你可以考慮把 \pi 換成 \begin{aligned} \pi&=\arccos(-1)\\ \pi&=\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\mathrm{d}x\\ \pi&=2\int_{-1}^1\sqrt{1-x^2}\mathrm{d}x\\ \pi&=\sqrt{6\sum_{n=1}^\infty\frac1{n^2}}\\ \pi&=4\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\\ \pi&=3\sum_{n=0}^\infty\frac1{(n+1)(2n+1)(4n+1)}\\ \pi&=\frac{1}{8\varpi^2}\Gamma\left(\frac14\right)^2\\ \pi&=\sqrt[4]{90\zeta(4)}\\ \pi&=\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}\left(\frac12\right)^n\\ \pi&=\frac1{\display style\frac{2\sqrt2}{9801}\display style\sum_{n=0}^{∞}{\frac{(4n)!}{(n!)^4}}\frac{26390n+1103}{396^{4n}}}\\ \pi&=\int_{0}^{1/\sqrt{2}} \frac{4\sqrt{2}-8x^{3}-4\sqrt{2}x^{4}-8x^{5}}{1-x^{8}}\,\mathrm{d}x\\ \pi &= \frac1{\display style\frac{1}{53360 \sqrt{640320}}\display style\sum_{n=0}^\infty (-1)^n \frac{(6n)!}{(n!)^3(3n)!} \times \frac{13591409 + 545140134n}{640320^{3n}}}\\ \pi&=12\lim\limits_{n\to\infty}\int_0^{2-\sqrt3}\frac{1-x^{4n}}{1+x^2}\mathrm{d}x\\ \pi&=\sqrt[4]{\frac{360}{17}\sum_{i=1}^\infty\large(\frac{\display style\sum_{j=1}^i\frac1j}{i}}\large)^2 \end{aligned}\\
以及\begin{aligned} \pi&=2\int_0^{+\infty}\frac{\sin x}x\mathrm dx\\ \pi&=2\int_{-\infty}^{+\infty}\sin(e^x)\mathrm dx\\ \pi&=\int_0^{+\infty}\frac{\sin\pi x}{x(1-x^2)}\mathrm dx\\ \pi&=\lim_{n \to \infty} \left\{ \sqrt{n} \cdot \sum_{i=1}^{\infty} \frac{2}{{\left( i - \dfrac{1}{2} \right)}^2 + n} \right\} \\ \pi&=\lim_{n\rightarrow\infty}\frac{1}{n}\prod^n_{i=1}(\frac{2i-1}{2i})^2 \\ \pi&=2\prod^\infty_{n=1}\frac{4n^2}{4n^2-1} \\ \pi&=\prod_{n=1}^\infty\left( \frac{(2n)^2}{(2n-1)(2n+1)} \right)\\ \pi&= \frac{8}{\display style\sum_{n=0}^\infty(n+3)\frac{(-1)^n(4n)!}{(4\sqrt{2})^{4n}(n!)^4}} \\ \pi&=\frac12\lim_{n\to\infty}\frac{e^{2n}(n!)^2}{n^{2n+1}} \\ \pi&=\frac{2}{\prod\limits_{n=2}^{\infty}\left( 1+\frac{(-1)(\display style\sum_{m=1}^{2^n}\lfloor \lfloor \frac{n}{1+\pi (m)} \rfloor^{\frac{1}{n}} \rfloor)}{2(1+\display style\sum_{m=1}^{2^n}\lfloor \lfloor \frac{n}{1+\pi (m)} \rfloor^{\frac{1}{n}} \rfloor)} \right)}\end{aligned}\\ 還有\begin{aligned}\pi&=\frac1{\display style\frac12-\display style\sum_{k=1}^\infty{(2k-3)!!(2k-1)!!\over2\cdot4^k(k!)^2}}\\ \pi&=\left (\frac{\display style\int_0^{+\infty}e^{-x^2}\sin x\mathrm{d}x}{\frac12e^{-\frac{1}{4}}\mathrm{erfi}\left(\frac{1}{2}\right)}\right )^2\\ \pi&=\sqrt{\frac{\display style\int_0^{+\infty}\frac{x^{2}}{(1+e^x)(1+e^{-x})}\mathrm{d}x}{\display style\lim_{x\to0}{\mathrm{d}\over\mathrm{d}x}\left[x\over e^x-1\right]}}\\ \pi&= 2\ln 2\int_0^{\frac{\pi }{2}} \frac{\sqrt{\frac{\ln ^2(\cos (x ))}{x ^2+\ln ^2(\cos (x ))}}}{\sqrt{x ^2+\ln ^2(\cos (x ))}} \, \mathrm{d}x\\ \pi&=\frac1{21-\display style\int_1^\infty{\{x\}\over x^2}\mathrm dx}\int_0^{2\pi}\zeta(e^{ix}+1)\mathrm dx \\ \pi&=\lim\limits_{x\to\infty}\left[\frac12\left(\frac{\Gamma\left(x\right)}{x^{x-\frac{1}{2}}e^{-x}~~\Re\left(x\right)}\right)^2\right]\end{aligned}\\ （逃
