackexchange.com/questions/263705

Leibniz formula for $π$
\[
\lim _{n \to \infty} \sum_{k=1}^n \frac{2(-1)^{k-1}}{2 k-1}=\frac{\pi}{2}
\]