Convergence of Improper Integrals Involving Continuous Decreasing Functions

Understanding the convergence or divergence of improper integrals, particularly those involving continuous decreasing functions, is a fundamental aspect of real analysis. In this article, we explore the conditions under which the integral (int_{1}^{infty} f(x) , dx) converges, given that (f(x)) is a continuous, decreasing function on the interval ([1, infty)) and (lim_{x to infty} f(x) 0).

Key Points and Theoretical Foundations

Continuous Decreasing Function: Since (f(x)) is a continuous and decreasing function, for any (x_1

Limit Condition: The condition (lim_{x to infty} f(x) 0) indicates that as (x) increases, (f(x)) approaches 0. However, this alone is not sufficient to guarantee the convergence of the integral. Additional information about the rate at which (f(x)) approaches 0 is necessary.

Convergence of the Integral

To check for the convergence of (int_{1}^{infty} f(x) , dx), we often compare it with integrals of standard forms that have known convergence properties. One useful comparison is with the function (g(x) frac{1}{x^p}) where (p > 1). The integral (int_{1}^{infty} frac{1}{x^p} , dx) converges for (p > 1) and diverges for (p leq 1).

Given (lim_{x to infty} f(x) 0), we can find constants (M > 0) and (N geq 1) such that for all (x geq N), we have (0 leq f(x) leq frac{1}{x^p}) for some (p > 1). By the comparison test, the integral (int_{1}^{infty} f(x) , dx) will converge if (f(x)) decreases to 0 quickly enough.

Examples and Further Analysis

Example 1: (f(x) frac{1}{x})

Consider the function (f(x) frac{1}{x}). The integral (int_{1}^{infty} frac{1}{x} , dx lim_{b to infty} int_{1}^{b} frac{1}{x} , dx lim_{b to infty} [ln x]_1^b lim_{b to infty} (ln b - ln 1) lim_{b to infty} ln b infty).

This integral diverges, indicating that not all functions that approach 0 as (x to infty) will converge.

Example 2: (f(x) frac{1}{1 ln x})

For the function (f(x) frac{1}{1 ln x}), we observe that (lim_{x to infty} f(x) 0). However, the rate at which it approaches 0 is not sufficient to ensure convergence. We can compare it with the integral (int_{1}^{infty} frac{1}{x} , dx), which as shown above, diverges.

Conclusion

In summary, the convergence of (int_{1}^{infty} f(x) , dx) depends critically on the rate at which (f(x)) approaches 0 as (x to infty). The statement that the integral is convergent is not guaranteed without additional information on the specific behavior of (f(x)).

Further Considerations

For a more general case, consider the function (f(x) frac{1}{x^u}) where (u) is a constant. The integral (int_{1}^{infty} frac{1}{x^u} , dx) converges if (u > 1) and diverges if (u leq 1). This example illustrates the impact of the exponent (u) on the convergence behavior of the integral.