Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 25: Problem 6
Show that the Laplace transform of the error function is given as \(F(s)=\int_{0}^{\infty} \operatorname{erf}(t) e^{-s t} \mathrm{~d} t=\frac{e^{-s^{2} / 4}}{s} \operatorname{erfc}\left(\frac{s}{2}\right)\) for \(s>0\).
Short Answer
Expert verified
By using integration by parts and known integrals, the Laplace transform of \ \operatorname{erf}(t) \) is shown to be \ \frac{e^{-s^2/4}}{s} \operatorname{erfc}(\frac{s}{2}) \.
Step by step solution
01
- Understand the Problem
We need to show that the Laplace transform of the error function \( \operatorname{erf}(t) \) results in the given expression. The Laplace transform of a function \( f(t) \) is defined as \( F(s) = \int_{0}^{\infty} f(t) e^{-st} \, dt \).
02
- Define the Error Function
Recall that the error function is defined as \[ \operatorname{erf}(t) = \frac{2}{\sqrt{\pi}} \int_{0}^{t} e^{-u^2} \, du \].
03
- Set Up the Laplace Transform
Substitute \( \operatorname{erf}(t) \) into the Laplace transform definition: \[ F(s) = \int_{0}^{\infty} \operatorname{erf}(t) e^{-st} \, dt \].
04
- Integrate by Parts
To simplify, we can integrate by parts. Let \( u = \operatorname{erf}(t) \) and \( dv = e^{-st} \, dt \). Then, \( du = \frac{2}{\sqrt{\pi}} e^{-t^2} \, dt \) and \( v = \frac{e^{-st}}{-s} \).
05
- Apply Integration by Parts Formula
The integration by parts formula is \[ \int u \, dv = uv - \int v \, du \]. Applying this, we get: \[ \int_{0}^{\infty} \operatorname{erf}(t) e^{-st} \, dt = \left. \frac{\operatorname{erf}(t) e^{-st}}{-s} \right|_{0}^{\infty} + \frac{2}{\sqrt{\pi}s} \int_{0}^{\infty} t e^{-\left(s^2/4 + t^2\right)} \, dt \].
06
- Evaluate Boundary Terms
Evaluate the boundary terms: \[ \lim_{t \to \infty} \operatorname{erf}(t) e^{-st} - \operatorname{erf}(0) \cdot \frac{1}{s} = 0 - 0 = 0 \] (since \( \operatorname{erf}(\infty) = 1 \) and \( e^{-st} \) goes to 0 as \( t \to \infty \)).
07
- Simplify Integral
Now simplify the remaining integral: \[ \frac{2}{s\sqrt{\pi}} \int_{0}^{\infty} e^{-u^2} e^{-(s/2)^2} e^{-u^2} \, du = \frac{2}{s\sqrt{\pi}} e^{-s^2/4} \int_{0}^{\infty} e^{-u^2} \, du \].
08
- Identify Related Function
Here, \( \int_{0}^{\infty} e^{-u^2} \, du \) is a known integral, equal to \( \frac{\sqrt{\pi}}{2} \operatorname{erfc}(\frac{s}{2}) \).
09
- Combine Results
Combine results to get: \[ F(s) = \frac{2}{s\sqrt{\pi}} e^{-s^2/4} \cdot \frac{\sqrt{\pi}}{2} \operatorname{erfc}(\frac{s}{2}) = \frac{e^{-s^2/4}}{s} \operatorname{erfc}(\frac{s}{2}) \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace Transform
The Laplace transform is a powerful mathematical tool used in engineering and physics. It converts a time-domain function into a complex frequency-domain function. The general formula for the Laplace transform of a function \( f(t) \) is given by:
\[ F(s) = \int_{0}^{\infty} f(t) e^{-st} \, dt \]
Here:
\[ F(s) = \int_{0}^{\infty} f(t) e^{-st} \, dt \]
Here:
- \( F(s) \) is the Laplace transform of \( f(t) \)
- \( s \) is a complex number frequency parameter
Error Function
The error function, denoted as \( \operatorname{erf}(t) \), is a special, non-elementary function commonly used in probability, statistics, and partial differential equations. It is defined as:
\[ \operatorname{erf}(t) = \frac{2}{\sqrt{\pi}} \int_{0}^{t} e^{-u^2} \, du \]
This function measures the probability that a value in a normally distributed dataset lies within a certain range. For large positive values of \( t \), \( \operatorname{erf}(t) \approx 1 \) and for large negative values, \( \operatorname{erf}(t) \approx -1 \).
