Calculate the improper integral: integral from -1 to ()

Calculate the improper integral: $$ \int_{-1}^\infty e^{-kx} \, dx $$

This is an improper integral at infinity, and its convergence depends on the parameter $k$.

Step 1: Express as a limit

$$\int_{-1}^\infty e^{-kx} \, dx = \lim_{R \to \infty} \int_{-1}^R e^{-kx} \, dx$$

Step 2: Find the antiderivative

We need to consider two cases based on whether $k = 0$ or $k \neq 0$.

Case 1: $k \neq 0$

$$\int e^{-kx} \, dx = \frac{e^{-kx}}{-k} + C = -\frac{1}{k}e^{-kx} + C$$

Verification:

$$\frac{d}{dx}\left[-\frac{1}{k}e^{-kx}\right] = -\frac{1}{k} \cdot (-k)e^{-kx} = e^{-kx}$$

Correct! ✓

Case 2: $k = 0$

$$\int e^{0} \, dx = \int 1 \, dx = x + C$$

Step 3: Evaluate for $k \neq 0$

$$\int_{-1}^R e^{-kx} \, dx = \left[-\frac{1}{k}e^{-kx}\right]_{-1}^R = -\frac{1}{k}e^{-kR} - \left(-\frac{1}{k}e^{k}\right)$$

$$= -\frac{1}{k}e^{-kR} + \frac{1}{k}e^k = \frac{1}{k}\left(e^k - e^{-kR}\right)$$

Step 4: Analyze the limit for different values of $k$

$$\int_{-1}^\infty e^{-kx} \, dx = \lim_{R \to \infty} \frac{1}{k}\left(e^k - e^{-kR}\right)$$

Case A: $k > 0$

As $R \to \infty$: $-kR \to -\infty$, so $e^{-kR} \to 0$

Therefore:

$$\int_{-1}^\infty e^{-kx} \, dx = \frac{1}{k}(e^k - 0) = \frac{e^k}{k}$$

The integral converges.

Case B: $k < 0$

As $R \to \infty$: $-kR \to +\infty$ (since $-k > 0$), so $e^{-kR} \to +\infty$

Therefore:

$$\lim_{R \to \infty} \frac{1}{k}\left(e^k - e^{-kR}\right) = \frac{1}{k}(e^k - \infty)$$

Since $k < 0$, we have $\frac{1}{k} < 0$, so:

$$\frac{1}{k} \cdot (-\infty) = +\infty$$

The integral diverges to $+\infty$.

Case C: $k = 0$

$$\int_{-1}^\infty e^0 \, dx = \int_{-1}^\infty 1 \, dx = \lim_{R \to \infty} [x]_{-1}^R = \lim_{R \to \infty} (R - (-1)) = \lim_{R \to \infty} (R + 1) = +\infty$$

The integral diverges to $+\infty$.

Final Answer

$$\boxed{\int_{-1}^\infty e^{-kx} \, dx = \begin{cases} \dfrac{e^k}{k} & \text{if } k > 0 \quad \text{(converges)}\\[10pt] +\infty & \text{if } k \leq 0 \quad \text{(diverges)} \end{cases}}$$

Summary of convergence

• Converges if and only if $k > 0$
• Diverges if $k \leq 0$
• When convergent, the value is $\frac{e^k}{k}$

Special cases

Case 1: $k = 1$

$$\int_{-1}^\infty e^{-x} \, dx = \frac{e^1}{1} = e \approx 2.718$$

Case 2: $k = 2$

$$\int_{-1}^\infty e^{-2x} \, dx = \frac{e^2}{2} \approx 3.695$$

Case 3: $k = \frac{1}{2}$

$$\int_{-1}^\infty e^{-x/2} \, dx = \frac{e^{1/2}}{1/2} = 2e^{1/2} = 2\sqrt{e} \approx 3.297$$

Case 4: $k = -1$ (diverges)

$$\int_{-1}^\infty e^{x} \, dx = +\infty$$

Case 5: $k = 0$ (diverges)

$$\int_{-1}^\infty 1 \, dx = +\infty$$

Why convergence depends on $k$

The behavior of $e^{-kx}$ as $x \to \infty$ depends on the sign of $k$:

If $k > 0$:

• $e^{-kx} = \frac{1}{e^{kx}} \to 0$ as $x \to \infty$
• The function decays exponentially
• The integral converges

If $k < 0$:

• $e^{-kx} = e^{|k|x} \to +\infty$ as $x \to \infty$
• The function grows exponentially
• The integral diverges

If $k = 0$:

• $e^{0} = 1$ (constant)
• The function doesn't decay
• The integral diverges

Graphical interpretation

For $k > 0$, the graph of $y = e^{-kx}$:

• At $x = -1$: $y = e^k > 1$
• At $x = 0$: $y = 1$
• As $x \to \infty$: $y \to 0$ (exponential decay)
• The area under the curve from $-1$ to $\infty$ is finite (equals $\frac{e^k}{k}$)

For $k < 0$, the graph of $y = e^{-kx}$:

• At $x = -1$: $y = e^k < 1$
• At $x = 0$: $y = 1$
• As $x \to \infty$: $y \to \infty$ (exponential growth)
• The area under the curve from $-1$ to $\infty$ is infinite

Behavior as $k$ varies (for $k > 0$)

The integral value $\frac{e^k}{k}$ as a function of $k > 0$:

• As $k \to 0^+$: $\frac{e^k}{k} \approx \frac{1 + k}{k} = \frac{1}{k} + 1 \to +\infty$
• As $k \to \infty$: $\frac{e^k}{k} \to +\infty$ (exponential growth dominates)
• Minimum occurs at $k = 1$: $\frac{e^1}{1} = e \approx 2.718$

Actually, let's find the minimum. Taking the derivative with respect to $k$:

$$\frac{d}{dk}\left[\frac{e^k}{k}\right] = \frac{ke^k - e^k}{k^2} = \frac{e^k(k-1)}{k^2}$$

Setting this to zero: $k - 1 = 0 \implies k = 1$

For $0 < k < 1$: derivative is negative (decreasing)

For $k > 1$: derivative is positive (increasing)

So the minimum value is indeed at $k = 1$, giving $e$.

Connection to probability

For $k > 0$, the function $ke^{-kx}$ (for $x \geq 0$) is the probability density function of the exponential distribution with rate parameter $k$.

The integral we computed is related to this distribution, though our lower limit is $-1$ rather than $0$.

Summary

1. The integral is improper at $x = \infty$
2. For $k \neq 0$, the antiderivative is $-\frac{1}{k}e^{-kx}$
3. Convergence depends critically on the sign of $k$: \begin{itemize}
4. $k > 0$: exponential decay $\implies$ converges to $\frac{e^k}{k}$
5. $k \leq 0$: no decay (or growth) $\implies$ diverges \end{itemize}
6. The minimum value (for $k > 0$) occurs at $k = 1$, giving $e$
7. This illustrates how parameters dramatically affect convergence

Key lessons:

• Exponential decay ($k > 0$) ensures convergence of integrals to infinity
• Exponential growth ($k < 0$) causes divergence
• The critical case is $k = 0$ (constant function), which diverges
• Parameters in exponentials have a decisive effect on integral behavior
• The sign of the exponent coefficient determines convergence
• This is a fundamental example in analysis and probability theory