Investigate the convergence of the integral: integral

Investigate the convergence of the integral: $$ \int_1^\infty \frac{1}{xe^x} \, dx $$

This integral is improper at infinity. We need to analyze the behavior of the integrand for large $x$.

Step 1: Analyze the integrand

The integrand is:

$$f(x) = \frac{1}{xe^x}$$

This can be written as:

$$f(x) = \frac{1}{x} \cdot \frac{1}{e^x} = \frac{1}{x} \cdot e^{-x}$$

Step 2: Behavior at infinity

As $x \to \infty$:

• $\frac{1}{x} \to 0$ (algebraic decay)
• $e^{-x} \to 0$ (exponential decay)
• $\frac{1}{xe^x} \to 0$ (product of both decays)

The exponential decay $e^{-x}$ is much faster than the algebraic decay $\frac{1}{x}$.

Step 3: Comparison with exponential function

For convergence, we can compare with $g(x) = e^{-x}$.

We know that:

$$\int_1^\infty e^{-x} \, dx = [-e^{-x}]_1^\infty = 0 - (-e^{-1}) = \frac{1}{e} \quad \text{(converges)}$$

Now, for $x \geq 1$:

$$\frac{1}{x} \leq 1$$

Therefore:

$$\frac{1}{xe^x} = \frac{1}{x} \cdot e^{-x} \leq 1 \cdot e^{-x} = e^{-x}$$

By the comparison test, since $0 \leq f(x) \leq g(x)$ for $x \geq 1$ and $\int_1^\infty g(x) \, dx$ converges, we conclude that $\int_1^\infty f(x) \, dx$ also converges.

Step 4: Direct verification (optional)

We can also verify by computing the integral directly.

Using integration by parts with $u = \frac{1}{x}$ and $dv = e^{-x} \, dx$:

$$du = -\frac{1}{x^2} \, dx, \quad v = -e^{-x}$$

$$\int \frac{1}{xe^x} \, dx = \int \frac{1}{x} \cdot e^{-x} \, dx = -\frac{e^{-x}}{x} - \int -e^{-x} \cdot \left(-\frac{1}{x^2}\right) dx$$

$$= -\frac{e^{-x}}{x} - \int \frac{e^{-x}}{x^2} \, dx$$

The remaining integral $\int \frac{e^{-x}}{x^2} \, dx$ also converges (by similar reasoning or repeated integration by parts).

Evaluating:

$$\int_1^R \frac{1}{xe^x} \, dx = \left[-\frac{e^{-x}}{x}\right]_1^R - \int_1^R \frac{e^{-x}}{x^2} \, dx$$

As $R \to \infty$:

$$-\frac{e^{-R}}{R} \to 0 \quad \text{(exponential decay dominates)}$$

Therefore, the integral converges to a finite value.

Conclusion

$$\boxed{\text{The integral } \int_1^\infty \frac{1}{xe^x} \, dx \text{ CONVERGES}}$$

Why it converges: Exponential decay

The key reason for convergence is the **exponential decay** factor $e^{-x}$.

General principle:

Exponential decay dominates all polynomial decay (or growth) at infinity:

$$\lim_{x \to \infty} \frac{x^n}{e^x} = 0 \quad \text{for any } n$$

This means:

$$\lim_{x \to \infty} x^n e^{-x} = 0 \quad \text{and the decay is very rapid}$$

In our case, even with the slow decay factor $\frac{1}{x}$, the exponential $e^{-x}$ ensures convergence.

Comparison with related integrals

Integral 1: $\int_1^\infty \frac{1}{x} \, dx$

This **diverges** (logarithmically): $\int_1^\infty \frac{dx}{x} = \ln x \big|_1^\infty = \infty$

Integral 2: $\int_1^\infty e^{-x} \, dx$

This **converges**: $\int_1^\infty e^{-x} \, dx = \frac{1}{e} \approx 0.368$

Integral 3: $\int_1^\infty \frac{1}{xe^x} \, dx$ (our integral)

This **converges** because the exponential factor dominates.

Integral 4: $\int_1^\infty \frac{x^n}{e^x} \, dx$ for any $n$

This **converges** for any finite $n$ (exponential decay always wins).

Quantitative bound

For $x \geq 1$:

$$\int_1^\infty \frac{1}{xe^x} \, dx \leq \int_1^\infty e^{-x} \, dx = \frac{1}{e} \approx 0.368$$

So the integral value is bounded above by $\frac{1}{e}$.

The actual value can be expressed using special functions (exponential integral), but the key point is that it's finite.

Rate of convergence

The integral converges exponentially fast. As we extend the upper limit from $R$ to $R+1$:

$$\int_R^{R+1} \frac{1}{xe^x} \, dx \leq \int_R^{R+1} e^{-x} \, dx = e^{-R} - e^{-(R+1)} = e^{-R}(1 - e^{-1}) \approx 0.632 \cdot e^{-R}$$

This decays exponentially with $R$.

General criterion

For integrals of the form:

$$\int_a^\infty P(x) e^{-x} \, dx$$

where $P(x)$ is a polynomial (or grows slower than exponentially), the integral **always converges** because exponential decay dominates all polynomial behavior.

Physical interpretation

This type of integral appears in physics and probability:

1. Statistical mechanics: Integrals involving $e^{-E/kT}$ (Boltzmann factors) weighted by other quantities.

2. Laplace transforms: The Laplace transform involves integrals of the form $\int_0^\infty f(t) e^{-st} \, dt$.

3. Probability: Exponential distributions and related continuous distributions.

The exponential factor ensures that "contributions" from large values are negligible.

Summary

1. The integrand is $\frac{1}{xe^x} = \frac{1}{x} \cdot e^{-x}$
2. Exponential decay $e^{-x}$ dominates algebraic factor $\frac{1}{x}$
3. By comparison test: $\frac{1}{xe^x} \leq e^{-x}$ for $x \geq 1$
4. Since $\int_1^\infty e^{-x} \, dx$ converges, our integral also converges
5. The integral converges exponentially fast
6. Value is bounded: $\int_1^\infty \frac{1}{xe^x} \, dx \leq \frac{1}{e}$

Key lessons:

• Exponential decay ensures convergence at infinity
• Exponential functions dominate all polynomial functions
• The comparison test is powerful when exponentials are involved
• $e^{-x}$ decays faster than any negative power of $x$
• For $\int P(x) e^{-x} \, dx$ with polynomial $P(x)$: always converges
• This is a fundamental result in analysis and probability theory