Evaluating the Divergence of a Definite Integral: A Deeper Analysis

Understanding the behavior of definite integrals, especially those that diverge, is crucial in advanced calculus and real analysis. In this article, we will explore a specific example of a definite integral and how its divergence can be proven through the comparison test. The integral in question is:

Evaluating the Integral

The integral we aim to evaluate and determine the divergence is:

(int_0^{frac{pi}{2}} frac{e^{cos{x}} sin{x}}{x} dx)

To evaluate this integral, we need to carefully analyze the integrand and its behavior over the interval ([0, frac{pi}{2}]).

Using the Comparison Test for Divergence

The comparison test is a powerful tool in analysis for determining whether a given integral diverges or converges. This test is based on the idea that if we can find another function which we know diverges and which is greater than or equal to our integrand over a given interval, then the original integral must also diverge.

The key step in using the comparison test here is to understand the behavior of the integral over the interval ([0, frac{pi}{2}]). Specifically, we need to show that the integral diverges by comparing it to a simpler, known divergent integral.

Step-by-Step Analysis

Let's first simplify the integrand:

(e^{cos{x}} sin{x})

For all (x in [0, frac{pi}{2}]), we can observe that:

(e^{cos{x}} sin{x} geq e^0 sin{0} cos{1})

This simplifies to:

(e^{cos{x}} sin{x} geq cos{1})

Thus, we have:

(frac{e^{cos{x}} sin{x}}{x} geq frac{cos{1}}{x})

Integrating the Comparison Function

Next, we need to consider the integral of (frac{cos{1}}{x}) over the interval ([0, frac{pi}{2}]). This integral can be broken down as follows:

(int_0^{frac{pi}{2}} frac{cos{1}}{x} dx)

Since (cos{1}) is a constant, we can factor it out:

(cos{1} int_0^{frac{pi}{2}} frac{1}{x} dx)

However, the integral (int_0^{frac{pi}{2}} frac{1}{x} dx) is a well-known improper integral that diverges. This can be shown by evaluating the limit:

(int_0^{frac{pi}{2}} frac{1}{x} dx lim_{t to 0^ } ln|x| Big |_t^{frac{pi}{2}})

Evaluating the expression, we get:

(lim_{t to 0^ } ln left(frac{pi}{2}right) - ln(t) infty)

This divergence is due to the logarithmic function (ln(t)) approaching negative infinity as (t) approaches zero.

Conclusion: Divergence via Comparison Test

Given that the integral (int_0^{frac{pi}{2}} frac{cos{1}}{x} dx) diverges, and since (frac{e^{cos{x}} sin{x}}{x} geq frac{cos{1}}{x}) for all (x in [0, frac{pi}{2}]), we can conclude by the comparison test that the integral in question also diverges.

In summary, the definite integral (int_0^{frac{pi}{2}} frac{e^{cos{x}} sin{x}}{x} dx) is divergent, and this divergence is established through the comparison test with a known divergent integral.

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