Corrected exercises: improper integrals (MATH 141)

a) The only problem is the infinite bound: $x^{-3}$ is continuous on $[1, \infty)$. So $\int_1^\infty \frac{dx}{x^3} = \lim_{t\to\infty}\int_1^t x^{-3}\,dx = \lim_{t\to\infty}\left[-\frac{1}{2x^2}\right]_1^t = \lim_{t\to\infty}\left(\frac{1}{2} - \frac{1}{2t^2}\right) = \frac{1}{2}$, because $\frac{1}{2t^2} \to 0$. The integral converges to $\frac{1}{2}$. Check of the antiderivative: $\frac{d}{dx}\left(-\frac{1}{2}x^{-2}\right) = x^{-3}$. The limit is written on the FIRST line. The copy that writes $\left[-\frac{1}{2x^2}\right]_1^\infty = 0 + \frac{1}{2}$ gets the number and loses the method mark, because $\infty$ is not a number that can be substituted: the definition of the integral is the limit, and the marker looks for it.

b) Here the infinite bound is the LOWER one: $\int_{-\infty}^{0} e^{3x}\,dx = \lim_{t\to-\infty}\int_t^0 e^{3x}\,dx = \lim_{t\to-\infty}\left[\frac{1}{3}e^{3x}\right]_t^0 = \lim_{t\to-\infty}\left(\frac{1}{3} - \frac{1}{3}e^{3t}\right) = \frac{1}{3}$. As $t \to -\infty$, $3t \to -\infty$ and $e^{3t} \to 0$. Converges to $\frac{1}{3}$. The slip to avoid is on the sign of the exponent: the same integrand on $[0, \infty)$ would diverge, since $e^{3t} \to \infty$ as $t \to +\infty$. Before taking the limit, say in words which way $t$ goes.

c) $\int_1^\infty \frac{dx}{\sqrt{x}} = \lim_{t\to\infty}\int_1^t x^{-1/2}\,dx = \lim_{t\to\infty}\left[2\sqrt{x}\right]_1^t = \lim_{t\to\infty}\left(2\sqrt{t} - 2\right) = \infty$. The limit is not finite, so the integral DIVERGES. Writing $= \infty$ is acceptable as long as the word diverges follows: an integral equal to infinity is a divergent integral, not a large number. The shaded region of the figure has infinite area although its height shrinks to $0$.

d) Substitution $u = \ln x$, $du = \frac{dx}{x}$; the bounds $2$ and $t$ become $\ln 2$ and $\ln t$. So $\int_2^t \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^{\ln t} u^{-2}\,du = \left[-\frac{1}{u}\right]_{\ln 2}^{\ln t} = \frac{1}{\ln 2} - \frac{1}{\ln t}$. As $t \to \infty$, $\ln t \to \infty$, so $\frac{1}{\ln t} \to 0$ and the integral converges to $\frac{1}{\ln 2}$, about $\frac{1}{0.69} \approx 1.44$. With one factor of $\ln x$ less, the same substitution gives $\int_2^t \frac{dx}{x\ln x} = \int_{\ln 2}^{\ln t}\frac{du}{u} = \ln(\ln t) - \ln(\ln 2) \to \infty$: DIVERGES, although so slowly that at $t = e^{10}$, about twenty-two thousand, the integral has only reached about $2.7$. A slow growth is still a divergence; no table of values decides a limit.

e) Both integrands tend to $0$, yet a) converges and c) diverges. So a limit of $0$ is NOT enough; it only rules out one thing: if $f(x)$ tended to a positive number $L$, then $f(x) \ge \frac{L}{2}$ from some point on, the area would grow at least linearly and the integral would diverge. What decides is HOW FAST $f$ tends to $0$: $\frac{1}{x^3}$ shrinks fast enough for the area over $[1, t]$ to stay below $\frac{1}{2}$, while $\frac{1}{\sqrt{x}}$ shrinks so slowly that the area keeps growing like $2\sqrt{t}$. The yardstick for how fast is fast enough is the family of $p$-integrals of the next exercise: $\int_1^\infty \frac{dx}{x^p}$ converges exactly when $p > 1$.

