Integration: Learning Objective 9

This is an improper integral. By definition, $\int_1^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_1^t f(x) \, dx$. Evaluate the definite integral first, then take the limit.

The integral is divergent because the limit does not exist as a finite number.

We evaluate the improper integral by converting it to a limit: \[ \int_1^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} \int_1^t \frac{1}{x} \, dx \] First, find the definite integral: \[ \int_1^t \frac{1}{x} \, dx = [\ln|x|]_1^t = \ln|t| - \ln|1| = \ln(t) - 0 = \ln(t) \] Now, take the limit as $t \to \infty$: \[ \lim_{t \to \infty} \ln(t) = \infty \] Since the limit is not a finite number, the integral is divergent.

Use the definition of an improper integral: $\lim_{t \to \infty} \int_1^t \frac{1}{x^2} \, dx$.

Convergent. The integral evaluates to 1.

We evaluate the improper integral by converting it to a limit: \[ \int_1^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t x^{-2} \, dx \] First, find the definite integral: \[ \int_1^t x^{-2} \, dx = \left[ \frac{x^{-1}}{-1} \right]_1^t = \left[ -\frac{1}{x} \right]_1^t = \left(-\frac{1}{t}\right) - \left(-\frac{1}{1}\right) = 1 - \frac{1}{t} \] Now, take the limit as $t \to \infty$: \[ \lim_{t \to \infty} \left(1 - \frac{1}{t}\right) = 1 - 0 = 1 \] Since the limit is a finite number, the integral is convergent and its value is 1.

This is the p-test for integrals. Evaluate the integral for a general $p$, considering the case $p=1$ separately (as done in Question 1). Analyze the limit of the result.

The integral is convergent if $p > 1$ and divergent if $p \le 1$.

We already know from Question 1 that for $p=1$, the integral diverges.

Now consider $p \neq 1$: \[ \int_1^{\infty} \frac{1}{x^p} \, dx = \lim_{t \to \infty} \int_1^t x^{-p} \, dx = \lim_{t \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_1^t = \lim_{t \to \infty} \frac{1}{1-p} \left[ \frac{1}{x^{p-1}} \right]_1^t \] \[ = \lim_{t \to \infty} \frac{1}{1-p} \left( \frac{1}{t^{p-1}} - 1 \right) \] The convergence of this limit depends on the behavior of $t^{p-1}$ as $t \to \infty$.

• If $p-1 > 0$ (i.e., $p>1$), then $t^{p-1} \to \infty$, so $\frac{1}{t^{p-1}} \to 0$. The limit becomes $\frac{1}{1-p}(0-1) = \frac{1}{p-1}$. The integral converges.
• If $p-1 < 0$ (i.e., $p<1$), then let $k=1-p > 0$. The term becomes $\frac{1}{t^{-k}} = t^k$, which goes to $\infty$ as $t \to \infty$. The integral diverges.

Combining all cases, the integral converges for $p>1$.

This improper integral requires integration by parts. After finding the antiderivative, evaluate the limit as the lower bound goes to $-\infty$. You may need L'Hôpital's Rule to evaluate the limit.

-1

First, we find the indefinite integral $\int xe^x dx$ using integration by parts. Let $u=x, dv=e^x dx$. Then $du=dx, v=e^x$. \[ \int xe^x dx = xe^x - \int e^x dx = xe^x - e^x \] Now, we evaluate the improper integral: \[ \int_{-\infty}^0 xe^x \,dx = \lim_{t \to -\infty} \int_{t}^0 xe^x \,dx = \lim_{t \to -\infty} [xe^x - e^x]_t^0 \] \[ = \lim_{t \to -\infty} [(0e^0 - e^0) - (te^t - e^t)] = \lim_{t \to -\infty} [-1 - te^t + e^t] \] We need to evaluate $\lim_{t \to -\infty} te^t$. This is an indeterminate form $(-\infty \cdot 0)$. Rewriting it for L'Hôpital's Rule: \[ \lim_{t \to -\infty} \frac{t}{e^{-t}} \overset{L'H}{=} \lim_{t \to -\infty} \frac{1}{-e^{-t}} \to \frac{1}{-\infty} = 0 \] Also, $\lim_{t \to -\infty} e^t = 0$. The final limit is: \[ -1 - 0 + 0 = -1 \]

Use the definition of an improper integral, then find the antiderivative of $e^{-y/2}$ using a simple u-substitution.

