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.)

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.

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 $\infty$.

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 $\epsilon \to 0$.

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

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

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,

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

Final Answer