積分則

• ${\displaystyle \mathcal{L}\left\{{\int }_0^{t}f\left(u\right)du\right\}=\frac{F\left(s\right)}{s}}$   ･･････(1)
• ${\displaystyle \mathcal{L}\left\{\frac{1}{t}f\left(t\right)\right\}={\int }_s^{\infty }F\left(\sigma \right)d\sigma }$  ･･････(2)

一般に

• ${\displaystyle \mathcal{L}\left\{{\int }_0^t{\int }_0^t{\int }_0^t\cdots {\int }_0^{t}f\left(u\right){\left(du\right)}^n\right\}}$ ${\displaystyle =\frac{F\left(s\right)}{s^n}}$   ･･････(3)
• ${\displaystyle \mathcal{L}\left\{\frac{1}{t^n}f\left(t\right)\right\}}$ ${\displaystyle ={\int }_s^{\text{∞}}{\int }_s^{\text{∞}}{\int }_s^{\text{∞}}{\cdots \int }_s^{\text{∞}}F\left(\sigma \right){\left(d\sigma \right)}^n}$   ･･････(4)

■証明

●(1)について

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

部分積分法を用いる．

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

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

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

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

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

●(2)について

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

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

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

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

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

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

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

●(3)について

${\displaystyle g_n\left(t\right)={{\displaystyle \int }}_0^t{{\displaystyle \int }}_0^t{{\displaystyle \int }}_0^t\cdots {{\displaystyle \int }}_0^{t}f\left(u\right){\left(du\right)}^n}$   ･･････(5)

${\displaystyle G_n\left(s\right)=\mathcal{L}\left\{g_n\left(t\right)\right\}}$   ･･････(6)

とおく．

${\displaystyle G_1\left(s\right)=\mathcal{L}\left\{g_1\left(t\right)\right\}}$ ${\displaystyle ={{\displaystyle \int }}_0^{t}f\left(u\right)du}$ ${\displaystyle =\frac{F\left(s\right)}{s}}$   （∵(1)より）  ･･････(7)

となる．

${\displaystyle G_n\left(s\right)}$ を以下のように式変形をする．

${\displaystyle G_n\left(s\right)=\mathcal{L}\left\{g_n\left(t\right)\right\}}$

${\displaystyle =\mathcal{L}\left\{{{\displaystyle \int }}_0^t{{\displaystyle \int }}_0^t{{\displaystyle \int }}_0^t\cdots {{\displaystyle \int }}_0^{t}f\left(u\right){\left(du\right)}^n\right\}}$

${\displaystyle =\mathcal{L}\left\{{\displaystyle {\int }_0^t{g}_{n-1}\left(u\right)du}\right\}}$

${\displaystyle =\frac{G_{n-1}\left(s\right)}{s}}$   ･･････(8)

となる．すなわち

${\displaystyle G_n\left(s\right)=\frac{G_{n-1}\left(s\right)}{s}}$   ･･････(9)

となる漸化式が得られる．この漸化式より

${\displaystyle G_n\left(s\right)=\frac{G_{n-1}\left(s\right)}{s}}$ ${\displaystyle =\frac{G_{n-2}\left(s\right)}{s^2}}$ ${\displaystyle =\cdots =\frac{G_1\left(s\right)}{s^{n-1}}}$   ･･････(10)

の関係式が得られる．

(10)，(7)より

${\displaystyle G_n\left(s\right)=\frac{1}{s^{n-1}}\cdot \frac{F\left(s\right)}{s}=\frac{F\left(s\right)}{s^n}}$   ･･････(11)

(6)に(5)を代入してからその結果を(11)に代入すると

${\displaystyle \mathcal{L}\left\{{{\displaystyle \int }}_0^t{{\displaystyle \int }}_0^t{{\displaystyle \int }}_0^t\cdots {{\displaystyle \int }}_0^{t}f\left(u\right){\left(du\right)}^n\right\}}$${\displaystyle =\frac{F\left(s\right)}{s^n}}$

となり，(3)が得られる．

●(4)について

${\displaystyle h_n\left(t\right)=\frac{1}{t^n}f\left(t\right)}$  ･･････(12)

${\displaystyle \mathcal{L}\left\{\frac{1}{t^n}f\left(t\right)\right\}=\mathcal{L}\left\{h_n\left(t\right)\right\}}$${\displaystyle =H_n\left(s\right)}$   ･･････(13)

とおく．

${\displaystyle H_1\left(s\right)=\mathcal{L}\left\{h_1\left(t\right)\right\}}$${\displaystyle =\mathcal{L}\left\{\frac{1}{t}f\left(t\right)\right\}}$${\displaystyle ={\displaystyle {\int }_s^{\infty }F\left(\sigma \right)d\sigma }}$  （∵(2)）  ･･････(14)

となる．

${\displaystyle H_n\left(s\right)=\mathcal{L}\left\{h_n\left(t\right)\right\}}$${\displaystyle =\mathcal{L}\left\{\frac{1}{t}{h}_{n-1}\left(t\right)\right\}}$${\displaystyle ={\displaystyle {\int }_s^{\infty }{H}_{n-1}\left(\sigma \right)d\sigma }}$

となる．すなわち

${\displaystyle H_n\left(s\right)={\displaystyle {\int }_s^{\infty }{H}_{n-1}\left(\sigma \right)d\sigma }}$  ･･････(15)

となる漸化式が得られる．この漸化式より

${\displaystyle H_n\left(s\right)={\displaystyle {\int }_s^{\infty }{H}_{n-1}\left(\sigma \right)d\sigma }}$

${\displaystyle ={\displaystyle {\int }_s^{\infty }{\displaystyle {\int }_s^{\infty }{H}_{n-2}\left(\sigma \right){\left(d\sigma \right)}^2}}}$

${\displaystyle \cdots }$

${\displaystyle ={{\displaystyle \int }}_s^{\infty }{{\displaystyle \int }}_s^{\infty }{{\displaystyle \int }}_s^{\infty }\cdots {{\displaystyle \int }}_s^{\infty }{H}_1\left(\sigma \right){\left(d\sigma \right)}^{n-1}}$  ･･････(17)

(17)，(14)より

${\displaystyle H_n\left(s\right)}$${\displaystyle ={{\displaystyle \int }}_s^{\infty }{{\displaystyle \int }}_s^{\infty }{{\displaystyle \int }}_s^{\infty }\cdots {{\displaystyle \int }}_s^{\infty }F\left(\sigma \right){\left(d\sigma \right)}^n}$  ･･････(18)

が得られる．

(13)と(18)より

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

となり，(4)の関係が導かれた．

 

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　最終更新日： 2026年9月28日