Chapter 18: Problem 41
Prove that $$ \int_{0}^{\infty} e^{-\alpha \hbar^{2}} d n \approx \int_{1}^{\infty} e^{-\alpha n^{2}} d n $$ if \(\alpha\) is very small. Hint: Prove that $$ \int_{0}^{1} e^{-\alpha n^{2}} d n \ll \int_{0}^{\infty} e^{-\alpha n^{2}} d n $$ by expanding the exponential in the first integral.
Short Answer
Step by step solution
Understanding the Problem
Expand the Exponential in the First Integral
Evaluate the Approximated Integrals
Compare the Integrals
Focus on the Relevant Integral
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Function
In the context of integrals, as we see in our exercise, the exponential function is used to model decay processes. Specifically, the function \( e^{-eta x^2} \) often arises in probability and statistics, and it is pivotal in Gaussian integrals. These integrals are used heavily in the fields of physics and statistics.
This exercise utilizes the exponential function \( e^{-eta x^2} \) to describe how the integral converges as a function's argument becomes very large. The beauty of the exponential function is its smooth nature and the predictable manner in which it declines. Applying this concept is essential when we approximate complex functions in various mathematical problems.
Series Expansion
In this exercise, we encounter the exponential function \( e^{-eta x^2} \). We employ a series expansion to approximate the function for small values of \( \beta \). By expanding, we gain the ability to simplify the exponential and evaluate it more easily.
For the expansion, we use:
- The exponential series: \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
- When \( x \) is close to zero, only the first few terms of the series significantly affect the result.
- In practical use, this means we can approximate: \( e^{-eta x^2} \approx 1 - \beta x^2 \).
Small Parameter Approximation
For our exercise, the small parameter is \( \alpha \), which becomes an essential part of approximating the integral correctly. With \( \alpha \) being a small number, the impact of terms such as \( \alpha n^2 \) in the series expansion of the exponential function is minimal compared to other terms.
Using small parameter approximation, we achieve:
- A simplified expression: When \( \alpha \) is small, \( e^{-\alpha n^2} \) is approximately equal to its first few terms, \( 1 - \alpha n^2 \).
- This reduction leads to an easier calculation of integrals, as seen in our exercise, where terms like \( \alpha n^2 \) have a minor contribution.
- An effective approach to understand the negligible impact of terms when \( \alpha \) approaches zero, allowing for better conclusions about integral convergence.