Laplace transform coming out to infinity

Question

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?

Answer 1

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

$$\mathcal{L}_t\left[\frac{\cos\left(\text{n}\cdot t\right)}{t}\right]_{\left(\text{s}\right)}:=\int_0^\infty\frac{\cos\left(\text{n}\cdot t\right)}{t}\cdot\exp\left(-\text{s}t\right)\space\text{d}x\tag1$$

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

$$\mathcal{L}_t\left[\frac{\cos\left(\text{n}\cdot t\right)}{t}\right]_{\left(\text{s}\right)}=\int_\text{s}^\infty\mathcal{L}_t\left[\cos\left(\text{n}\cdot t\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$

Using the table of selected Laplace transforms, we find:

$$\mathcal{L}_t\left[\cos\left(\text{n}\cdot t\right)\right]_{\left(\sigma\right)}=\frac{\sigma}{\sigma^2+\text{n}^2}\tag3$$

So, we get:

$$\mathcal{L}_t\left[\frac{\cos\left(\text{n}\cdot t\right)}{t}\right]_{\left(\text{s}\right)}=\int_\text{s}^\infty\frac{\sigma}{\sigma^2+\text{n}^2}\space\text{d}\sigma\tag4$$

Let's substitute $\text{u}:=\sigma^2+\text{n}^2$, so we get:

$$\mathcal{L}_t\left[\frac{\cos\left(\text{n}\cdot t\right)}{t}\right]_{\left(\text{s}\right)}=\frac{1}{2}\cdot\lim_{\epsilon\to\infty}\int_{\text{s}^2+\text{n}^2}^{\epsilon^2+\text{n}^2}\frac{1}{\text{u}}\space\text{d}\text{u}=\frac{1}{2}\cdot\lim_{\epsilon\to\infty}\left[\ln\left|\text{u}\right|\right]_{\text{s}^2+\text{n}^2}^{\epsilon^2+\text{n}^2}=$$ $$\frac{1}{2}\cdot\lim_{\epsilon\to\infty}\left(\ln\left|\epsilon^2+\text{n}^2\right|-\ln\left|\text{s}^2+\text{n}^2\right|\right)=\frac{1}{2}\cdot\lim_{\epsilon\to\infty}\left|\frac{\epsilon^2+\text{n}^2}{\text{s}^2+\text{n}^2}\right|\to\infty\tag5$$