a) At $x = 0$ the integrand $\frac{1}{\sqrt{x}}$ is undefined and tends to $+\infty$; at $x = 4$ it is $\frac{1}{2}$. So the integral is improper at $0$ only: $\int_0^4 \frac{dx}{\sqrt{x}} = \lim_{t\to0^+}\int_t^4 x^{-1/2}\,dx = \lim_{t\to0^+}\left[2\sqrt{x}\right]_t^4 = \lim_{t\to0^+}\left(4 - 2\sqrt{t}\right) = 4$. Converges. The region is infinitely tall and yet has area $4$: it becomes thin fast enough near the axis. Compare with Exercise 1 c), where the SAME function on $[1, \infty)$ diverged: the behaviour at $0$ and the behaviour at $\infty$ are two separate questions.

b) At $x = 3$ the denominator $\sqrt{3 - x}$ vanishes, so the problem is at the UPPER bound and the limit is taken from the left: $\int_2^3 \frac{dx}{\sqrt{3 - x}} = \lim_{t\to3^-}\int_2^t (3 - x)^{-1/2}\,dx = \lim_{t\to3^-}\left[-2\sqrt{3 - x}\right]_2^t = \lim_{t\to3^-}\left(-2\sqrt{3 - t} + 2\right) = 2$. The minus sign of the antiderivative comes from the chain rule: $\frac{d}{dx}\left(-2(3 - x)^{1/2}\right) = -2 \cdot \frac{1}{2}(3 - x)^{-1/2} \cdot (-1) = (3 - x)^{-1/2}$. Forgetting it gives $-2$, a negative value for a positive integrand, which alone should stop the pen.

c) $\frac{1}{x^2} \to \infty$ as $x \to 0^+$: $\int_0^1 \frac{dx}{x^2} = \lim_{t\to0^+}\int_t^1 x^{-2}\,dx = \lim_{t\to0^+}\left[-\frac{1}{x}\right]_t^1 = \lim_{t\to0^+}\left(-1 + \frac{1}{t}\right) = \infty$. DIVERGES. Next to a), the lesson is the one of Exercise 1, but at the other end: near $0$, $\frac{1}{x^2}$ blows up much faster than $\frac{1}{\sqrt{x}}$, and that is what makes the area infinite. Exercise 3 turns this into a rule.

d) $\ln x \to -\infty$ as $x \to 0^+$, so the integral is improper at $0$. By parts, $u = \ln x$, $dv = dx$, so $du = \frac{dx}{x}$ and $v = x$: $\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x$, and indeed $\frac{d}{dx}(x\ln x - x) = \ln x + 1 - 1 = \ln x$. Then $\int_t^1 \ln x\,dx = (0 - 1) - (t\ln t - t) = -1 - t\ln t + t$. The last limit is the crux: $t\ln t$ is of the form $0 \cdot (-\infty)$, which is indeterminate and is NOT $0$ by decree. Write it as a quotient, $t\ln t = \frac{\ln t}{1/t}$, of the form $\frac{-\infty}{\infty}$, and apply L'Hôpital's rule: $\lim_{t\to0^+}\frac{1/t}{-1/t^2} = \lim_{t\to0^+}(-t) = 0$. So $\int_0^1 \ln x\,dx = -1 - 0 + 0 = -1$. Converges. The sign is the figure: on $(0, 1)$, $\ln x < 0$, so the whole region lies BELOW the axis and the integral is minus its area. The shaded region has area $1$, although it has no bottom.

e) $\tan x = \frac{\sin x}{\cos x}$ and $\cos x \to 0^+$ as $x \to \frac{\pi}{2}^-$, so the integrand is unbounded at the upper bound. With $\int \tan x\,dx = -\ln\lvert\cos x\rvert + C$ (substitution $u = \cos x$): $\int_0^{\pi/2}\tan x\,dx = \lim_{t\to\pi/2^-}\left[-\ln(\cos x)\right]_0^t = \lim_{t\to\pi/2^-}\left(-\ln(\cos t) + \ln 1\right)$. Since $\cos t \to 0^+$, $\ln(\cos t) \to -\infty$ and $-\ln(\cos t) \to +\infty$. DIVERGES. This one looks exactly like an ordinary definite integral from a trigonometry table; the only warning is that $\tan$ is not defined at $\frac{\pi}{2}$, which is why checking the endpoints comes before anything else.

a) For $p \ne 1$: $\int_1^t x^{-p}\,dx = \left[\frac{x^{1-p}}{1 - p}\right]_1^t = \frac{t^{1-p} - 1}{1 - p}$. If $p > 1$, the exponent $1 - p$ is negative, so $t^{1-p} = \frac{1}{t^{p-1}} \to 0$ and the limit is $\frac{-1}{1 - p} = \frac{1}{p - 1}$: converges. If $p < 1$, the exponent is positive, $t^{1-p} \to \infty$: diverges. For $p = 1$ the power rule does not apply: $\int_1^t \frac{dx}{x} = \ln t \to \infty$, diverges. Conclusion: $\int_1^\infty \frac{dx}{x^p} = \frac{1}{p - 1}$ if $p > 1$, divergent if $p \le 1$. The case $p = 1$ is the one students forget, and it is the one on the boundary: $\frac{1}{x}$ is the fastest-decaying power that still diverges.

