# 8.8 | Improper Integrals

• 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 strong and diverges otherwise.

Find the area bounded by the graphs of

1. $$y = \frac{1}{x^2}, x = 1, x = 2,$$ and the $$x-axis$$.
2. $$y = \frac{1}{x^2}, x = 1, x = 3,$$ and the $$x-axis$$.
3. $$y = \frac{1}{x^2}, x = 1, x = 10,$$ and the $$x-axis$$.

$$\begin{array}{c|c} b & \displaystyle \int_1^b \frac{1}{x^2}\; dx \\ \hline 2 & \frac{1}{2} \\ 3 & \frac{2}{3} \\ \vdots & \vdots \\ 10 & \frac{9}{10} \end{array}$$
1. 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}
2. 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}
3. 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}
Find the area of the region under the curve
$$y = \frac{1}{x^2}$$
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}

Find the area under the curve
$$y = \frac{1}{x}$$
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.
Determine all values of $$p$$ for which the improper integral
$$\int_1^{\infty} \frac{1}{x^p} \; dx$$
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, $$\int_1^{\infty} \frac{1}{x^p} \; dx = \begin{cases} \frac{-1}{1-p} & 1 < p \\ \text{divergent} & 1 \geq p \end{cases}$$
Let $$f$$ and $$g$$ be continuous and suppose that $$f(x) \geq g(x) \geq 0$$ for all $$x \geq 0;$$ that is, $$f$$ dominates $$g$$ on $$[a, \infty).$$

1. If $$\displaystyle \int_a^ \infty f(x) \; dx$$ is convergent, then so is $$\displaystyle \int_a^ \infty g(x) \; dx$$.
2. If $$\displaystyle \int_a^ \infty g(x) \; dx$$ is divergent, then so is $$\displaystyle \int_a^ \infty f(x) \; dx$$.
Determine if the improper integral converges or diverges. If it converges, find its value.
$$\int_1^\infty 8x^2 e^{-x^3} \; dx$$

$$\int 8x^2 e^{-x^3} \; dx = -\frac{8}{3} e^{-x^3} + C$$
\begin{aligned} \int_1^b 8x^2 e^{-x^3} \; dx &= -\frac{8}{3} e^{-x^3}\Big|_1^b \\ & = -\frac{8}{3} e^{-b^3} +\frac{8}{3}e^{-1} \end{aligned}
\begin{aligned} \int_1^\infty 8x^2 e^{-x^3} \; dx &= \lim_{b \rightarrow \infty} \int_1^b 8x^2 e^{-x^3} \; dx \\ & = \lim_{b \rightarrow \infty} \left( -\frac{8}{3e^{b^3}} +\frac{8}{3e} e \right) \\ & = \frac{8}{3e} \end{aligned}
Thus, the improper integral converges to $$8/(3e)$$.