Expert answers to "Use the definition of Laplace Transforms to show..."

Jason Farmer

Answered question

2021-02-12

Use the definition of Laplace Transforms to show that:
$L {t^{n}} = \frac{n!}{s^{n + 1}}, n = 1, 2, 3, \dots$

Nicole Conner

Skilled2021-02-13Added 97 answers

Step 1
Given:
The laplace transformation,

L {t^{n}} = \frac{n!}{s^{n + 1}}

Step 2
Proof:
Consider

L (t^{n})

and simplify it as follows.

L {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t

L {t^{n}} = \int_{0}^{\infty} e^{- s t} (t^{n}) d t

Apply integration by parts,

u v = u v - i n t u v

Here,

u = t^{n} d v = e^{- s t}

d u = n t^{n + 1} v = \frac{- e^{- s t}}{s}

L {t^{n}} = {[\frac{- t^{n} \cdot e^{- s t}}{s}]}_{0}^{\infty} - \int_{0}^{\infty} n t^{n - 1} (\frac{e^{- s t}}{s}) d t

= 0 + \frac{n}{s} \int_{0}^{\infty} t^{n - 1} (e^{- s t}) d t

= \frac{n}{s} L (t^{n - 1})

Step 3
Now substitute the values 1, 2, 3... for n.
When n=1

L {t^{1}} = \frac{1}{s} L (1)

= \frac{1}{s} (\frac{1}{s})

= \frac{1}{s^{2}}

When n=2

L {t^{2}} = \frac{2}{s} (L (t^{1}))

= \frac{2}{s} (\frac{1}{s^{2}})

= \frac{2!}{s^{3}}

Step 4
When n = 3,

L {t^{3}} = \frac{3}{s} L (t^{3 - 1})

= \frac{3}{s} (L (t^{2}))

= \frac{3}{s} (\frac{2}{s^{3}})

= \frac{6}{s^{3 + 1}}

= \frac{3!}{s^{3 + 1}}

Step 5
Thus, for n = n,

L (t^{n}) = \frac{n!}{s^{n + 1}}

Hence it is proved.

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