 # 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
$\frac{\mathrm{cos}t}{t}$
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?
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Well, we are trying to find (solving a more general problem):
${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\cdot \mathrm{exp}\left(-\text{s}t\right)\text{d}x$
Using the 'frequency-domain integration' property of the Laplace transform, we can write:
${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}={\int }_{\text{s}}^{\mathrm{\infty }}{\mathcal{L}}_{t}{\left[\mathrm{cos}\left(\text{n}\cdot t\right)\right]}_{\left(\sigma \right)}\text{d}\sigma$
Using the table of selected Laplace transforms, we find:
${\mathcal{L}}_{t}{\left[\mathrm{cos}\left(\text{n}\cdot t\right)\right]}_{\left(\sigma \right)}=\frac{\sigma }{{\sigma }^{2}+{\text{n}}^{2}}$
So, we get:
${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}={\int }_{\text{s}}^{\mathrm{\infty }}\frac{\sigma }{{\sigma }^{2}+{\text{n}}^{2}}\phantom{\rule{1em}{0ex}}\text{d}\sigma$
Let's substitute $\text{u}\phantom{\rule{0.222em}{0ex}}={\sigma }^{2}+{\text{n}}^{2}$ , so we get:
${\mathcal{L}}_{t}{\left[\frac{\mathrm{cos}\left(\text{n}\cdot t\right)}{t}\right]}_{\left(\text{s}\right)}=\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}{\int }_{{\text{s}}^{2}+{\text{n}}^{2}}^{{ϵ}^{2}+{\text{n}}^{2}}\frac{1}{\text{u}}\phantom{\rule{1em}{0ex}}\text{d}\text{u}=\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}{\left[\mathrm{ln}|\text{u}|\right]}_{{\text{s}}^{2}+{\text{n}}^{2}}^{{ϵ}^{2}+{\text{n}}^{2}}=$
$\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}\left(\mathrm{ln}|{ϵ}^{2}+{\text{n}}^{2}|-\mathrm{ln}|{\text{s}}^{2}+{\text{n}}^{2}|\right)=\frac{1}{2}\cdot \underset{ϵ\to \mathrm{\infty }}{lim}|\frac{{ϵ}^{2}+{\text{n}}^{2}}{{\text{s}}^{2}+{\text{n}}^{2}}|\to \mathrm{\infty }$