Laplace Transform of sint/t | Laplace of sin(at)/t

The Laplace transform of sint/t is equal to tan^-1(1/s), that is, L{sint/t} = tan^-1(1/s). The formulas for the Laplace of sint/t and the Laplace of sin(at)/t are given as follows:

$\boxed{\mathcal{L}\left\{\dfrac{\sin t}{t} \right\}=\tan^{-1}\left(\dfrac{1}{s} \right)}$

and

$\boxed{\mathcal{L}\left\{\dfrac{\sin at}{t} \right\}=\tan^{-1}\left(\dfrac{a}{s} \right)}$

Lets learn how to find Laplace of sin(t)/t.

Laplace of sint/t

Question: What is the Laplace of sint/t?

Answer:

First observe that sint/t is a function of the form f(t)/t. Thus, to compute its Laplace transform, we will use the division by t formula given below:

$\mathcal{L} \left\{\dfrac{f(t)}{t} \right\} = \displaystyle \int_s^\infty \mathcal{L}(f(t)) ds$.

We have: f(t) = sint.

So L{f(t)} = L{sin t} = $\dfrac{1}{s^2+1}$.

Now, using the above formula, it follows that

$\mathcal{L} \left\{\dfrac{\sin t}{t} \right\}$ $= \displaystyle \int_s^\infty \mathcal{L}(\sin t) ds$

As the Laplace of sint is equal to 1/(s²+1), we have:

$\mathcal{L} \left\{\dfrac{\sin t}{t} \right\}$ $=\displaystyle \int_s^\infty \dfrac{1}{s^2+1} ds$

= $\left[ \tan^{-1} s\right]_s^\infty$

= tan^-1∞ – tan^-1s

= π/2 – tan^-1s

= cot^-1s, since tan^-1s + cot^-1s = π/2.

= tan^-1(1/s), this is because cot^-1s = tan^-1(1/s)

Therefore, the Laplace transform of sint/t is equal to tan^-1(1/s). In other words, we have L{sint/t} = tan^-1(1/s).

More on Laplace Transformation:

Laplace transform of t² by definition

Laplace of sinat/t

Answer: The Laplace transform of sin(at)/t is equal to tan^-1(a/s).

Explanation:

Using the formula $\mathcal{L} \left\{\dfrac{f(t)}{t} \right\} = \displaystyle \int_s^\infty \mathcal{L}(f(t)) ds$ with f(t) = sinat, we obtain that

$\mathcal{L} \left\{\dfrac{\sin at}{t} \right\}$ $= \displaystyle \int_s^\infty \mathcal{L}(\sin at) ds$

= $\displaystyle \int_s^\infty \dfrac{a}{s^2+a^2} ds$

= $\left[ \tan^{-1} \dfrac{s}{a}\right]_s^\infty$

= tan^-1∞ – tan^-1(s/a)

= π/2 – tan^-1(s/a)

= cot^-1(s/a)

= tan^-1(a/s).

Therefore, the Laplace transform of sinat/t is equal to tan^-1(a/s). In other words, we have L{sinat/t} = tan^-1(a/s).

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