Type 1: Infinite Intervals
- If \(f\) is continuous on \([a,\infty)\), then \(\int_{a}^{\infty}f(x)dx=\lim_{t\to\infty}\int_{a}^{t}f(x)dx.\)
- If \(f\) is continuous on \((-\infty,a]\), then \(\int_{-\infty}^{a}f(x)dx=\lim_{t\to-\infty}\int_{t}^{a}f(x)dx.\)
- If \(f\) is continuous on \((-\infty,\infty)\), then break up the integral at a convenient value of \(a\), i.e., write \(\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx\), and then use (1) and (2)
Type II. Discontinuous Functions
- If \(f\) is continuous on \([a, b)\) and discontinuous at \(b\), then \(\int_{a}^{b}f(x)dx=\lim_{t\rightarrow b^{-}}\int_{a}^{t}f(x)dx.\)
- If \(f\) is continuous on \((a, b]\) and discontinuous at \(a\), then \(\int_{a}^{b}f(x)dx=\lim_{t\rightarrow a^{+}}\int_{t}^{b}f(x)dx.\)
- If f is continuous on \([a, b]\), except at a number \(c\) in \((a,b)\), then break up the integral at \(c\), i.e., write \(\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx\), and then use (1) and (2)
The improper integral is said to:
- CONVERGE if the limit in (1) and (2) exists, or if both limits in (3) exist.
- DIVERGE if the limit in (1) or (2) does not exist, or if at least one of the limits in (3) does not exist.
Remember:
The improper integral \(\int_1^\infty\frac1{x^p}dx\), where \(p\) is a real number converges when \(p>1\), and diverges when \(p\leq1\).
Example:
Determine whether the following integrals converge or diverge.
\(\mathrm{(a)}\quad\int_2^\infty e^{3x}dx\)
\(\mathrm{(b)}\quad\int_{-\infty}^\infty\frac{1}{x^2+1}dx\)
\(\mathrm{(c)}\quad\int_0^1x\ln xdx\)
\(\begin{aligned} \mathrm{(a)} \\ & \int_2^\infty e^{3x}dx=\lim_{t\to\infty}\int_2^te^{3x}dx=\lim_{t\to\infty}\left.\left(\frac{e^{3x}}{3}\right)\right|_2^t \\ & =\lim_{t\to\infty}\left[\left(\frac{e^{3t}}{3}\right)-\left(\frac{e^6}{3}\right)\right]=+\infty \\ & \mathrm{Thus,~}\int_2^\infty e^{3x}dx\mathrm{~diverges.} \end{aligned}\)
\(\begin{aligned} \mathrm{(b)} \\ & \int_{-\infty}^\infty\frac{1}{x^2+1}dx=\int_{-\infty}^0\frac{1}{x^2+1}dx+\int_0^\infty\frac{1}{x^2+1}dx \\ & =\lim_{t\to-\infty}\int_t^0\frac{1}{x^2+1}dx+\lim_{s\to\infty}\int_0^s\frac{1}{x^2+1}dx \\ & =\lim_{t\to-\infty}\left(\arctan(x)\right)|_t^0+\lim_{s\to\infty}\left(\arctan(x)\right)|_0^s \\ & =\lim_{t\to-\infty}\left(\arctan(0)-\arctan(t)\right)+\lim_{s\to\infty}\left(\arctan(s)-\arctan(0)\right) \\ & =\left(0+\frac{\pi}{2}\right)+\left(\frac{\pi}{2}-0\right)=\pi \end{aligned}\)
Thus, \(\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx\) converges.
Note that arctan(x) tends to \(\pi/2\) as \(x\) approaches \(+\infty\) and to \(-\pi/2\) as \(x\) approaches \(-\infty\).
(c)
Compute the indefinite integral using integration by parts where \(u=\ln x\mathrm{~and~}d\nu=xdx.\)
\(\int x\ln xdx=\frac{x^2\ln x}{2}-\int\left(\frac{x^2}{2}\right)\left(\frac{1}{x}dx\right)=\frac{x^2\ln x}{2}-\frac{1}{2}\int xdx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C\)
Now, solve the improper integral.
\(\begin{aligned} & \int_0^1x\ln xdx=\lim_{t\to0^+}\int_t^1x\ln xdx \\ & =\lim_{t\to0^+}\left.\left(\frac{x^2\ln x}{2}-\frac{x^2}{4}\right)\right|_t^1 \\ & =\lim_{t\to0^+}\left(\frac{(1)^2\ln(1)}{2}-\frac{(1)^2}{4}\right)-\left(\frac{t^2\ln t}{2}-\frac{t^2}{4}\right) \\ & =\left(0-\frac{1}{4}\right)-\left(0-0\right)=-\frac{1}{4} \\ & \mathrm{Thus,~}\int_0^1x\ln xdx\text{ converges.} \end{aligned}\)
Note that, applying L’Hôpital’s Rule to solve the limit,
\(\lim_{t\to0^+}t^2\ln t=\lim_{t\to0^+}\frac{\ln t}{\frac{1}{t^2}}=\lim_{t\to0^+}\frac{\ln t}{t^{-2}}\overset{\mathrm{H}}{\operatorname*{=}}\lim_{t\to0^+}\frac{\frac{1}{t}}{-2t^{-3}}=\lim_{t\to0^+}\frac{\frac{1}{t}}{\frac{-2}{t^3}}\)
\(=\lim_{t\to0^+}\left(\frac{1}{t}\right)\left(\frac{t^3}{-2}\right)=\lim_{t\to0^+}\left(-\frac{t^2}{2}\right)=0\)