Trying to find the Laplace transform of $\displaystyle{\frac{{{\cos{{t}}}}}{{{t}}}}$ It

Question

Trying to find the Laplace transform of $\displaystyle{\frac{{{\cos{{t}}}}}{{{t}}}}$ It

Mary Bates 2022-04-08 Answered

Trying to find the Laplace transform of

\frac{\cos t}{t}

It comes out as infinity, but that doesn't make any sense.
Does this mean that this function doesn't have a Laplace transform or is something wrong here?

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Answers (1)

Jax Burns

Answered 2022-04-09 Author has 13 answers

Well, we are trying to find (solving a more general problem):

L_{t} {[\frac{\cos (n \cdot t)}{t}]}_{(s)} = \int_{0}^{\infty} \frac{\cos (n \cdot t)}{t} \cdot \exp (- s t) d x

Using the 'frequency-domain integration' property of the Laplace transform, we can write:

L_{t} {[\frac{\cos (n \cdot t)}{t}]}_{(s)} = \int_{s}^{\infty} L_{t} {[\cos (n \cdot t)]}_{(σ)} d σ

Using the table of selected Laplace transforms, we find:

L_{t} {[\cos (n \cdot t)]}_{(σ)} = \frac{σ}{σ^{2} + n^{2}}

So, we get:

L_{t} {[\frac{\cos (n \cdot t)}{t}]}_{(s)} = \int_{s}^{\infty} \frac{σ}{σ^{2} + n^{2}} d σ

Let's substitute

u = σ^{2} + n^{2}

, so we get:

L_{t} {[\frac{\cos (n \cdot t)}{t}]}_{(s)} = \frac{1}{2} \cdot lim_{ϵ \to \infty} \int_{s^{2} + n^{2}}^{ϵ^{2} + n^{2}} \frac{1}{u} d u = \frac{1}{2} \cdot lim_{ϵ \to \infty} {[\ln | u |]}_{s^{2} + n^{2}}^{ϵ^{2} + n^{2}} =

\frac{1}{2} \cdot lim_{ϵ \to \infty} (\ln | ϵ^{2} + n^{2} | - \ln | s^{2} + n^{2} |) = \frac{1}{2} \cdot lim_{ϵ \to \infty} | \frac{ϵ^{2} + n^{2}}{s^{2} + n^{2}} | \to \infty

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