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阿贝尔-普兰纳公式

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hbghlyj Posted at 2022-10-17 19:08:33 |Read mode
Abel–Plana formula:对于满足一定条件的$f$,
\[\sum_{n=0}^{\infty}f(n)-\int_{0}^{\infty}f(x)\,dx=\frac{f(0)}2+i\int_{0}^{\infty}\frac{f(it)-f(-it)}{e^{2\pi t}-1}\, dt\]
这给出了一个离散和式与它的积分之间的关系。
对于偶函数,$f(it)=f(-it)$,所以右边的积分是0.
例如,
$f(x)=\frac1{x^2+1}$,则\begin{aligned}
f(0)&=1\\
\sum _{n=0}^{\infty } f(n)&=\frac{1+\pi  \coth \pi}2\\
\int_0^{\infty }f(x) \, dx&=\frac\pi2
\end{aligned}
把$\sin x/x$的解析延拓记为$f(x)$,
\begin{aligned}
f(0)&=1\\
\sum _{n=0}^{\infty } f(n)&=\frac{1+\pi}2\\
\int_0^{\infty }f(x)\, dx&=\frac\pi2
\end{aligned}

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 Author| hbghlyj Posted at 2022-10-21 06:30:31
左边的例子$f(x)=\frac1{x^2+1}$不满足公式.

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2025-4-22 23:40 GMT+8

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