Q11Advanced Engineering Mathematics

Question

Find \int_{0}^{\infty} \frac{e ^{- t} - e ^{- x}}{t} d t

Answer 1

We are tasked with rigorously evaluating the improper definite integral $I = \int_{0}^{\infty} \frac{e ^{- a t} - e ^{- b t}}{t} d t$ . While this can be solved using double integrals (Feynman's trick), it is significantly more elegant to solve it using the advanced properties of the Laplace Transform. Specifically, we will utilize the theorem that links an integral from 0 to infinity to the limit of a Laplace transform as the frequency variable $s$ approaches zero.

1. The Division by t Property

First, we define the numerator function as $f (t) = e^{- a t} - e^{- b t}$ . The standard Laplace transform of this function is a straightforward application of the linearity property and basic transform tables:

F (s) = L {e^{- a t} - e^{- b t}} = \frac{1}{s + a} - \frac{1}{s + b}

Next, we apply the "Division by $t$ " property of the Laplace transform. This critical property states that dividing a function by $t$ in the time domain corresponds directly to integrating its transform from $s$ to infinity in the frequency (Laplace) domain, provided the limit of $f (t) / t$ as $t \to 0$ exists.

L {\frac{f ( t )}{t}} = \int_{s}^{\infty} F (u) d u

Applying this property to our specific function yields:

L {\frac{e ^{- a t} - e ^{- b t}}{t}} = \int_{s}^{\infty} (\frac{1}{u + a} - \frac{1}{u + b}) d u

We integrate this expression term by term using natural logarithms:

= [ln (u + a) - ln (u + b)]_{s}^{\infty} = [ln (\frac{u + a}{u + b})]_{s}^{\infty}

To properly evaluate the upper limit as $u \to \infty$ , we divide the numerator and denominator inside the logarithm by $u$ : $ln (\frac{1 + a / u}{1 + b / u})$ . As $u \to \infty$ , the terms $a / u$ and $b / u$ vanish to zero, leaving $ln (1) = 0$ .

= 0 - ln (\frac{s + a}{s + b}) = ln (\frac{s + b}{s + a})

2. Applying the Limit to Evaluate the Integral

We have now established that $L {\frac{e ^{- a t} - e ^{- b t}}{t}} = \int_{0}^{\infty} e^{- s t} \frac{e ^{- a t} - e ^{- b t}}{t} d t = ln (\frac{s + b}{s + a})$ .

The problem asks for the evaluation of the integral without the $e^{- s t}$ term. We can eliminate this term by taking the mathematical limit as $s \to 0$ on both sides of the equation. This is valid provided the integral converges.

s \to 0 lim \int_{0}^{\infty} e^{- s t} \frac{e ^{- a t} - e ^{- b t}}{t} d t = s \to 0 lim ln (\frac{s + b}{s + a})

\int_{0}^{\infty} \frac{e ^{- a t} - e ^{- b t}}{t} d t = ln (\frac{0 + b}{0 + a}) = ln (\frac{b}{a})

This final elegant result is famously known in mathematical literature as Frullani's Integral. If the specific problem had parameters like $a = 1, b = x$ , the final answer would simply be $ln (x /1) = ln (x)$ .

Q11Advanced Engineering Mathematics

Question

Answer

1. The Division by t Property

2. Applying the Limit to Evaluate the Integral