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# 3.1: Improper Integrals

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An improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or $$\infty$$ or $$-\infty$$ or, in some cases, as both endpoints approach limits. Such an integral is often written symbolically just like a standard definite integral, in some cases with infinity as a limit of integration.

Definition: Improper Integrals

If $$f(x)$$ is continuous on $$\left(a,b\right]$$ and not continuous at $$x = a$$, then we define the improper integral:

$\int _a^b f(x)\, dx = \lim_{m \rightarrow a^+} \int_m^b f(x)\, dx$

Example 1

\begin{align} \int_0^1\dfrac{1}{\sqrt{1-x^2}}dx &= \lim_{m\to{1^-}} \int_0^m \dfrac{1}{\sqrt{1-x^2}}dx \\ &= \left( \lim_{m\to1^-} \sin^{-1}x\right]_0^m \\ &= \lim_{m{\to}1^-} (\sin^{-1}m-\sin^{-1}0) \\&=\dfrac{\pi}{2}. \end{align}

## Improper Integrals Involving Infinity

We define the improper integral with an infinity as a one of the limits.

Definition: Improper Integrals Involving Infinity

If $$f(x)$$ is continuous on $$\left(a,b\right]$$ and not continuous at $$x = a$$, then we define

$\int _a^\infty f(x)\, dx = \lim_{m \rightarrow \infty} \int_a^m f(x)\, dx.$

Example 2

\begin{align} \int_{1}^{\infty} \dfrac{1}{x^2} dx &= \lim_{m\to\infty} \int_{1}^{m} \dfrac{1}{x^2} dx \\ &= \lim_{m\to\infty}\left( -\dfrac{1}{x}\right]_1^m \\ &= \lim_{m\to\infty}\big[-\dfrac{1}{m}+1\big] \\ &=1. \end{align}

Exercises

1. $\int_{0}^{\infty}xe^{-x} dx$

2. $\int_{-\infty}^{\infty} \dfrac{1}{1+x^2} dx$

3. $\int_{0}^{\infty} \sin x dx$

4. Determine for which values of $$p$$ the integral $$\int _1^m \dfrac{1}{x^p}dx$$ is defined.

5. Use the formula for arc length to show that the circumference of the semi-circle, $$y=\sqrt{r^2-x^2}$$ is $$\pi \, r$$.

## Vertical Asymptotes

Definition

Let $$f(x)$$ have a vertical asymptote at $$x = c$$. Then

1. $\int_{a}^{b} f(x) dx = \lim_{m\to{c^-}}\int_{a}^{m} f(x) dx$

2. $\int_{c}^{a} f(x) dx =\lim_{m\to{c^+}}\int_{m}^{a} f(x) dx$

3. $\int_{a}^{b} f(x) dx = \lim_{m\to{c^-}} \int_{a}^{m} f(x)dx = \lim_{M\to{c^+}} \int_{M}^{b}f(x) dx.$

Example 3

Evaluate

$\int _0^1 \dfrac{1}{\sqrt{x}} \, dx.$

Solution

\begin{align} \int_{0}^{1}\dfrac{1}{\sqrt{x}} dx &= \lim_{m\to{0^-}}\int_{m}^{1}\dfrac{1}{\sqrt{x}} dx \\ &= \lim_{m\to{0^-}}\left(2\sqrt{x}\right]_m^1 \\ &=\lim_{m\to{0^-}}(2-2\sqrt{m}) \\ &= 2. \end{align}

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.