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?

Mary Bates
2022-04-08
Answered

Trying to find the Laplace transform of

$\frac{\mathrm{cos}t}{t}$

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?

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?

You can still ask an expert for help

Jax Burns

Answered 2022-04-09
Author has **13** answers

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

${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}(\text{n}\cdot t)}{t}\right]}_{\left(\text{s}\right)}{\textstyle \phantom{\rule{0.222em}{0ex}}}={\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{cos}(\text{n}\cdot t)}{t}\cdot \mathrm{exp}(-\text{s}t)\text{d}x$

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

${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}(\text{n}\cdot t)}{t}\right]}_{\left(\text{s}\right)}={\int}_{\text{s}}^{\mathrm{\infty}}{\mathcal{L}}_{t}{\left[\mathrm{cos}(\text{n}\cdot t)\right]}_{\left(\sigma \right)}\text{d}\sigma$

Using the table of selected Laplace transforms, we find:

$\mathcal{L}}_{t}{\left[\mathrm{cos}(\text{n}\cdot t)\right]}_{\left(\sigma \right)}=\frac{\sigma}{{\sigma}^{2}+{\text{n}}^{2}$

So, we get:

${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}(\text{n}\cdot t)}{t}\right]}_{\left(\text{s}\right)}={\int}_{\text{s}}^{\mathrm{\infty}}\frac{\sigma}{{\sigma}^{2}+{\text{n}}^{2}}{\textstyle \phantom{\rule{1em}{0ex}}}\text{d}\sigma$

Let's substitute$\text{u}{\textstyle \phantom{\rule{0.222em}{0ex}}}={\sigma}^{2}+{\text{n}}^{2}$ , so we get:

${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}(\text{n}\cdot t)}{t}\right]}_{\left(\text{s}\right)}=\frac{1}{2}\cdot \underset{\u03f5\to \mathrm{\infty}}{lim}{\int}_{{\text{s}}^{2}+{\text{n}}^{2}}^{{\u03f5}^{2}+{\text{n}}^{2}}\frac{1}{\text{u}}{\textstyle \phantom{\rule{1em}{0ex}}}\text{d}\text{u}=\frac{1}{2}\cdot \underset{\u03f5\to \mathrm{\infty}}{lim}{\left[\mathrm{ln}\left|\text{u}\right|\right]}_{{\text{s}}^{2}+{\text{n}}^{2}}^{{\u03f5}^{2}+{\text{n}}^{2}}=$

$\frac{1}{2}\cdot \underset{\u03f5\to \mathrm{\infty}}{lim}(\mathrm{ln}|{\u03f5}^{2}+{\text{n}}^{2}|-\mathrm{ln}|{\text{s}}^{2}+{\text{n}}^{2}|)=\frac{1}{2}\cdot \underset{\u03f5\to \mathrm{\infty}}{lim}\left|\frac{{\u03f5}^{2}+{\text{n}}^{2}}{{\text{s}}^{2}+{\text{n}}^{2}}\right|\to \mathrm{\infty}$

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

Using the table of selected Laplace transforms, we find:

So, we get:

Let's substitute

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