sinx和cosx的无穷乘积

利用

\[\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin \pi x} \]

得到

\[\sin \pi x=\frac{\pi}{-x\Gamma(x)\Gamma(-x)} \]

由Weierstrass公式易得

\[\frac1{\Gamma(x)}=xe^{\gamma x}\prod_{n=1}^\infty\left(1+\frac xn\right)e^{-\frac xn} \]

易知

\[\sin \pi x=\pi x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right) \]

于是

\[\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right) \]

注意到

\[\cos \pi x=\frac{\sin 2 \pi x}{2 \sin \pi x}=\frac{2 \pi x \cdot \prod\limits_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{n^{2}}\right)}{2 \pi x \cdot \prod\limits_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right)}=\frac{\prod\limits_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{n^{2}}\right)}{\prod\limits_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right)} \]

\[\prod_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{(2 n-1)^{2}}\right) \prod_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{(2 n)^{2}}\right)=\prod_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{(2 n-1)^{2}}\right) \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right) \]

于是

\[\cos \pi x=\prod_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{(2 n-1)^{2}}\right)\\ \cos x=\prod_{n=1}^{\infty}\left(1-\frac{4 x^{2}}{\pi^2(2 n-1)^{2}}\right) \]

---

\[\sum_{n=0}^{\infty} \frac{1}{n^{2}+1}=\frac{1}{2}[\pi \coth(\pi)+1] \]

解：利用

\[\frac{\sin \pi z}{\pi z}=\prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right) \]

上式两边取对数有

\[\ln\left(\frac{\sin \pi z}{\pi z}\right)=\ln\left(\prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right)\right)=\sum_{n=1}^{\infty} \ln \left(1-\frac{z^{2}}{n^{2}}\right) \]

两边求导得到

\[\frac{\pi z}{\sin \pi z}\left(\frac{\cos \pi z}{z}-\frac{\sin \pi z}{\pi z^{2}}\right)=\sum_{n=1}^{\infty} \frac{-2 z}{n^{2}\left(1-\frac{z^{2}}{n^{2}}\right)} \]

化简得到

\[\sum_{n=1}^{\infty} \frac{1}{n^{2}-z^{2}}=\frac{1}{2 z^{2}}-\frac{\pi \cot \pi z}{2 z} \]

本题取$ z=i $即有

\[ \sum_{n=1}^{\infty} \frac{1}{n^{2}+1}=\frac{1}{2}[\pi \coth(\pi)-1] \]

\[\sum_{n=0}^{\infty} \frac{1}{n^{2}+1}=\frac{1}{2}[\pi \coth(\pi)+1] \]

---

\[\sum_{n=1}^{\infty}\arctan\frac{1}{n^{2}}=\arctan\left(\frac{1-\cot\frac{\pi}{\sqrt{2}}\tanh\frac{\pi}{\sqrt{2}}}{1+\cot\frac{\pi}{\sqrt{2}}\tanh\frac{\pi}{\sqrt{2}}}\right) \]

解：利用

\[\sin\pi z=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right) \]

注意到

\[\sin z=-i\sinh (iz) \]

用 \(iz\) 替换 \(z\)，于是

\[\prod_{n=1}^{\infty}\left(1+\frac{z^{2}}{n^{2}}\right)=\frac{\sinh\pi z}{\pi z} \]

故

\[\begin{aligned} \sum_{n=1}^{\infty}\arctan\frac{1}{n^{2}}&=\frac{1}{2 i}\ln\prod_{n=1}^{\infty}\frac{1+\frac{i}{n^{2}}}{1-\frac{i}{n^{2}}}\\ &=\frac{1}{2 i}\ln\left(e^{-i\pi /2}\frac{\sinh\pi e^{i\pi /4}}{\sinh\pi e^{-i\pi /4}}\right) \\ & =\frac{1}{2 i}\ln\left(\frac{\sinh\frac{\pi}{\sqrt{2}}\cos\frac{\pi}{\sqrt{2}}+i\cosh\frac{\pi}{\sqrt{2}}\cos\frac{\pi}{\sqrt{2}}}{\cosh\frac{\pi}{\sqrt{2}}\sin\frac{\pi}{\sqrt{2}}+i\sinh\frac{\pi}{\sqrt{2}}\cos\frac{\pi}{\sqrt{2}}}\right) \\ &=\arctan\left(\frac{1-\cot\frac{\pi}{\sqrt{2}}\tanh\frac{\pi}{\sqrt{2}}}{1+\cot\frac{\pi}{\sqrt{2}}\tanh\frac{\pi}{\sqrt{2}}}\right) \end{aligned} \]

参考

参考资料1
参考资料2
参考资料3