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積分則

${\displaystyle \mathcal{L}\left\{{\int }_0^{t}f\left(t\right)dt\right\}=\frac{F\left(s\right)}{s}+\frac{f^{\left(-1\right)}\left(0\right)}{s}}$

ただし，${\displaystyle f^{\left(-1\right)}\left(0\right)={\int }_{-\infty }^{0}f\left(t\right)dt}$

${\displaystyle \mathcal{L}\left\{\frac{1}{t}f\left(t\right)\right\}={\int }_s^{\infty }F\left(s\right)ds}$

一般に

${\displaystyle \mathcal{L}\left\{{\int }_0^t{\int }_0^t{\int }_0^t\cdots {\int }_0^{t}f\left(t\right){\left(dt\right)}^n\right\}}$${\displaystyle =\frac{F\left(s\right)}{s^n}+\frac{f^{\left(-1\right)}\left(0\right)}{s^n}+\frac{f^{\left(-2\right)}\left(0\right)}{s^{n-1}}+}$${\displaystyle \cdots +\frac{f^{\left(-n\right)}\left(0\right)}{s}}$

ただし

${\displaystyle {\displaystyle f^{\left(-n\right)}\left(0\right)={\int }_{-\text{∞}}^0{\int }_{-\text{∞}}^0{\int }_{-\text{∞}}^0}}$${\displaystyle \cdots {\int }_{-\text{∞}}^{0}f\left(t\right){\left(dt\right)}^n}$

${\displaystyle \mathcal{L}\left\{\frac{1}{t^n}f\left(t\right)\right\}={\int }_s^{\text{∞}}{\int }_s^{\text{∞}}{\int }_s^{\text{∞}}}$${\displaystyle {\cdots \int }_s^{\text{∞}}{F\left(s\right)\left(ds\right)}^n}$

■証明

${\displaystyle \mathcal{L}\left\{{\displaystyle {\int }_0^{t}f\left(t\right)dt}\right\}={\displaystyle {\int }_0^{\infty }{e}^{-st}\left\{{\displaystyle {\int }_0^{t}f\left(t\right)dt}\right\}dt}}$

${\displaystyle ={\displaystyle {\int }_0^{\infty }{\left(-\frac{1}{s}{e}^{-st}\right)}^{\text{'}}\left\{{\displaystyle {\int }_0^{t}f\left(t\right)dt}\right\}dt}}$

${\displaystyle ={\left[-\frac{1}{s}{e}^{-st}{\displaystyle {\int }_0^{t}f\left(t\right)dt}\right]}_0^{\text{∞}}}$${\displaystyle -{\displaystyle {\int }_0^{\text{∞}}\left(-\frac{1}{s}{e}^{-st}\right)\frac{d}{dt}\left\{{\displaystyle {\int }_0^{t}f\left(t\right)dt}\right\}dt}}$

${\displaystyle ={\left[-\frac{1}{s}{e}^{-st}{\displaystyle {\int }_0^{t}f\left(t\right)dt}\right]}_0^{\text{∞}}}$${\displaystyle +\frac{1}{s}{\displaystyle {\int }_0^{\text{∞}}{e}^{-st}f\left(t\right)dt}}$

${\displaystyle =\frac{1}{s}{\displaystyle {\int }_0^{\infty }{e}^{-st}f\left(t\right)dt}}$

${\displaystyle =\frac{F\left(s\right)}{s}}$

■証明

${\displaystyle {\displaystyle {\int }_s^{\infty }F\left(s\right)}ds}$${\displaystyle ={\displaystyle {\int }_s^{\infty }{\displaystyle {\int }_0^{\infty }{e}^{-st}f\left(t\right)dt}}ds}$

${\displaystyle ={\displaystyle {\int }_0^{\infty }\left\{{\displaystyle {\int }_s^{\infty }{e}^{-st}f\left(t\right)}ds\right\}dt}}$

${\displaystyle ={\displaystyle {\int }_0^{\infty }f\left(t\right){\displaystyle {\int }_s^{\infty }{e}^{-st}}dsdt}}$

${\displaystyle ={\displaystyle {\int }_0^{\infty }f\left(t\right){\left[-\frac{1}{t}{e}^{-st}\right]}_s^{\infty }dt}}$

${\displaystyle ={\displaystyle {\int }_0^{\infty }f\left(t\right)\left\{0-\left(-\frac{1}{t}{e}^{-st}\right)\right\}dt}}$

${\displaystyle ={\displaystyle {\int }_0^{\infty }{e}^{-st}\left\{\frac{f\left(t\right)}{t}\right\}dt}}$

${\displaystyle =\mathcal{L}\left\{\frac{1}{t}f\left(t\right)\right\}}$

　

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　最終更新日： 2023年6月6日

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