Questions: Determine whether the improper integral diverges or converges. ∫ from 0 to ∞ of (1 / (e^(4x) + e^(-4x))) dx converges diverges Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.)

Determine whether the improper integral diverges or converges.

∫ from 0 to ∞ of (1 / (e^(4x) + e^(-4x))) dx

converges
diverges

Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.)

Transcript text: Determine whether the improper integral diverges or converges. \[ \int_{0}^{\infty} \frac{1}{e^{4 x}+e^{-4 x}} d x \] converges diverges Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.) $\square$

Solution

Solution Steps

To determine whether the improper integral converges or diverges, we can analyze the behavior of the integrand as $ x $ approaches the limits of integration (0 and $\infty$). If the integral converges, we can evaluate it using numerical integration methods.

Step 1: Analyze the Integral

We need to evaluate the improper integral

\[ \int_{0}^{\infty} \frac{1}{e^{4x} + e^{-4x}} \, dx. \]

To determine whether this integral converges or diverges, we will analyze the behavior of the integrand as $ x $ approaches 0 and $ \infty $.

Step 2: Behavior as $ x \to 0 $

As $ x \to 0 $, we can approximate the integrand:

\[ e^{4x} \approx 1 \quad \text{and} \quad e^{-4x} \approx 1. \]

Thus,

\[ e^{4x} + e^{-4x} \approx 2, \]

leading to

\[ \frac{1}{e^{4x} + e^{-4x}} \approx \frac{1}{2}. \]

The integral near 0 behaves like

\[ \int_{0}^{\epsilon} \frac{1}{2} \, dx = \frac{\epsilon}{2}, \]

which converges as $ \epsilon \to 0 $.

Step 3: Behavior as $ x \to \infty $

As $ x \to \infty $, we have

\[ e^{4x} \to \infty \quad \text{and} \quad e^{-4x} \to 0. \]

Thus,

\[ e^{4x} + e^{-4x} \approx e^{4x}, \]

leading to

\[ \frac{1}{e^{4x} + e^{-4x}} \approx \frac{1}{e^{4x}}. \]

The integral behaves like

\[ \int_{M}^{\infty} \frac{1}{e^{4x}} \, dx, \]

which converges since

\[ \int \frac{1}{e^{4x}} \, dx = -\frac{1}{4e^{4x}}. \]

Step 4: Evaluate the Integral

Since the integral converges, we can evaluate it:

\[ \int_{0}^{\infty} \frac{1}{e^{4x} + e^{-4x}} \, dx. \]

Using the substitution $ u = e^{4x} $, we have $ du = 4e^{4x} \, dx $ or $ dx = \frac{du}{4u} $. The limits change from $ x = 0 $ to $ x = \infty $ which corresponds to $ u = 1 $ to $ u = \infty $:

\[ \int_{1}^{\infty} \frac{1}{u + \frac{1}{u}} \cdot \frac{du}{4u} = \frac{1}{4} \int_{1}^{\infty} \frac{1}{u^2 + 1} \, du. \]

The integral

\[ \int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u), \]

evaluated from 1 to $\infty$ gives:

\[ \left[ \tan^{-1}(u) \right]_{1}^{\infty} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \]

Thus,

\[ \int_{0}^{\infty} \frac{1}{e^{4x} + e^{-4x}} \, dx = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16}. \]