3.7 Improper integrals (Page 3/6)

Calculus volume 2 Page 3 / 6

Evaluate $\int_{−3}^{+ \infty} e^{− x} d x .$ State whether the improper integral converges or diverges.

$e^{3},$ converges

Integrating a discontinuous integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form $\int_{a}^{b} f (x) d x,$ where $f (x)$ is continuous over $[a, b)$ and discontinuous at $b .$ Since the function $f (x)$ is continuous over $[a, t]$ for all values of $t$ satisfying $a < t < b,$ the integral $\int_{a}^{t} f (x) d x$ is defined for all such values of $t .$ Thus, it makes sense to consider the values of $\int_{a}^{t} f (x) d x$ as $t$ approaches $b$ for $a < t < b .$ That is, we define $\int_{a}^{b} f (x) d x = lim_{t \to b^{-}} \int_{a}^{t} f (x) d x,$ provided this limit exists. [link] illustrates $\int_{a}^{t} f (x) d x$ as areas of regions for values of $t$ approaching $b .$

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis. — As t approaches b from the left, the value of the area from a to t approaches the area from a to b .

We use a similar approach to define $\int_{a}^{b} f (x) d x,$ where $f (x)$ is continuous over $(a, b]$ and discontinuous at $a .$ We now proceed with a formal definition.

Definition

Let $f (x)$ be continuous over $[a, b) .$ Then,

$\int_{a}^{b} f (x) d x = lim_{t \to b^{-}} \int_{a}^{t} f (x) d x .$
Let $f (x)$ be continuous over $(a, b] .$ Then,

$\int_{a}^{b} f (x) d x = lim_{t \to a^{+}} \int_{t}^{b} f (x) d x .$

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
If $f (x)$ is continuous over $[a, b]$ except at a point $c$ in $(a, b),$ then

$\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x,$

provided both $\int_{a}^{c} f (x) d x$ and $\int_{c}^{b} f (x) d x$ converge. If either of these integrals diverges, then $\int_{a}^{b} f (x) d x$ diverges.

The following examples demonstrate the application of this definition.

Integrating a discontinuous integrand

Evaluate $\int_{0}^{4} \frac{1}{\sqrt{4 - x}} d x,$ if possible. State whether the integral converges or diverges.

The function $f (x) = \frac{1}{\sqrt{4 - x}}$ is continuous over $[0, 4)$ and discontinuous at 4. Using [link] from the definition, rewrite $\int_{0}^{4} \frac{1}{\sqrt{4 - x}} d x$ as a limit:

\begin{matrix} \int_{0}^{4} \frac{1}{\sqrt{4 - x}} d x & = lim_{t \to 4^{-}} \int_{0}^{t} \frac{1}{\sqrt{4 - x}} d x & Rewrite as a limit. \\ = lim_{t \to 4^{-}} (−2 \sqrt{4 - x}) |_{\begin{matrix} 0 \end{matrix}}^{\begin{matrix} t \end{matrix}} & Find the antiderivative. \\ = lim_{t \to 4^{-}} (−2 \sqrt{4 - t} + 4) & Evaluate the antiderivative. \\ = 4 . & Evaluate the limit. \end{matrix}

The improper integral converges.

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Integrating a discontinuous integrand

Evaluate $\int_{0}^{2} x ln x d x .$ State whether the integral converges or diverges.

Since $f (x) = x ln x$ is continuous over $(0, 2]$ and is discontinuous at zero, we can rewrite the integral in limit form using [link] :

\begin{matrix} \int_{0}^{2} x ln x d x & = lim_{t \to 0^{+}} \int_{t}^{2} x ln x d x & Rewrite as a limit. \\ = lim_{t \to 0^{+}} (\frac{1}{2} x^{2} ln x - \frac{1}{4} x^{2}) |_{\begin{matrix} t \end{matrix}}^{\begin{matrix} 2 \end{matrix}} & \begin{matrix} Evaluate \int^{​} x ln x d x using integration by parts \\ with u = ln x and d v = x . \end{matrix} \\ = lim_{t \to 0^{+}} (2 ln 2 - 1 - \frac{1}{2} t^{2} ln t + \frac{1}{4} t^{2}) . & Evaluate the antiderivative. \\ = 2 ln 2 - 1 . & \begin{matrix} Evaluate the limit. lim_{t \to 0^{+}} t^{2} ln t is indeterminate. \\ To evaluate it, rewrite as a quotient and apply \\ L’Hôpital’s rule. \end{matrix} \end{matrix}

The improper integral converges.

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Integrating a discontinuous integrand

Evaluate $\int_{−1}^{1} \frac{1}{x^{3}} d x .$ State whether the improper integral converges or diverges.

Since $f (x) = 1 / x^{3}$ is discontinuous at zero, using [link] , we can write

\int_{−1}^{1} \frac{1}{x^{3}} d x = \int_{−1}^{0} \frac{1}{x^{3}} d x + \int_{0}^{1} \frac{1}{x^{3}} d x .

If either of the two integrals diverges, then the original integral diverges. Begin with $\int_{−1}^{0} \frac{1}{x^{3}} d x :$

\begin{matrix} \int_{−1}^{0} \frac{1}{x^{3}} d x & = lim_{t \to 0^{-}} \int_{−1}^{t} \frac{1}{x^{3}} d x & Rewrite as a limit. \\ = lim_{t \to 0^{-}} (- \frac{1}{2 x^{2}}) |_{\begin{matrix} −1 \end{matrix}}^{\begin{matrix} t \end{matrix}} & Find the antiderivative. \\ = lim_{t \to 0^{-}} (- \frac{1}{2 t^{2}} + \frac{1}{2}) & Evaluate the antiderivative. \\ = + \infty . & Evaluate the limit. \end{matrix}

Therefore, $\int_{−1}^{0} \frac{1}{x^{3}} d x$ diverges. Since $\int_{−1}^{0} \frac{1}{x^{3}} d x$ diverges, $\int_{−1}^{1} \frac{1}{x^{3}} d x$ diverges.

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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