$\operatorname{sinc}x$无穷级数

hbghlyj

wikipedia
$\displaystyle \operatorname {sinc} x:={\frac {\sin x}{x}}.$(又见这帖)
$$\sum_{n=1}^\infty \operatorname{sinc}(n) = \frac{\pi - 1}{2}$$
$$\sum_{n=1}^\infty \operatorname{sinc}^2(n) = \frac{\pi - 1}{2}$$$$\sum_{n=1}^\infty (-1)^{n+1}\,\operatorname{sinc}(n) = \frac{1}{2}$$$$\sum_{n=1}^\infty (-1)^{n+1}\,\operatorname{sinc}^2(n)= \frac{1}{2}$$$$\sum_{n=1}^\infty (-1)^{n+1}\,\operatorname{sinc}^3(n) = \frac{1}{2}$$相关:
kuing.cjhb.site/forum.php?mod=viewthread&tid=755

战巡

回复 1# hbghlyj


没那么多时间，只做第一个，其他类推，都差不多的

\[\frac{\sin(n)}{n}=\int_0^{\infty}\sin(n)e^{-nx}dx=-\frac{i}{2}\int_0^\infty(e^{in}-e^{-in})e^{-nx}dx\]
那么
\[\sum_{n=1}^\infty\frac{\sin(n)}{n}=-\frac{i}{2}\sum_{n=1}^\infty\int_0^\infty(e^{n(i-x)}-e^{-n(i+x)})dx\]
这里就懒得证明什么可交换性了，反正也不会太困难
有
\[=\frac{i}{2}\int_0^\infty\sum_{n=1}^\infty(e^{-n(x+i)}-e^{-n(x-i)})dx\]
\[=\frac{i}{2}\int_0^\infty\left(\frac{1}{e^{x+i}-1}-\frac{1}{e^{x-i}-1}\right)dx\]
\[=\int_0^\infty\frac{\sin(1)}{2(\cosh(x)-\cos(1))}dx\]
\[=-\arctan\left(\coth(\frac{x}{2})\tan(\frac{1}{2})\right)|_{0}^\infty=\frac{\pi-1}{2}\]