关于$f(t)$的Laplace逆变换的注记
A Note on the Inverse Laplace Transformation of $f(t)$
摘要：
令$\mathcal{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t)dt$表示$f$的拉普拉斯变换。众所周知，如果$f(t)$是区间$t:[0,\infty)$上的分段连续函数，并且对于$t > N$是指数阶，则$\lim_{s\to\infty}F(s) = 0$，其中$F(s) = \mathcal{L}\{f(t)\}$。在本文中，我们证明了鲜为人知的逆不成立，即，如果$F(s)$是一个连续的功能方面的$s$为$\lim_{s\to\infty}F(s) = 0$，那么它并不遵循$F(s)$是拉普拉斯变换的分段连续函数的指数级。
英文摘要：
Let $\mathcal{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t)dt$ denote the Laplace transform of $f$. It is well-known that if $f(t)$ is a piecewise continuous function on the interval $t:[0,\infty)$ and of exponential order for $t > N$; then $\lim_{s\to\infty}F(s) = 0$, where $F(s) = \mathcal{L}\{f(t)\}$. In this paper we prove that the lesser known converse does not hold true; namely, if $F(s)$ is a continuous function in terms of $s$ for which $\lim_{s\to\infty}F(s) = 0$, then it does not follow that $F(s)$ is the Laplace transform of a piecewise continuous function of exponential order.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
