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Magazine Chapter 100: Problem 3
Bestimme durch Reihenintegration eine Stammfunktion zu \(\mathrm{e}^{x} / x\) auf \(\mathbf{R}^{+}\).
Short Answer
Expert verified
The antiderivative is \( \sum_{n=1}^{\infty} \frac{x^n}{n \cdot n!} + C \).
Step by step solution
01
Understand the Function
The function we need to find an antiderivative for is \( \frac{e^x}{x} \) over \( \mathbb{R}^+ \). This function does not have a simple antiderivative in terms of elementary functions.
02
Define Power Series Representation
Since \( e^x \) can be expressed as a power series \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \), we can consider the integrand \( \frac{e^x}{x} \) as \( \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
03
Simplify the Series Expression
Simplify the series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) to \( \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \) by dividing each term by \( x \). This gives us \( \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \).
04
Integrate Term by Term
Integrate each term of the series \( \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \). This results in a new series of integrals: \( \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \mathrm{dx} \).
05
Obtain the Antiderivative
The antiderivative of this series is \( \sum_{n=0}^{\infty} \frac{x^n}{n \cdot n!} \). Each term is integrated using \( \int x^{n-1} \mathrm{dx} = \frac{x^n}{n} \).
06
Write Out the Antiderivative
Thus, the antiderivative as a series representation of \( \frac{e^x}{x} \) on \( \mathbb{R}^+ \) is \( \sum_{n=1}^{\infty} \frac{x^n}{n \cdot n!} + C \), where \( C \) is the integration constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
An antiderivative, often referred to as an indefinite integral, is a function whose derivative is the original function. In the context of our exercise, we seek a function whose derivative yields \( \frac{e^x}{x} \). Finding the antiderivative is like reverse-engineering differentiation.
Standard functions such as polynomials, trigonometric, and exponential functions usually have straightforward antiderivatives. However, some functions, like the one given in the exercise, \( \frac{e^x}{x} \), don't have simple antiderivatives. For such cases, we approach the problem differently by using tools like power series to find a solution.
Standard functions such as polynomials, trigonometric, and exponential functions usually have straightforward antiderivatives. However, some functions, like the one given in the exercise, \( \frac{e^x}{x} \), don't have simple antiderivatives. For such cases, we approach the problem differently by using tools like power series to find a solution.
Power Series
A power series is an infinite series of the form \( \sum_{n=0}^\infty a_n x^n \), where \( a_n \) represents the coefficients and \( x \) is the variable. Power series allow us to express complex functions in a form that is easier to work with.
In this case, the exponential function \( e^x \) can be represented as a power series:
In this case, the exponential function \( e^x \) can be represented as a power series:
- \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)
- \( \frac{e^x}{x} = \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \)
Exponentiation
Exponentiation refers to the operation of raising a number or expression to a power. In the context of this exercise, it relates to the exponentiation in the power series of \( e^x \).
When dealing with series, exponentiation affects how we simplify and handle the terms. By expressing \( e^x \) as \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), each term includes an exponentiated \( x \). Rewriting \( \frac{e^x}{x} \) effectively means lowering the exponent by dividing by \( x \), making each term \( \frac{x^{n-1}}{n!} \).
This manipulation is central to transitioning from the original function to a form suitable for term-by-term integration.
When dealing with series, exponentiation affects how we simplify and handle the terms. By expressing \( e^x \) as \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), each term includes an exponentiated \( x \). Rewriting \( \frac{e^x}{x} \) effectively means lowering the exponent by dividing by \( x \), making each term \( \frac{x^{n-1}}{n!} \).
This manipulation is central to transitioning from the original function to a form suitable for term-by-term integration.
Term-by-term Integration
Term-by-term integration is a process where we integrate each individual term of a series rather than attempting to integrate the series as a whole. This method is especially useful when dealing with power series.
In our problem, once we have restructured \( \frac{e^x}{x} \) into the series \( \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \), we can easily integrate each term to find the antiderivative:
In our problem, once we have restructured \( \frac{e^x}{x} \) into the series \( \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \), we can easily integrate each term to find the antiderivative:
- \( \int x^{n-1} \mathrm{dx} = \frac{x^n}{n} \)
- \( \sum_{n=1}^{\infty} \frac{x^n}{n \cdot n!} + C \)