b) For $p \le 0$ the integrand $x^{-p}$ is continuous on $[0, 1]$ and nothing is improper. For $p > 0$ it blows up at $0$. For $p \ne 1$: $\int_t^1 x^{-p}\,dx = \frac{1 - t^{1-p}}{1 - p}$. If $0 < p < 1$, $1 - p > 0$ and $t^{1-p} \to 0$ as $t \to 0^+$: the limit is $\frac{1}{1 - p}$, converges. If $p > 1$, $t^{1-p} = \frac{1}{t^{p-1}} \to \infty$: diverges. For $p = 1$: $\int_t^1 \frac{dx}{x} = -\ln t \to \infty$, diverges. Conclusion: $\int_0^1 \frac{dx}{x^p} = \frac{1}{1 - p}$ if $p < 1$, divergent if $p \ge 1$. Same computation as a), with the opposite verdict, because $t$ now goes to $0$ instead of $\infty$.

c) On $(1, \infty)$ a larger exponent makes the function SMALLER: $\frac{1}{x^2} < \frac{1}{x} < \frac{1}{\sqrt{x}}$, the red curve lowest. At infinity the danger is a function that does not die fast enough, so a large $p$ helps. On $(0, 1)$ the order is REVERSED: $\frac{1}{x^2} > \frac{1}{x} > \frac{1}{\sqrt{x}}$, the red curve highest. Near $0$ the danger is a function that explodes too fast, so a large $p$ hurts. The three curves meet at $(1, 1)$, which is why $1$ is the natural place to split, and $p = 1$ sits on the boundary of both rules and fails both.

d) $\int_1^\infty \frac{dx}{x^{1.01}}$: $p = 1.01 > 1$ on $[1, \infty)$, converges to $\frac{1}{0.01} = 100$. However close to $1$, a $p$ above $1$ converges; the value is simply large. $\int_0^1 \frac{dx}{x\sqrt{x}}$: $x\sqrt{x} = x^{3/2}$, so $p = \frac{3}{2} \ge 1$ on $(0, 1]$: diverges. $\int_1^\infty \frac{dx}{\sqrt[3]{x^2}}$: $p = \frac{2}{3} \le 1$ on $[1, \infty)$: diverges. $\int_0^1 \frac{dx}{\sqrt[3]{x^2}}$: the same $p = \frac{2}{3}$, now on $(0, 1]$ where $p < 1$ is required: converges to $\frac{1}{1 - 2/3} = 3$. The last two lines are the whole chapter in miniature: one function, two intervals, two verdicts. Always say WHICH rule is applied, with its interval.

e) $\int_0^\infty$ is improper at both ends, so it must be split, for instance at $1$: $\int_0^\infty \frac{dx}{x^p} = \int_0^1 \frac{dx}{x^p} + \int_1^\infty \frac{dx}{x^p}$, and it converges only if BOTH pieces converge. For $p > 0$, the first needs $p < 1$ and the second $p > 1$: no $p$ satisfies both, and $p = 1$ fails both. For $p \le 0$ the first piece is an ordinary integral but the second diverges since $p \le 0 < 1$. So $\int_0^\infty \frac{dx}{x^p}$ diverges for every real $p$. No pure power has a finite area on $(0, \infty)$: a convergent integral there needs an integrand that behaves like one power near $0$ and like another near $\infty$, as in Exercise 7.

a) $f(x) = (x - 1)^{-2/3}$ is continuous everywhere except at $x = 1$, where the denominator vanishes and $f(x) \to +\infty$ from both sides. At $x = 0$, $f(0) = \frac{1}{\sqrt[3]{1}} = 1$, and at $x = 3$, $f(3) = \frac{1}{\sqrt[3]{4}}$: the endpoints are harmless, which is exactly what makes this integral dangerous. Split at $1$: $\int_0^3 f(x)\,dx = \lim_{t\to1^-}\int_0^t f(x)\,dx + \lim_{s\to1^+}\int_s^3 f(x)\,dx$, two INDEPENDENT limits, with two different letters.