\[ \operatorname{erf}(t) = \frac{2}{\sqrt{\pi}} \int_{0}^{t} e^{-u^2} \, du \]
This function measures the probability that a value in a normally distributed dataset lies within a certain range. For large positive values of \( t \), \( \operatorname{erf}(t) \approx 1 \) and for large negative values, \( \operatorname{erf}(t) \approx -1 \).
Integration by Parts
Integration by parts is a technique used to integrate products of functions. The formula is derived from the product rule of differentiation and is given by:
\[ \int u \, dv = uv - \int v \, du \]
For this case, we set
\[ \int_{0}^{\infty} \operatorname{erf}(t) e^{-st} \, dt = \left. \frac{\operatorname{erf}(t) e^{-st}}{-s} \right|_{0}^{\infty} + \frac{2}{\sqrt{\pi}s} \int_{0}^{\infty} t e^{-\left(s^2/4 + t^2\right)} \, dt \]
\[ \int u \, dv = uv - \int v \, du \]
For this case, we set
- \( u = \operatorname{erf}(t) \)
- \( dv = e^{-st} \, dt \)
- Then, \( du = \frac{2}{\sqrt{\pi}} e^{-t^2} \, dt \) and \( v = \frac{e^{-st}}{-s} \)
\[ \int_{0}^{\infty} \operatorname{erf}(t) e^{-st} \, dt = \left. \frac{\operatorname{erf}(t) e^{-st}}{-s} \right|_{0}^{\infty} + \frac{2}{\sqrt{\pi}s} \int_{0}^{\infty} t e^{-\left(s^2/4 + t^2\right)} \, dt \]
Boundary Terms
Boundary terms arise when applying the integration by parts formula. These terms are evaluated at the limits of integration. In our example, we need to evaluate:
\[ \left. \frac{\operatorname{erf}(t) e^{-st}}{-s} \right|_{0}^{\infty} \]
This simplifies to:
\[ \left. \frac{\operatorname{erf}(t) e^{-st}}{-s} \right|_{0}^{\infty} \]
This simplifies to:
- At \( t = \infty \), \( e^{-st} \rightarrow 0 \)
- \( \operatorname{erf}(\infty) = 1 \)
- At \( t = 0 \), \( \operatorname{erf}(0) = 0 \)
Complementary Error Function
The complementary error function, denoted as \( \operatorname{erfc}(t) \), is closely related to the error function and is defined as:
\[ \operatorname{erfc}(t) = 1 - \operatorname{erf}(t) \]
It represents the probability that a normally distributed variable falls outside a certain range. The function has useful properties for physical and engineering problems, especially in communication theory for noise analysis. Combining the results:
\[ F(s) = \frac{2}{s\sqrt{\pi}} e^{-s^2/4} \cdot \frac{\sqrt{\pi}}{2} \operatorname{erfc}\left(\frac{s}{2}\right) = \frac{e^{-s^2/4}}{s} \operatorname{erfc}\left(\frac{s}{2}\right) \]
\[ \operatorname{erfc}(t) = 1 - \operatorname{erf}(t) \]
It represents the probability that a normally distributed variable falls outside a certain range. The function has useful properties for physical and engineering problems, especially in communication theory for noise analysis. Combining the results:
\[ F(s) = \frac{2}{s\sqrt{\pi}} e^{-s^2/4} \cdot \frac{\sqrt{\pi}}{2} \operatorname{erfc}\left(\frac{s}{2}\right) = \frac{e^{-s^2/4}}{s} \operatorname{erfc}\left(\frac{s}{2}\right) \]
