Commutativity of integration and Taylor expansion of the integrand in an integral

When we talk about the commutativity of integration and the Taylor expansion of an integrand, we are delving into some fascinating aspects of calculus and analysis. Let's break this down step by step to clarify how these concepts interact and why they are important.

Understanding Commutativity in Integration

Commutativity in the context of integration refers to the ability to interchange the order of integration and summation or integration and differentiation under certain conditions. This is particularly relevant when we have an integral of a function that can be expressed as a series, such as a Taylor series.

The Taylor Expansion

The Taylor expansion allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For a function \( f(x) \), the Taylor series around the point \( a \) is given by:

• \( f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \ldots \)

This series converges to \( f(x) \) within a certain radius of convergence. Now, if we have an integral of a function that can be expressed as a Taylor series, we can consider the integral of the series itself.

Interchanging Integration and Summation

Under certain conditions, we can interchange the integral and the summation of a Taylor series. This is often justified by the Dominated Convergence Theorem or Fubini's Theorem, which provide the necessary conditions for such operations. Specifically, if the series converges uniformly on the interval of integration, we can write:

• \( \int_a^b f(x) \, dx = \int_a^b \left( \sum_{n=0}^{\infty} c_n (x - a)^n \right) dx = \sum_{n=0}^{\infty} c_n \int_a^b (x - a)^n \, dx \)

Here, \( c_n \) are the coefficients from the Taylor series expansion of \( f(x) \). This interchange is crucial because it allows us to compute the integral of a function by integrating each term of its Taylor series separately.

Example for Clarity

Let’s consider a simple example: the function \( f(x) = e^x \). The Taylor series expansion around \( x = 0 \) is:

• \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)

If we want to integrate \( e^x \) from 0 to 1, we can write:

• \( \int_0^1 e^x \, dx = \int_0^1 \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) dx \)

By interchanging the integral and the summation (assuming uniform convergence), we get:

• \( = \sum_{n=0}^{\infty} \frac{1}{n!} \int_0^1 x^n \, dx \)

Calculating the integral \( \int_0^1 x^n \, dx = \frac{1}{n+1} \), we find:

• \( = \sum_{n=0}^{\infty} \frac{1}{n! (n+1)} \)

This series converges to \( e - 1 \), which matches the direct evaluation of the integral \( \int_0^1 e^x \, dx \). This example illustrates how the commutativity of integration and the Taylor expansion can simplify complex calculations.

Final Thoughts

In summary, the interplay between the commutativity of integration and the Taylor expansion is a powerful tool in calculus. It allows us to evaluate integrals of functions that may be difficult to handle directly by leveraging the properties of series. Understanding the conditions under which we can interchange these operations is key to applying these concepts effectively in mathematical analysis.