正方形の中心までの距離の期待値

(1)
有名問題です。OEISのA103712などを見ると良いです。
領域 $-1\leq x \leq 1, -1 \leq y \leq 1$から一様ランダムに点$(x, y)$を選び、距離$r = \sqrt{x^2 + y^2}$について、期待値$E_1$を求める。
第1象限だけ考えて4倍すればよいので、

\begin{aligned} E_1 &= \frac{1}{4} \int_{-1}^{1}\int_{-1}^{1} \sqrt{x^2+y^2}\,dx\,dy \\ &= \int_{0}^{1}\int_{0}^{1} \sqrt{x^2+y^2}\,dx\,dy \\ \end{aligned}

そのまま解いても良いが、ここで、$0\le\theta\le\dfrac{\pi}{4}$では$r\cos\theta=1$、$\dfrac{\pi}{4}\le\theta\le\dfrac{\pi}{2}$では$r\sin\theta=1$、また$\mathrm{d}x\mathrm{d}y = r\mathrm{d}r\mathrm{d}\theta$であるから、

\begin{aligned} E_1 &= \int_{0}^{\pi/4} \int_{0}^{1/\cos\theta} r^2 \,dr\,d\theta + \int_{\pi/4}^{\pi/2} \int_{0}^{1/\sin\theta} r^2 \,dr\,d\theta \\ &= \frac{1}{3} \int_{0}^{\pi/4} \frac{1}{\cos^3\theta} \,d\theta + \frac{1}{3} \int_{\pi/4}^{\pi/2} \frac{1}{\sin^3\theta} \,d\theta \\ &= \frac{1}{3} \int_{0}^{\pi/4} \frac{1}{\cos^3\theta} \,d\theta+ \frac{1}{3} \int_{\pi/4}^{\pi/2} \frac{1}{\cos^3\left(\frac{\pi}{2}-\theta\right)} \,d\theta \\&\xrightarrow{\ \phi=\frac{\pi}{2}-\theta\ }\frac{1}{3} \int_{0}^{\pi/4} \frac{1}{\cos^3\theta} \,d\theta+ \frac{1}{3} \int_{0}^{\pi/4} \frac{1}{\cos^3\phi} \,d\phi \\ &= \frac{2}{3} \int_{0}^{\pi/4} \frac{1}{\cos^3\theta} \,d\theta\\ &= \frac{2}{3} \int_{0}^{\pi/4} \frac{\cos\theta}{(1-\sin^2\theta)^2}\,d\theta \\&\xrightarrow{\,u=\sin\theta\,}\frac{2}{3} \int_{0}^{1/\sqrt{2}} \frac{1}{(1-u^2)^2}\,du \\&= \frac{2}{3} \int_{0}^{1/\sqrt{2}} \frac{1}{(1-u)^2(1+u)^2}\,du \\&= \frac{2}{3} \int_{0}^{1/\sqrt{2}} \left(\frac{1}{4}\frac{1}{1-u}+ \frac{1}{4}\frac{1}{(1-u)^2}+ \frac{1}{4}\frac{1}{1+u}+ \frac{1}{4}\frac{1}{(1+u)^2}\right) du \\&= \frac{1}{6} \left[-\ln(1-u) + \frac{1}{1-u}+ \ln(1+u) - \frac{1}{1+u}\right]_{0}^{1/\sqrt{2}} \\&= \frac{1}{6} \left[\ln\frac{1+u}{1-u} + \frac{2u}{1-u^2}\right]_{0}^{1/\sqrt{2}} \\&= \frac{1}{6} \left[\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) + 2\sqrt{2}\right] \\&= \frac{1}{3}\left(\sqrt{2} + \ln(1+\sqrt{2})\right) (\fallingdotseq 0.765) \end{aligned}