How to find the Laplace transform of 1−cos⁡(t)t2? F(S)=?

Name: How to find the Laplace transform of 1−cos⁡(t)t2? F(S)=?
Uploaded: 2022-02-23T17:22:57+00:00
Description: How to find the Laplace transform of 1−cos⁡(t)t2? F(S)=?

LetF(s)=∫0∞f(t)e−stdt=∫0∞1−cos⁡tt2e−stdtThe function f(t) satisfies the bound f(t)=O(1∧t−2), thus it is absolutely integrable and we can apply Leibniz's integral to obtainFs)=∫0∞(1−cos⁡t)e−stdt=1s−ss2+1Integrating and using the condition F′(∞)=0, we haveF′(s)=log⁡s−log⁡s2+1Thus we haveF(s)=∫{log⁡s−log⁡s2+1}dsThe first term is easily integrated to yield slog⁡s−s. For the second term, note that−∫log⁡s2+1ds=−slog⁡s2+1+∫s2s2+1ds=−slog⁡s2+1+s−arctan⁡s+CCombining, we obtainF(s)=slog⁡s−slog⁡s2+1−arctan⁡s+CBut since F(∞)=0, we must have C=π2 and thereforeF(s)=slog⁡s−slog⁡s2+1−arctan⁡s+π2=slog⁡(ss2+1)+arctan⁡(1s)

If L{f(t)}=F(s) and Laplace of the function g(t)=f(t)t exists, then L{f(t)t}=∫s∞F(u)du ∫s∞F(u)du=∫s∞∫0∞f(t)e−utdtdu =∫0∞f(t)∫s∞e−utdudt =∫0∞f(t)e−ut−t|s∞dt =∫0∞f(t)te−stdt =L{f(t)t}

LetF(s)=∫0∞f(t)e−stdt=∫0∞1−cos⁡tt2e−stdtThe function f(t) satisfies the bound f(t)=O(1∧t−2), thus it is absolutely integrable and we can apply Leibniz's integral to obtainFs)=∫0∞(1−cos⁡t)e−stdt=1s−ss2+1Integrating and using the condition F′(∞)=0, we haveF′(s)=log⁡s−log⁡s2+1Thus we haveF(s)=∫{log⁡s−log⁡s2+1}dsThe first term is easily integrated to yield slog⁡s−s. For the second term, note that−∫log⁡s2+1ds=−slog⁡s2+1+∫s2s2+1ds=−slog⁡s2+1+s−arctan⁡s+CCombining, we obtainF(s)=slog⁡s−slog⁡s2+1−arctan⁡s+CBut since F(∞)=0, we must have C=π2 and thereforeF(s)=slog⁡s−slog⁡s2+1−arctan⁡s+π2=slog⁡(ss2+1)+arctan⁡(1s)

If L{f(t)}=F(s) and Laplace of the function g(t)=f(t)t exists, then L{f(t)t}=∫s∞F(u)du ∫s∞F(u)du=∫s∞∫0∞f(t)e−utdtdu =∫0∞f(t)∫s∞e−utdudt =∫0∞f(t)e−ut−t|s∞dt =∫0∞f(t)te−stdt =L{f(t)t}

Question: "How to find the Laplace transform of 1−cos⁡(t)t2?..."

Marla Payton

Answered question

2021-12-08

How to find the Laplace transform of

\frac{1 - \cos (t)}{t^{2}}

?
F(S)=?

Answer & Explanation

Maricela Alarcon

Beginner2021-12-09Added 28 answers

Let
$F (s) = \int_{0}^{\infty} f (t) e^{- s t} d t = \int_{0}^{\infty} \frac{1 - \cos t}{t^{2}} e^{- s t} d t$
The function f(t) satisfies the bound $f (t) = O (1 \land t^{- 2})$ , thus it is absolutely integrable and we can apply Leibniz's integral to obtain
$F s) = \int_{0}^{\infty} (1 - \cos t) e^{- s t} d t = \frac{1}{s} - \frac{s}{s^{2} + 1}$
Integrating and using the condition $F' (\infty) = 0$ , we have
$F^{'} (s) = \log s - \log \sqrt{s^{2} + 1}$
Thus we have
$F (s) = \int {\log s - \log \sqrt{s^{2} + 1}} d s$
The first term is easily integrated to yield $s \log s - s$ . For the second term, note that
$- \int \log \sqrt{s^{2} + 1} d s = - s \log \sqrt{s^{2} + 1} + \int \frac{s^{2}}{s^{2} + 1} d s$
$= - s \log \sqrt{s^{2} + 1} + s - \arctan s + C$
Combining, we obtain
$F (s) = s \log s - s \log \sqrt{s^{2} + 1} - \arctan s + C$
But since $F (\infty) = 0$ , we must have $C = \frac{π}{2}$ and therefore
$F (s) = s \log s - s \log \sqrt{s^{2} + 1} - \arctan s + \frac{π}{2}$
$= s \log (\frac{s}{\sqrt{s^{2} + 1}}) + \arctan (\frac{1}{s})$

Vivian Soares

Beginner2021-12-10Added 36 answers

If $L {f (t)} = F (s)$ and Laplace of the function $g (t) = \frac{f (t)}{t}$ exists, then
$L {\frac{f (t)}{t}} = \int_{s}^{\infty} F (u) d u$
$\int_{s}^{\infty} F (u) d u = \int_{s}^{\infty} \int_{0}^{\infty} f (t) e^{- u t} d t d u$
$= \int_{0}^{\infty} f (t) \int_{s}^{\infty} e^{- u t} d u d t$
$= \int_{0}^{\infty} f (t) \frac{e^{- u t}}{- t} |_{s}^{\infty} d t$
$= \int_{0}^{\infty} \frac{f (t)}{t} e^{- s t} d t$
$= L {\frac{f (t)}{t}}$

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