Improper integrals: infinite limits, infinite integrands and comparison

Worked solutions

Problem 1

1. $\displaystyle\int_1^b x^{-3}\,dx = \Big[-\frac{1}{2x^2}\Big]_1^b = \frac12 - \frac{1}{2b^2}$Replace $\infty$ with $b$ and integrate.
2. $\dfrac{1}{2b^2} \to 0$ as $b \to \infty$Take the limit.
3. $\tfrac12$The integral converges: the area under $\tfrac1{x^3}$ from $1$ out to infinity is finite.

Problem 2

1. For $p \ne 1$: $\displaystyle\int_1^b x^{-p}\,dx = \frac{b^{1-p} - 1}{1 - p}$The power rule.
2. If $p > 1$, $b^{1-p} \to 0$ and the integral is $\dfrac{-1}{1 - p} = \dfrac{1}{p - 1}$; if $p < 1$, $b^{1-p} \to \infty$The exponent $1 - p$ is negative in the first case and positive in the second.
3. For $p = 1$: $\displaystyle\int_1^b \frac{dx}x = \ln b \to \infty$The borderline case diverges, slowly.
4. Converges to $\tfrac{1}{p - 1}$ for $p > 1$, diverges for $p \le 1$; $p = \tfrac32$ gives $2$, $p = 1$ diverges$\tfrac1x$ and $\tfrac1{x^{1.01}}$ look almost the same, but only the second has a finite area.

Problem 3

1. $\dfrac{1}{\sqrt x}$ blows up at $x = 0$, so $\displaystyle\int_0^1 \frac{dx}{\sqrt x} = \lim_{\varepsilon\to0^+}\Big[2\sqrt x\Big]_\varepsilon^1$The problem is at the left end.
2. $= \lim_{\varepsilon\to0^+}\big(2 - 2\sqrt\varepsilon\big) = 2$Converges.
3. For $p \ne 1$: $\displaystyle\int_\varepsilon^1 x^{-p}\,dx = \frac{1 - \varepsilon^{1-p}}{1 - p}$The same power rule as Problem 2.
4. $\varepsilon^{1-p} \to 0$ when $p < 1$ and $\to \infty$ when $p > 1$; $p = 1$ gives $-\ln\varepsilon \to \infty$Near $0$ the roles of the exponents are reversed.
5. $2$; $\displaystyle\int_0^1 x^{-p}\,dx$ converges to $\tfrac{1}{1 - p}$ exactly when $p < 1$Near $0$ a small power blows up slowly enough, and near $\infty$ a large power decays fast enough. $p = 1$ fails at both ends.

Problem 4

1. $\dfrac{1}{x^2}$ blows up at $x = 0$, inside $[-1, 2]$Check the interval for a zero in a denominator before integrating.
2. $\displaystyle\int_{-1}^2 = \int_{-1}^0 + \int_0^2$, and both pieces must convergeSplit at the singularity.
3. $\displaystyle\int_\varepsilon^2 \frac{dx}{x^2} = \frac1\varepsilon - \frac12 \to \infty$The right piece diverges: this is $p = 2 \ge 1$ near $0$.
4. No: the integral divergesOne divergent piece is enough. The left piece diverges as well.

Problem 5

1. $\ln x \to -\infty$ as $x \to 0^+$, so take $\displaystyle\lim_{\varepsilon\to0^+}\int_\varepsilon^1 \ln x\,dx$An infinite integrand at the left end.
2. $\displaystyle\int \ln x\,dx = x\ln x - x$By parts with $u = \ln x$, $dv = dx$.
3. $\displaystyle\int_\varepsilon^1 \ln x\,dx = (0 - 1) - (\varepsilon\ln\varepsilon - \varepsilon)$Evaluate at the limits.
4. $\varepsilon\ln\varepsilon = \dfrac{\ln\varepsilon}{1/\varepsilon} \to 0$L'Hôpital: $\dfrac{1/\varepsilon}{-1/\varepsilon^2} = -\varepsilon \to 0$. The same limit justifies $0\log0 = 0$ in entropy.
5. $-1$Negative, because $\ln x < 0$ on $(0, 1)$: the region lies below the axis.

