Problem 1 Compute the following integrals: (a) \int_{0}^{\infty}e^{-\sqrt{x}}dx; (b) \int_{0}^{1}x\ln x dx; (c) \int_{0}^{\infty}e^{-3x}dx; (d) \int_{0}^{\infty}xe^{-x^2}dx; (e) \int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}; (f) \int_{1}^{e}\frac{dx}{x\sqrt{\ln x}}.
Solution (a) (c) (d) are improper integrals of type I, whereas (b) (e) (f) are of type II. The antiderivatives respectively are -2 e^{-\sqrt(x)} (\sqrt(x) + 1), \frac{1}{4}x^2(2\ln x - 1), -\frac{1}{3}e^{-3x}, -\frac{1}{2}e^{-x^2}, \sin^{-1}x, 2\sqrt{\ln x}. By the definition of improper integrals, the answers are 2, -1/4, 1/3, 1/2, \pi/2, 2.
Comparison test Let f, g be functions on [a,b) and 0 \le f(x) \le g(x) for all x\in[a,b). If \int_a^b g(x) dx converges, then \int_a^b f(x) dx converges.
Absolute convergence If \int_a^b |f(x)|dx converges, then \int_a^b f(x) dx converges.
Comparison test for absolute convergence Let f, g be functions on [a,b) and |f(x)| \le g(x) for all x\in[a,b). If \int_a^b g(x) dx converges, then \int_a^b f(x) dx converges.
Limit comparison test Let f, g be functions on [a,b), f(x) \ge 0 and \lim_{x\to b^-} f(x)/g(x) exists and is nonzero. Then \int_a^b f(x) dx converges if and only if \int_a^b f(x) dx converges.
Problem 2 For each of the following integrals decide whether it converges or not. (a) \int_{0}^{\infty} e^{\sin x}dx; (b) \int_{1}^{\infty} \sin x e^{-x}dx; (c) \int_{0}^{1}\frac{1}{e^x+e^{-x}-2}dx; (d) \int_{1}^{\infty}(e^{1/x}-1)\sin x \sin 1/x dx.
Solution (sketch) (a) Because e^{\sin x} \ge 1/e and \int_0^\infty 1/e dx diverges, by the comparison test, the improper integral diverges. (b) Because |\sin x e^{-x}| \le e^{-x} and \int_1^\infty e^{-x} converges, by the comparison test for absolute convergence, the improper integral converges. (c) Because \lim_{x\to 0^+} \frac{1}{e^x + e^{-x} - 2} / \frac{1}{x^2} = 1 (why?) and \int_1^\infty 1/x^2 dx converges, by the limit comparison test, the improper integral converges. (d) Because |(e^{1/x}-1)\sin x \sin 1/x| \le (e^{1/x}-1)\sin 1/x, by the comparison test for absolute convergence, the convergence of \int_1^\infty (e^{1/x}-1)\sin 1/x dx would imply the convergence of the improper integral. Because \lim_{x\to\infty} \left((e^{1/x}-1)\sin 1/x\right) / \frac{1}{x^2} = 1 (why?) and \int_1^\infty 1/x^2 dx converges, by the limit comparison test, the improper integral \int_1^\infty (e^{1/x}-1)\sin 1/x dx converges.
Remark In problem 2(c), when x ~ 0, since e^x ~ 1 + x + x^2 / 2, e^{-x} ~ 1 - x + x^2 / 2, e^x + e^{-x} - 2 ~ x^2. This motivates one to compare \frac{1}{e^x + e^{-x} - 2} with 1/x^2. In problem 2(d), when x~\infty, we know that e^{1/x} ~ 1 + 1/x and \sin(1/x) ~ 1/x. This motivates one to compare (e^{1/x}-1)\sin(1/x) with 1/x^2.
Problem 3 Suppose \lim_{x \rightarrow \infty}f(x) exists and suppose that \int_{0}^{\infty}f(x)dx converges. Show that \lim_{x \rightarrow \infty}f(x)=0.