$2e^{-2}$

First, write the integral as a limit: \[ \int_{4}^{\infty} e^{-y/2} \,dy = \lim_{t \to \infty} \int_{4}^{t} e^{-y/2} \,dy \] The antiderivative of $e^{-y/2}$ is $\frac{e^{-y/2}}{-1/2} = -2e^{-y/2}$. \[ = \lim_{t \to \infty} [-2e^{-y/2}]_4^t = \lim_{t \to \infty} (-2e^{-t/2} - (-2e^{-4/2})) \] \[ = \lim_{t \to \infty} (-2e^{-t/2} + 2e^{-2}) \] As $t \to \infty$, the term $-t/2 \to -\infty$, so $e^{-t/2} \to 0$. \[ = 0 + 2e^{-2} = 2e^{-2} \]

The function $f(x) = \frac{1}{x-1}$ has an infinite discontinuity at $x=1$, which is inside the interval $[0,3]$. You must split the integral into two improper integrals: $\int_0^1 \frac{1}{x-1} \, dx + \int_1^3 \frac{1}{x-1} \, dx$. The original integral converges only if both parts converge.

The integral is divergent.

We split the integral at the point of discontinuity, $x=1$: \[ \int_0^3 \frac{1}{x-1} \, dx = \int_0^1 \frac{1}{x-1} \, dx + \int_1^3 \frac{1}{x-1} \, dx \] Let's evaluate the first part: \[ \int_0^1 \frac{1}{x-1} \, dx = \lim_{t \to 1^-} \int_0^t \frac{1}{x-1} \, dx = \lim_{t \to 1^-} [\ln|x-1|]_0^t \] \[ = \lim_{t \to 1^-} (\ln|t-1| - \ln|-1|) = \lim_{t \to 1^-} \ln|t-1| \] As $t \to 1^-$, $t-1 \to 0^+$, so $\ln|t-1| \to -\infty$.

Since one of the parts diverges, the entire integral is divergent. There is no need to evaluate the second part.

This integral is improper because $\ln(x)$ has an infinite discontinuity at $x=0$. Express it as a limit and use integration by parts to find the antiderivative.

-1

First, write the integral as a limit: \[ \int_0^1 \ln(x) \, dx = \lim_{t \to 0^+} \int_t^1 \ln(x) \, dx \] We find the antiderivative of $\ln(x)$ using integration by parts: $\int \ln(x) dx = x\ln(x) - x$. \[ = \lim_{t \to 0^+} [x\ln(x) - x]_t^1 = \lim_{t \to 0^+} [(1\ln(1)-1) - (t\ln(t)-t)] \] \[ = \lim_{t \to 0^+} [ (0-1) - t\ln(t) + t ] = -1 - \lim_{t \to 0^+} t\ln(t) + \lim_{t \to 0^+} t \] We need to find $\lim_{t \to 0^+} t\ln(t)$. This is an indeterminate form $0 \cdot (-\infty)$. We use L'Hôpital's Rule: \[ \lim_{t \to 0^+} t\ln(t) = \lim_{t \to 0^+} \frac{\ln t}{1/t} \overset{L'H}{=} \lim_{t \to 0^+} \frac{1/t}{-1/t^2} = \lim_{t \to 0^+} (-t) = 0 \] The final result is: \[ -1 - 0 + 0 = -1 \]

Use the Comparison Theorem for Improper Integrals. Compare the integrand with a simpler function, like $\frac{1}{x}$, whose integral over the same interval is known to diverge.

The integral diverges by the Comparison Theorem, using the divergent integral $\int_1^{\infty} \frac{1}{x} \, dx$ as the lower bound.

We use the Comparison Theorem. Let $f(x) = \frac{1+e^{-x}}{x}$ and $g(x) = \frac{1}{x}$.

1. Establish the inequality: For $x \ge 1$, we know that $e^{-x}$ is a positive value. Therefore, $1+e^{-x} > 1$.
Dividing by $x$ (which is positive) preserves the inequality: \[ \frac{1+e^{-x}}{x} > \frac{1}{x} \] 2. Analyze the comparison integral: As shown in Question 1, the integral of the smaller function, $\int_1^{\infty} \frac{1}{x} \, dx$, is divergent.

3. Conclusion: Since our integral $\int_1^{\infty} \frac{1+e^{-x}}{x} \, dx$ is greater than a known divergent integral, it must also be divergent by the Comparison Theorem.