b) An antiderivative: $F(x) = 3(x - 1)^{1/3}$, since $F'(x) = 3 \cdot \frac{1}{3}(x - 1)^{-2/3} = f(x)$, and the cube root is defined for negative numbers. Left piece: $\lim_{t\to1^-}\left[3(x - 1)^{1/3}\right]_0^t = \lim_{t\to1^-}\left(3(t - 1)^{1/3} - 3(-1)^{1/3}\right) = 0 - 3(-1) = 3$. Right piece: $\lim_{s\to1^+}\left[3(x - 1)^{1/3}\right]_s^3 = 3\sqrt[3]{2} - 0 = 3\sqrt[3]{2}$. Both converge, so $\int_0^3 \frac{dx}{(x - 1)^{2/3}} = 3 + 3\sqrt[3]{2}$, about $3 + 3.78 = 6.78$ since $\sqrt[3]{2} \approx 1.26$. Each piece converges because near $1$ the integrand behaves like $\frac{1}{u^{2/3}}$ with $u = \lvert x - 1\rvert$, and $p = \frac{2}{3} < 1$ is the convergent case of Exercise 3 b). The infinitely tall spike of the figure has finite area.

c) The error is the first equals sign: $\frac{1}{x - 1}$ is not continuous at $x = 1$, which lies inside $[0, 3]$, so the FTC does not apply and the value $\ln 2$ means nothing. Split at $1$ and look at the right piece: $\int_1^3 \frac{dx}{x - 1} = \lim_{s\to1^+}\left[\ln(x - 1)\right]_s^3 = \lim_{s\to1^+}\left(\ln 2 - \ln(s - 1)\right) = +\infty$, since $\ln(s - 1) \to -\infty$. One piece diverges, so $\int_0^3 \frac{dx}{x - 1}$ DIVERGES; the other piece need not even be computed. It diverges too, to $-\infty$, and that is the tempting mistake: $+\infty$ on the right and $-\infty$ on the left do NOT cancel into a finite number. $\infty - \infty$ is not a value, and the shortcut silently computed exactly that.

d) Shortcut: $\int_0^{\pi}\sec^2 x\,dx = \left[\tan x\right]_0^{\pi} = \tan\pi - \tan 0 = 0 - 0 = 0$. Impossible: $\sec^2 x = \frac{1}{\cos^2 x} \ge 1$ wherever it is defined, so if the integral existed it would be at least $\int_0^\pi 1\,dx = \pi$. A positive integrand with a zero integral is the alarm bell. The cause: $\cos\frac{\pi}{2} = 0$, so $\sec^2 x$ has a vertical asymptote at $\frac{\pi}{2}$, inside $[0, \pi]$. Left piece: $\int_0^{\pi/2}\sec^2 x\,dx = \lim_{t\to\pi/2^-}\left(\tan t - 0\right) = +\infty$. DIVERGES, so the whole integral diverges. The antiderivative $\tan x$ is itself discontinuous at $\frac{\pi}{2}$: an antiderivative that jumps is another sign that the FTC was used where it does not hold.

e) $x^{-1/3}$ is unbounded at $x = 0$, inside $[-1, 2]$. Antiderivative $F(x) = \frac{3}{2}x^{2/3}$, where $x^{2/3} = \sqrt[3]{x^2}$ is defined and continuous for all $x$: $F'(x) = x^{-1/3}$ for $x \ne 0$. Left piece: $\lim_{t\to0^-}\left(\frac{3}{2}t^{2/3} - \frac{3}{2}(-1)^{2/3}\right) = 0 - \frac{3}{2} = -\frac{3}{2}$, negative as it must be, since $\sqrt[3]{x} < 0$ on $(-1, 0)$. Right piece: $\lim_{s\to0^+}\left(\frac{3}{2}2^{2/3} - \frac{3}{2}s^{2/3}\right) = \frac{3}{2}\sqrt[3]{4}$. Both converge ($p = \frac{1}{3} < 1$), so $\int_{-1}^{2}\frac{dx}{\sqrt[3]{x}} = \frac{3}{2}\left(\sqrt[3]{4} - 1\right)$, about $\frac{3}{2}(1.59 - 1) \approx 0.88$. The shortcut $F(2) - F(-1)$ gives the same number, but only BECAUSE both pieces turned out to converge, which the shortcut never checked. On a copy it scores the value and loses the method: the marks are for the split and the two limits, since the same shortcut produced nonsense in c) and d).