Problem 6

1. $\displaystyle\int_{-\infty}^\infty = \int_{-\infty}^0 + \int_0^\infty$Both ends are infinite, so split and take two separate limits.
2. $\displaystyle\int_0^b \frac{dx}{1 + x^2} = \arctan b \to \frac\pi2$$\arctan$ is the antiderivative, and it levels off at $\tfrac\pi2$.
3. $\displaystyle\int_a^0 \frac{dx}{1 + x^2} = -\arctan a \to \frac\pi2$ as $a \to -\infty$The integrand is even, so the left half matches the right.
4. $\pi$Both pieces converge. Dividing by $\pi$ gives the Cauchy density $\tfrac{1}{\pi(1 + x^2)}$, which integrates to $1$ (see Problem 10).

Problem 7

1. Let $u = x^2$, $du = 2x\,dx$The $x$ in front is exactly what the substitution needs.
2. $\displaystyle\int_0^b xe^{-x^2}\,dx = \frac12\int_0^{b^2} e^{-u}\,du = \frac12\big(1 - e^{-b^2}\big)$Change the limits with the variable: $u$ runs from $0$ to $b^2$.
3. $e^{-b^2} \to 0$Take the limit.
4. $\tfrac12$The Gaussian $e^{-x^2}$ on its own has no elementary antiderivative; with the extra $x$ it does.

Problem 8

1. $x^3 + 1 > x^3$, so $\sqrt{x^3 + 1} > x^{3/2}$ and $\dfrac{1}{\sqrt{x^3 + 1}} < \dfrac{1}{x^{3/2}}$ for $x \ge 1$A bigger denominator makes a smaller fraction.
2. $\displaystyle\int_1^\infty \frac{dx}{x^{3/2}} = \frac{1}{3/2 - 1} = 2$The $p$-test with $p = \tfrac32 > 1$ (Problem 2).
3. The integrand is positive and below a function whose integral convergesComparison in the direction that proves convergence.
4. It converges, and its value is less than $2$Numerically it is about $1.895$. There is no elementary antiderivative, so the comparison is the proof.

Problem 9

1. $\dfrac{1}{x(x + 1)} = \dfrac1x - \dfrac{1}{x + 1}$Partial fractions, by the cover-up rule.
2. $\displaystyle\int_1^b \Big(\frac1x - \frac1{x + 1}\Big)\,dx = \Big[\ln\frac{x}{x + 1}\Big]_1^b = \ln\frac{b}{b + 1} - \ln\frac12$Combine the logs before taking any limit.
3. $\dfrac{b}{b + 1} \to 1$, so $\ln\dfrac{b}{b + 1} \to 0$The two logs grow together and their difference settles down.
4. $\ln2 \approx 0.693$The integrand behaves like $\tfrac1{x^2}$ for large $x$, so convergence was expected; the split pieces $\tfrac1x$ and $\tfrac1{x + 1}$ each diverge on their own.

Problem 10

1. Split at $0$: $\displaystyle\int_0^b \frac{x}{1 + x^2}\,dx = \frac12\ln(1 + b^2)$Substitute $u = 1 + x^2$.
2. $\tfrac12\ln(1 + b^2) \to \infty$ as $b \to \infty$The right half diverges, and by symmetry the left half goes to $-\infty$.
3. The integral divergesA two-sided integral needs both halves to converge on their own.
4. A symmetric cutoff gives $\displaystyle\int_{-b}^{b} \frac{x}{1 + x^2}\,dx = 0$, but $\displaystyle\int_{-b}^{2b} \frac{x}{1 + x^2}\,dx = \frac12\ln\frac{1 + 4b^2}{1 + b^2} \to \ln2$Different ways of letting the limits go to infinity give different answers, which is why neither one is the value.
5. It diverges, so the Cauchy distribution has no meanThe mean would be $\tfrac1\pi\int x\cdot\tfrac{1}{1 + x^2}\,dx$, this integral divided by $\pi$. Sample averages of Cauchy draws never settle down.