Calculate the Laplace transform of the function $\frac{\sin t}{t}$ and then use it to compute the integral
$$ \int_0^\infty \frac{e^{-at} - e^{-bt}}{t} \, dt, $$
where $a, b \in \mathbb{R}$.
Attempt: I need to use the following theorem:
Let $F(z)$ be the Laplace transform of the function $f(t)$. Then it holds that:
$$ \mathcal{L} \left\{ \int_0^t f(u) \, du \right\} = \frac{F(z)}{z} \quad \text{and} \quad \mathcal{L} \left\{ \frac{f(t)}{t} \right\} = \int_z^\infty F(u) \, du. $$
To solve the given integral, I start by finding the Laplace transform of the function $\frac{\sin t}{t}$. Utilizing the given theorem:
$$ \mathcal{L} \left\{ \frac{f(t)}{t} \right\} = \int_z^\infty F(u) \, du $$
where $F(z)$ is the Laplace transform of $f(t)$. Here, $f(t) = \sin t$ and its Laplace transform is $F(z) = \frac{1}{z^2 + 1}$. Therefore,
$$ \mathcal{L} \left\{ \frac{\sin t}{t} \right\} = \int_z^\infty \frac{1}{u^2 + 1} \, du = \int_z^\infty \frac{1}{u^2 + 1} \, du = \left[ \tan^{-1}(u) \right]_z^\infty = \frac{\pi}{2} - \tan^{-1}(z) $$
Next, I try using the Laplace transform to compute the integral:
$$ \int_0^\infty \frac{e^{-at} - e^{-bt}}{t} \, dt $$
Taking the Laplace transform of $\frac{e^{-at} - e^{-bt}}{t}$, I use the fact that the Laplace transform of $e^{-at}$ is $\frac{1}{z+a}$ and for $e^{-bt}$, it is $\frac{1}{z+b}$. Thus,
$$ \mathcal{L} \left\{ \frac{e^{-at} - e^{-bt}}{t} \right\} = \int_z^\infty \left( \frac{1}{u+a} - \frac{1}{u+b} \right) \, du $$
Are my steps till now ok and how do I from here evaluate the integral?