Existence of Laplace transform of $\frac{\cos (at)}{t}$

Question

Find the Laplace transform of $\dfrac{\sin(at)}{t}$. Does $\mathcal L\left[\dfrac{\cos(at)}{t}\right]$ exist?

I've found the Laplace transform $\mathcal L\left[\dfrac{\sin(at)}{t}\right]$. But I'm having trouble with the division of $t$ method for $\dfrac{\cos(at)}{t}$ and couldn't find the right answer.

Answer 1

Use:

$$\frac{\cos\left(\text{k}t\right)}{t}=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}\left(\text{k}t\right)^{2\text{n}}}{t\left(2\text{n}\right)!}$$

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So, we get:

$$\mathcal{L}_t\left[\frac{\cos\left(\text{k}t\right)}{t}\right]_{\left(\text{s}\right)}:=\int_0^\infty e^{-\text{s}t}\times\frac{\cos\left(\text{k}t\right)}{t}\space\text{d}t=\int_0^\infty e^{-\text{s}t}\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}\left(\text{k}t\right)^{2\text{n}}}{t\left(2\text{n}\right)!}\space\text{d}t=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\int_0^\infty e^{-\text{s}t}\times\frac{\left(\text{k}t\right)^{2\text{n}}}{t}\space\text{d}t=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}\text{k}^{2\text{n}}}{\left(2\text{n}\right)!}\int_0^\infty e^{-\text{s}t}\times\frac{t^{2\text{n}}}{t}\space\text{d}t=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}\text{k}^{2\text{n}}}{\left(2\text{n}\right)!}\int_0^\infty e^{-\text{s}t}t^{2\text{n}-1}\space\text{d}t$$

Now, use when $\Re\left[\text{s}\right]>0\space\wedge\space\color{red}{\Re\left[\text{n}\right]>0}$:

$$\int_0^\infty e^{-\text{s}t}t^{2\text{n}-1}\space\text{d}t=\text{s}^{-2\text{n}}\Gamma\left(2\text{n}\right)$$