Questions: Determine whether the improper integral converges or diverges. ∫ from -∞ to ∞ (5 dx) / (e^x + e^(-x))

Solution Steps

To determine whether the improper integral converges or diverges, we can first simplify the integrand. Notice that the denominator \( e^x + e^{-x} \) can be rewritten using hyperbolic functions. Specifically, it can be expressed as \( 2\cosh(x) \). The integral then becomes:

\[ \int_{-\infty}^{\infty} \frac{5}{2\cosh(x)} \, dx \]

This integral can be split into two parts from \(-\infty\) to \(0\) and from \(0\) to \(\infty\). We can then evaluate each part separately. The function \(\frac{1}{\cosh(x)}\) is known to be integrable over the entire real line, which suggests that the integral converges.

We start with the improper integral:

\[ \int_{-\infty}^{\infty} \frac{5 \, dx}{e^{x} + e^{-x}} \]

We can rewrite the denominator using the hyperbolic cosine function:

\[ e^{x} + e^{-x} = 2\cosh(x) \]

Thus, the integral becomes:

\[ \int_{-\infty}^{\infty} \frac{5}{2\cosh(x)} \, dx \]

The integral can be expressed as:

\[ \frac{5}{2} \int_{-\infty}^{\infty} \frac{1}{\cosh(x)} \, dx \]

It is known that:

\[ \int_{-\infty}^{\infty} \frac{1}{\cosh(x)} \, dx = \pi \]

Therefore, we can substitute this result into our integral:

\[ \frac{5}{2} \cdot \pi = \frac{5\pi}{2} \]

Since the value of the integral is finite, we conclude that the improper integral converges.

Final Answer