Calculate the Laplace transform of the function $\frac{\sin t}{t}$ and then use it to compute the integral

Question

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?

Answer 1

I'll show how to compute $$ I(a,b):=\int_0^{\infty}\frac{e^{-at}-e^{-bt}}{t}\,dt \tag{1} $$ using the Laplace transform of $\frac{\sin t}{t}$. Let's start with the identity $$ \int_0^{\infty}\frac{\sin(\omega t)}{\omega}\,d\omega=\frac{\pi}{2}\qquad(t>0). \tag{2} $$ Plugging $(2)$ into $(1)$, and changing the order of integration, we get \begin{align} I(a,b)&=\frac{2}{\pi}\int_0^{\infty}\int_0^{\infty}\frac{e^{-at}-e^{-bt}}{t}\frac{\sin(\omega t)}{\omega}\,d\omega\,dt \\ &=\frac{2}{\pi}\int_0^{\infty}\int_0^{\infty}(e^{-at}-e^{-bt})\frac{\sin(\omega t)}{\omega t}\,dt\,d\omega \\ &=\frac{2}{\pi}\int_0^{\infty}\left(\mathcal{L}\left\{\frac{\sin(\omega t)}{\omega t}\right\}\!(a)-\mathcal{L}\left\{\frac{\sin(\omega t)}{\omega t}\right\}\!(b)\right)d\omega. \tag{3} \end{align} Using the Laplace transform of $\frac{\sin t}{t}$ that you derived, together with the scaling property $\mathcal{L}\{f(\omega t)\}(s)=\frac{1}{\omega}\mathcal{L}\{f(t)\}\!\left(\frac{s}{\omega}\right)$, we can rewrite $(3)$ as \begin{align} I(a,b)&=\frac{2}{\pi}\int_0^{\infty}\frac{1}{\omega}\left(\arctan\left(\frac{b}{\omega}\right)-\arctan\left(\frac{a}{\omega}\right)\right)d\omega \\ &=\frac{2}{\pi}\int_0^{\infty}\int_a^b\frac{1}{\omega^2+x^2}\,dx\,d\omega \\ &=\frac{2}{\pi}\int_a^b\int_0^{\infty}\frac{1}{\omega^2+x^2}\,d\omega\,dx \\ &=\frac{2}{\pi}\int_a^b\frac{\pi}{2x}\,dx \\ &=\ln\left(\frac{b}{a}\right). \tag{4} \end{align} A much simpler derivation of this result is found in https://math.stackexchange.com/a/564237/1163258.