a) Complete the square: $x^2 + 2x + 5 = (x + 1)^2 + 4$. With $u = x + 1$, $\int\frac{dx}{(x + 1)^2 + 4} = \frac{1}{2}\arctan\frac{x + 1}{2} + C$; check: $\frac{d}{dx}\frac{1}{2}\arctan\frac{x + 1}{2} = \frac{1}{2} \cdot \frac{1/2}{1 + (x + 1)^2/4} = \frac{1}{(x + 1)^2 + 4}$. Split at $c = -1$, where the arctangent is $0$. Right half: $\lim_{s\to\infty}\frac{1}{2}\arctan\frac{s + 1}{2} - 0 = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$. Left half: $0 - \lim_{t\to-\infty}\frac{1}{2}\arctan\frac{t + 1}{2} = -\frac{1}{2}\left(-\frac{\pi}{2}\right) = \frac{\pi}{4}$. Both converge: the integral is $\frac{\pi}{2}$. Splitting at $c = 0$ instead gives $\frac{1}{2}\left(\frac{\pi}{2} - \arctan\frac{1}{2}\right) + \frac{1}{2}\left(\arctan\frac{1}{2} + \frac{\pi}{2}\right)$, the same $\frac{\pi}{2}$: $c$ only moves an $\arctan\frac{1}{2}$ from one half to the other. Choose the $c$ that makes the arithmetic clean.

b) With $u = 1 + x^2$, $\int\frac{x}{1 + x^2}\,dx = \frac{1}{2}\ln(1 + x^2) + C$. Symmetric limit: $\int_{-t}^{t}\frac{x\,dx}{1 + x^2} = \frac{1}{2}\ln(1 + t^2) - \frac{1}{2}\ln(1 + t^2) = 0$ for every $t$, so the limit is $0$. But the definition asks for each half on its own: $\int_0^\infty \frac{x\,dx}{1 + x^2} = \lim_{s\to\infty}\frac{1}{2}\ln(1 + s^2) = +\infty$. One half diverges, so $\int_{-\infty}^{\infty}\frac{x}{1 + x^2}\,dx$ DIVERGES. On the figure, the positive half and the negative half each have infinite area (the integrand behaves like $\frac{1}{x}$ at infinity, the $p = 1$ case); the symmetric limit only says that the two infinities were grown at the same speed.

c) $\int_{-t}^{2t}\frac{x\,dx}{1 + x^2} = \frac{1}{2}\ln(1 + 4t^2) - \frac{1}{2}\ln(1 + t^2) = \frac{1}{2}\ln\frac{1 + 4t^2}{1 + t^2}$. As $t \to \infty$, $\frac{1 + 4t^2}{1 + t^2} \to 4$, so the limit is $\frac{1}{2}\ln 4 = \ln 2$. Letting the bounds go to infinity in another proportion gives another answer; with $-t$ and $kt$ one gets $\ln k$, so ANY real number can be produced. That is why the definition insists on two independent limits: an integral whose value depends on how the bounds move has no value at all, and the $0$ of b) was an artefact of symmetry.

d) With $u = x^2$, $\int x e^{-x^2}\,dx = -\frac{1}{2}e^{-x^2} + C$. Right half: $\lim_{s\to\infty}\left(-\frac{1}{2}e^{-s^2} + \frac{1}{2}\right) = \frac{1}{2}$. Left half: $\lim_{t\to-\infty}\left(-\frac{1}{2} + \frac{1}{2}e^{-t^2}\right) = -\frac{1}{2}$. Both halves CONVERGE, so the integral converges and equals $\frac{1}{2} - \frac{1}{2} = 0$. The integrand is odd, like in b), and the answer is $0$, but here the symmetry argument is legitimate because it comes AFTER checking that one half converges. The rule: odd integrand on a symmetric infinite interval gives $0$ only once convergence of one half is proved.

e) $u = e^x$, $du = e^x\,dx$, and $e^{2x} = u^2$, so $\frac{e^x\,dx}{1 + e^{2x}} = \frac{du}{1 + u^2}$. The bounds: as $x \to -\infty$, $u = e^x \to 0^+$; as $x \to +\infty$, $u \to +\infty$. The integral becomes $\int_0^\infty\frac{du}{1 + u^2} = \lim_{s\to\infty}\arctan s - \arctan 0 = \frac{\pi}{2}$. Converges. Two infinite bounds became one finite bound and one infinite bound: the lower one is not improper at all in $u$, because $\frac{1}{1 + u^2}$ is continuous at $0$. Writing $u$ from $-\infty$ to $\infty$ by habit would give $\pi$, twice the answer.