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

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?

• Careful - the very last integral diverges! Commented Jul 10 at 18:01
• Frullani's theorem can be used to check the result. Commented Jul 10 at 20:18
• Also, your solution seems to calculate two separate problems rather than use the former to prove the latter. (It is, however, not at all evident to me what the original problem may have intended as a deduction.) Commented Jul 10 at 20:22
• I was asked to solve this problem with mentioned theorem in my attempt. So I need to use this
– user1339343
Commented Jul 10 at 22:47
• Commented Jul 12 at 19:18

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.
• (+1) Haven't seen this version of "$x\cdot 1=x$" before. Commented Jul 12 at 22:24