If $f(t)=\frac{e^{bt}-e^{at}}{t}$ how to compute $\mathcal{L} \{ f(t) \}$

Question

I have a problem trying to solve $$ \int_0^{+\infty}\frac{e^{(b-s)t}}{t} $$ I know that I can use the Ei (exponential integral function) but after that I don't know what's exactly this means.

I begin with the definition $ \mathcal{L} \{ f(t) \}:=\int_0^{+\infty}e^{-st}f(t)\,dt $

In my laplace transform table says that $\mathcal{L} \{ f(t) \}=\ln \frac{s-a}{s-b}$

How I know that is true?

Answer 1

You are trying to compute the following integral: $$\int_0^\infty \frac{e^{-t(s-b)}-e^{-t(s-a)}}{t}\,dt$$ By Frullani's integral, your integral is just $$\ln\frac{s-b}{s-a}$$

Answer 2

Don't know what Frullani's integral is? No problem. Your integral can instead be converted to a double integral first by noting that $$\int_{s - b}^{s - a} e^{-xt} \, dx = \frac{e^{-(s - b)t} - e^{-(s - a) t}}{t}.$$ Then \begin{align} \mathcal{L} \{f(t)\} &= \int_0^\infty \frac{e^{-(s - b)t} - e^{-(s - a) t}}{t} \, dt\\ &= \int_0^\infty \int_{s - b}^{s - a} e^{-xt} \, dx \, dt\\ &= \int_{s - b}^{s - a} \int_0^\infty e^{-xt} \, dt \, dx\\ &= \int_{s - b}^{s - a} \frac{dx}{x}\\ &= \ln \left (\frac{s - a}{s - b} \right ). \end{align} Note when $a \neq b$, for convergence we require $s > \max \{a,b\}$.

Answer 3

Here, we will use Feynman's Trick for differentiating under the integral.

Let $F(s)$ be defined by the integral

$$\begin{align} F(s)&=\int_0^\infty \frac{e^{-(s-b)t}-e^{-(s-a)t}}{t}\,dt\tag1 \end{align}$$

Differentiating under the integral in $(1)$ reveals

$$\begin{align} F'(s)&=\int_0^\infty \left(e^{-(s-a)t}-e^{-(s-b)t}\right)\,dt\\\\ &=\frac1{s-a}-\frac1{s-b}\tag 2 \end{align}$$

Next, integrating both sides of $(2)$ we obtain

$$\int_{s}^\infty F'(u)\,du=\int_s^\infty \left(\frac1{u-a}-\frac1{u-b}\right)\,du$$

from which we see that

$$F(s)=\log\left(\frac{s-a}{s-b}\right)$$