- If \(f\) is continuous on \([a, \infty)\), then
$$
\int_a^{\infty} f(x) \; dx = \lim_{b \rightarrow \infty} \int_a^b f(x) \; dx
$$
provided that the limit exists. - If \(f\) is continuous on \((-\infty, b]\), then
$$
\int_{-\infty}^b f(x) \; dx = \lim_{a \rightarrow -\infty} \int_a^b f(x) \; dx
$$
provided that the limit exists. - If \(f\) is continuous on \((-\infty, \infty)\), then
$$
\int_{-\infty}^{\infty} f(x) \; dx = \int_{-\infty}^c f(x) \; dx + \int_c^{\infty} f(x) \; dx
$$
where \(c\) is any real number and provided that both improper integrals on the right-hand side converge.
In the first two cases, if the limit on the right-hand side exists, then we say the improper integral on the left-hand side converges and diverges otherwise. In the last case, if the both integrands on the right-hand side converge then the improper integral on the left-hand side converges and diverges otherwise.
-
If \(f(x)\) has an essential discontinuity at \(x = a\), but continuous otherwise on \([a,b]\), then
$$
\int_a^b f(x) \; dx = \lim_{t \rightarrow a^+} \int_t^b f(x) \; dx
$$
provided that the limit exists. -
If \(f(x)\) has an essential discontinuity at \(x = b\), but continuous otherwise on \([a,b]\), then
$$
\int_a^b f(x) \; dx = \lim_{t \rightarrow b^-} \int_a^t f(x) \; dx
$$
provided that the limit exists. -
If \(f(x)\) has an essential discontinuity at \(x = c\) in \((a,b)\), but continuous otherwise on \([a,b]\), then
$$
\int_a^b f(x) \; dx = \int_a^c f(x) \; dx + \int_c^b f(x) \; dx
$$
provided that both improper integrals on the right-hand side exists.
We extend the terminology of convergence and divergence from the case when the improper integral arises from one of the limits of integration being infinite. So, in particular, the improper integral in the last case converges provided that both integrals on the right
$$
\int_0^\infty \frac{x}{(x^2+2)^2} \; dx
$$
By definition,
$$
\begin{equation}
\label{eq:1.1}
\int_0^\infty \frac{x}{(x^2+2)^2} \; dx = \lim_{b \rightarrow \infty} \int_0^b \frac{x}{(x^2+2)^2} \; dx.
\end{equation}
$$
By \(u\)-substitution, with you \(u = x^2+2\), we obtain
$$
\begin{equation}
\label{eq:1.2}
\int \frac{x}{(x^2+2)^2} \; dx = \int \frac{1}{2 u^2} \; dx = \frac{-1}{2u} + C = \frac{-1}{2(x^2+2)}+C.
\end{equation}
$$
Using \(\eqref{eq:1.2}\), we obtain
$$
\begin{equation}
\label{eq:1.3}
\int_0^b \frac{x}{(x^2+2)^2} \; dx = \frac{-1}{2(x^2+2)} \Big|_0^b = \frac{-1}{2(b^2+2)} +\frac{1}{4}.
\end{equation}
$$
Substituting \(\eqref{eq:1.3}\) into \(\eqref{eq:1.1}\), we obtain
$$
\begin{equation}
\int_0^\infty \frac{x}{(x^2+2)^2} \; dx = \lim_{b \rightarrow \infty} \left(\frac{-1}{2(b^2+2)} +\frac{1}{4} \right) = \frac{1}{4}.
\end{equation}
$$
So the integral converges to 1/4.
- \(y = \frac{1}{x^2}, x = 1, x = 2,\) and the \(x-axis\).
- \(y = \frac{1}{x^2}, x = 1, x = 3,\) and the \(x-axis\).
- \(y = \frac{1}{x^2}, x = 1, x = 10,\) and the \(x-axis\).
\begin{array}{c|c}
b & \displaystyle \int_1^b \frac{1}{x^2}\; dx \\ \hline
2 & \frac{1}{2} \\
3 & \frac{2}{3} \\
\vdots & \vdots \\
10 & \frac{9}{10}
\end{array}
$$
- The area of the region is given by
$$
\begin{aligned}
A & = \int_1^2 \frac{1}{x^2} \; dx \\
& = – \frac{1}{x}\Big|_1^2 \\
& = – \frac{1}{2} + 1 = \frac{1}{2}
\end{aligned}
$$ - The area of the region is given by
$$
\begin{aligned}
A & = \int_1^3 \frac{1}{x^3} \; dx \\
& = – \frac{1}{x}\Big|_1^3 \\
& = – \frac{1}{3} + 1 = \frac{2}{3}
\end{aligned}
$$ - The area of the region is given by
$$
\begin{aligned}
A & = \int_1^{10} \frac{1}{x^3} \; dx \\
& = – \frac{1}{x}\Big|_1^{10} \\
& = – \frac{1}{10} + 1 = \frac{9}{10}
\end{aligned}
$$
$$
y = \frac{1}{x^2}
$$
for \(x \geq 1\).
$$
\begin{aligned}
A & = \int_1^{\infty} \frac{1}{x^2} \; dx \\
& = \lim_{b \rightarrow \infty}\int_1^b \frac{1}{x^2} \; dx \\
& = \lim_{b \rightarrow \infty}\left(- \frac{1}{x}\Big|_1^b\right) \\
& = \lim_{b \rightarrow \infty} \left(- \frac{1}{b} + 1 \right) = 1
\end{aligned}
$$
$$
y = \frac{1}{x}
$$
for \(x \geq 1\).
$$
\begin{aligned}
A & = \int_1^{\infty} \frac{1}{x} \; dx \\
& = \lim_{b \rightarrow \infty}\int_1^b \ln|x| \; dx \\
& = \lim_{b \rightarrow \infty}\left(\ln x \Big|_1^b\right) \\
& = \lim_{b \rightarrow \infty} \left(\ln b + \ln 1 \right) = \infty
\end{aligned}
$$
So the area under the curve is infinite.
$$
\int_1^{\infty} \frac{1}{x^p} \; dx
$$
converges and its value.
$$
\begin{aligned}
\int_1^{\infty} \frac{1}{x} \; dx
& = \lim_{b \rightarrow \infty}\int_1^b \frac{1}{x^p} \; dx \\
& = \lim_{b \rightarrow \infty} \left(\frac{x^{1-p}}{1-p}\Big|_1^b \right) \\
& = \lim_{b \rightarrow \infty} \left(\frac{b^{1-p}}{1-p}- \frac{1}{1-p} \right) \\
& = \begin{cases} \frac{-1}{1-p} & 1-p<0 \\ \text{divergent} & 1-p \geq 0 \end{cases}\\
& = \begin{cases} \frac{-1}{1-p} & 1 < p \\ \text{divergent} & 1 \geq p \end{cases}
\end{aligned}
$$
So,
$$
\int_1^{\infty} \frac{1}{x^p} \; dx = \begin{cases} \frac{-1}{1-p} & 1 < p \\ \text{divergent} & 1 \geq p \end{cases}
$$