Question

Laplace transform of sint/t

Answer 1

Answer:

Yes! Here is your answer!

Step-by-step explanation:

For L(d/dt(sint/t)), we first calculate laplace of derivative of function sint/t as s×L(sint/t)-(sin0)/0. Now sin0/0 can be calculated using limiting value of sint/t at t=0. As t tends to 0 sint~=t so limiting value=1. So the answer becomes sL(sint/t)-1.

Now we can calculate L(sint/t) as  ∫∞1/(2+1)  where 1/(s^2+1) is laplace transform of sint.

The integration results in tan^-1(∞)-tan^-1(s) or π/2-tan^-1(s).

So L(d/dt(sint/t))=πs/2-s×tan^-1(s)-1.

Answer 2

The Laplace transform of sint/t is $\frac{\pi}{2}$.

Given:

The expression is sint/t.

To find:

The Laplace transform of sint/t.

Formula used:

The Laplace transform is $F(s)=\int_{0}^{\infty} f(t) e^{-s t}dt$.

Step-by-step explanation:

The Laplace transform of sint/t is given by

L(sint/t) $=\int_{0}^{\infty} \frac{\sin t}{t} d t &$

Apply Laplace formula.

$=\lim _{s \rightarrow 0} \int_{0}^{\infty} e^{-s t} \frac{\sin t}{t} d t$

$=\lim _{s \rightarrow 0} \mathcal{L}\left[\frac{\sin t}{t}\right]$

$&=\lim _{s \rightarrow 0} \int_{s}^{\infty} \frac{d u}{u^{2}+1}$

Integrate the above.

$=\left.\lim _{s \rightarrow 0} \arctan u\right|_{s} ^{\infty}$

Substitute the upper and lower limits.

$&=\lim _{s \rightarrow 0}\left[\frac{\pi}{2}-\arctan (s)\right]$

After application of limit, we get

$=\frac{\pi}{2} .$

Hence, the Laplace transform of sint/t is $\frac{\pi}{2}$.

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