Use the definition of Laplace Transforms to show that: displaystyle{L}{leftlbrace{t}^{n}rightrbrace}=frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},ldots

Use the definition of Laplace Transforms to show that: displaystyle{L}{leftlbrace{t}^{n}rightrbrace}=frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},ldots

Question
Laplace transform
asked 2021-02-12
Use the definition of Laplace Transforms to show that:
\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},\ldots\)

Answers (1)

2021-02-13
Step 1
Given:
The laplace transformation,
\(\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}\)
Step 2
Proof:
Consider \(L(t^n)\) and simplify it as follows.
\(\displaystyle{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}\)
\(\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}{\left({t}^{n}\right)}{\left.{d}{t}\right.}\)
Apply integration by parts, \(\displaystyle{u}{v}'={u}{v}-\int{u}'{v}\)
Here,
\(\displaystyle{u}={t}^{n}{d}{v}={e}^{{-{s}{t}}}\)
\(\displaystyle{d}{u}={n}{t}^{{{n}+{1}}}\ \ \ {v}=\frac{{-{e}^{{-{s}{t}}}}}{{s}}\)
\(\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}={{\left[\frac{{-{t}^{n}\cdot{e}^{{-{s}{t}}}}}{{s}}\right]}_{{0}}^{\infty}}-{\int_{{0}}^{\infty}}{n}{t}^{{{n}-{1}}}{\left(\frac{{{e}^{{-{s}{t}}}}}{{s}}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={0}+\frac{n}{{s}}{\int_{{0}}^{\infty}}{t}^{{{n}-{1}}}{\left({e}^{{-{s}{t}}}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle=\frac{n}{{s}}{L}{\left({t}^{{{n}-{1}}}\right)}\)
Step 3
Now substitute the values 1, 2, 3... for n.
When n=1
\(\displaystyle{L}{\left\lbrace{t}^{1}\right\rbrace}=\frac{1}{{s}}{L}{\left({1}\right)}\)
\(\displaystyle=\frac{1}{{s}}{\left(\frac{1}{{s}}\right)}\)
\(\displaystyle=\frac{1}{{s}^{2}}\)
When n=2
\(\displaystyle{L}{\left\lbrace{t}^{2}\right\rbrace}=\frac{2}{{s}}{\left({L}{\left({t}^{1}\right)}\right)}\)
\(\displaystyle=\frac{2}{{s}}{\left(\frac{1}{{s}^{2}}\right)}\)
\(\displaystyle=\frac{{{2}!}}{{s}^{3}}\)
Step 4
When n = 3,
\(\displaystyle{L}{\left\lbrace{t}^{3}\right\rbrace}=\frac{3}{{s}}{L}{\left({t}^{{{3}-{1}}}\right)}\)
\(\displaystyle=\frac{3}{{s}}{\left({L}{\left({t}^{2}\right)}\right)}\)
\(\displaystyle=\frac{3}{{s}}{\left(\frac{2}{{s}^{3}}\right)}\)
\(\displaystyle=\frac{6}{{{s}^{{{3}+{1}}}}}\)
\(\displaystyle=\frac{{{3}!}}{{{s}^{{{3}+{1}}}}}\)
Step 5
Thus, for n = n,
\(\displaystyle{L}{\left({t}^{n}\right)}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}\)
Hence it is proved.
0

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