Solution Assume for the sake of contradiction that L = \lim_{x\to \infty} f(x) is nonzero. Because \lim_{x\to\infty} |L|/f(x) exists and is nonzero, and \int_0^\infty |L|dx diverges, by the limit comparison test, \int_0^\infty f(x) dx diverges. This yields a contradiction.
Fact 1 \int_0^1 1/x^k converges if and only if k < 1.
Fact 2 \int_1^\infty 1/x^k converges if and only if k > 1.
Problem 4 For which values of k does the integral converge? (a) \int_{0}^{\pi/2}\frac{dx}{\sin^k x}; (b) \int_{2}^{\infty}\frac{1}{x\ln^k x}dx; (c) \int_{2}^{\infty}\frac{x^{1/x}-1}{\ln^k x}dx.
Solution (a) Because \lim_{x\to 0} \frac{1}{\sin^k x} / \frac{1}{x^k} = 1, by the limit comparison test and Fact 1, \int_0^{\pi/2} 1/\sin^k x dx converges if and only if \int_0^{\pi / 2} 1/x^k dx converges if and only if k < 1. (b) Under the substitution u = \ln x, \int_2^\infty \frac{1}{x\ln^k x}dx = \int_{\ln 2}^\infty 1/u^k du, which converges if and only if k > 1 by Fact 2. (c) Because \lim_{x\to\infty} \frac{x^{1/x} - 1}{\ln^k x} / \frac{1}{x\ln^{k-1} x} = 1 (why?), by the limit comparison test, the improper integral converges if and only if \int_2^\infty \frac{1}{x\ln^{k-1}x} converges if and only if k - 1 > 1 by part (b).
Remark In problem 4(c), when x ~ \infty, x^{1/x} = e^{\ln x / x} ~ 1 + \ln x / x. This motivates one to compare \frac{x^{1/x}-1}{\ln^k x} with \frac{1}{x\ln^{k-1}x}.
Problem 5 Decide whether the following integral converges or not. (a) \int_{0}^{1/2}\frac{1}{x^x-1-\sin (x \ln x)}dx; (b) \int_{1}^{\infty}\left(x^\frac{1}{x (\ln x)^3}-1\right)\sin xdx.
Idea (a) When x ~ 0, x^x = e^{x\ln x} ~ 1 + x\ln x + \frac{1}{2}x^2\ln^2 x and \sin(x\ln x) ~ x\ln x. This motivates one to compare \frac{1}{x^x - 1 - \sin(x\ln x)} with \frac{1}{x^2\ln^2 x}. (b) When x ~ \infty, x^{\frac{1}{x \ln^3 x}} = e^{\frac{1}{x \ln^2 x}} ~ \frac{1}{x\ln^2 x}. This motivates one to compare x^\frac{1}{x (\ln x)^3}-1 with \frac{1}{x\ln^2 x}.
Solution (a) Because \lim_{x\to 0} \frac{1}{x^x-1\sin(x\ln x)}/\frac{1}{x^2\ln^2 x} = \frac{1}{2} (why?), by the limit comparison test, the improper integral converges if \int_0^{1/2} \frac{1}{x^2 \ln^2x} dx converges. Under the substitution, u=\ln x, \int_0^{1/2} \frac{1}{x^2 \ln^2x} dx = \int_{-\infty}^{\ln 1/2} \frac{1}{e^u u^2}du. Note that \lim_{x\to-\infty}\frac{1}{e^u u^2} = \infty. By problem 3, \int_{-\infty}^{\ln 1/2} \frac{1}{e^u u^2}du does not converge. (b) Because |\left(x^\frac{1}{x (\ln x)^3}-1\right)\sin x| \le x^\frac{1}{x (\ln x)^3}-1, by the comparison test for absolute convergence, the convergence of \int_1^\infty x^\frac{1}{x (\ln x)^3}-1 dx would imply the convergence of the improper integral. Because \lim_{x\to\infty} \left(x^\frac{1}{x (\ln x)^3}-1\right) / \frac{1}{x\ln^2 x} = 1 (why?) and \int_1^\infty \frac{1}{x \ln^2 x}dx converges by problem 4(b), by the limit comparison test, \int_1^\infty x^\frac{1}{x (\ln x)^3}-1 dx converges.