# Recitation 12

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.