Examples

Improper integral - infinite bound

Show that the improper integral ${\displaystyle {\int }_2^{\infty }\frac{dx}{x^3}}$ converges. What is its value?

Solution

(1) Improper integral definition:

$${\int }_2^{\infty }\frac{dx}{x^3}\gg \gg \underset{R\to \infty }{\lim }{\int }_2^R\frac{dx}{x^3}$$

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(2) Replace infinity with a new symbol $R$:

$${\int }_2^R\frac{dx}{x^3}\gg \gg -\frac{1}{2}{x}^{-2}{|}_2^R\gg \gg \frac{1}{8}-\frac{1}{2R^2}$$

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(3) Take limit as $R\to \infty$:

$$\begin{array}{c}\underset{R\to \infty }{\lim }{\int }_2^R\frac{dx}{x^3}\gg \gg \underset{R\to \infty }{\lim }\frac{1}{8}-\frac{1}{2R^2}\gg \gg \frac{1}{8}\\ \\ \gg \gg \text{converges}\end{array}$$

Improper integral - infinite integrand

Show that the improper integral ${\displaystyle {\int }_0^9\frac{dx}{\sqrt{x}}}$ converges. What is its value?

Solution

(1) Improper integral definition:

$${\int }_0^9\frac{dx}{\sqrt{x}}\gg \gg \underset{R\to 0^{+}}{\lim }{\int }_R^9\frac{dx}{\sqrt{x}}$$

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(2) Switch $0$ to $R$ and integrate:

$$\begin{array}{c}{\int }_R^9\frac{dx}{\sqrt{x}}\gg \gg {\int }_R^9{x}^{-1/2}dx\\ \\ \gg \gg 2x^{+1/2}{|}_R^9\gg \gg 6-2\sqrt{R}\end{array}$$

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(3) Take limit as $R\to 0^{+}$:

$$\begin{array}{c}\underset{R\to 0^{+}}{\lim }{\int }_R^9\frac{dx}{\sqrt{x}}\gg \gg \underset{R\to 0^{+}}{\lim }6-2\sqrt{R}\gg \gg 6\\ \\ \gg \gg \text{converges}\end{array}$$

Example - Improper integral - infinity inside the interval

Does the integral ${\displaystyle {\int }_{-1}^{+1}\frac{1}{x}dx}$ converge or diverge?

Solution

WRONG APPROACH:

It is tempting to compute the integral incorrectly, like this:

$${\int }_{-1}^{+1}\frac{1}{x}dx\gg \gg \ln |x|{|}_{-1}^{+1}\gg \gg \ln |2|-\ln |-2|=0$$

But this is wrong! There is an infinite integrand at $x=0$. We must break it into parts!

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RIGHT APPROACH:

(1) Break apart at $x=0$ discontinuity:

$${\int }_{-1}^{+1}\frac{1}{x}dx\gg \gg {\int }_{-1}^0\frac{1}{x}dx+{\int }_0^{+1}\frac{1}{x}dx$$

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(2) Improper integral definition, both terms separately:

$$\gg \gg \underset{R\to 0^{-}}{\lim }{\int }_{-1}^R\frac{1}{x}dx+\underset{R\to 0^{+}}{\lim }{\int }_R^{+1}\frac{1}{x}dx$$

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(3) Integrate:

$$\begin{array}{c}{\int }_{-1}^R\frac{1}{x}dx\gg \gg \ln |R|-\ln |-1|\gg \gg \ln |R|\\ \\ {\int }_R^{+1}\frac{1}{x}dx\gg \gg \ln |1|-\ln |R|\gg \gg -\ln R\end{array}$$

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(4) Limits:

$$\begin{array}{c}\underset{R\to 0^{-}}{\lim }\ln |R|=-\infty ,\underset{R\to 0^{+}}{\lim }-\ln R=+\infty \\ \\ \gg \gg \text{diverges}\end{array}$$

Neither limit is finite. For ${\displaystyle {\int }_{-1}^{+1}\frac{1}{x}dx}$ to exist we’d need both of these limits to be finite. So the original integral diverges.

Comparison to p-integrals

Determine whether the integral converges:

(a) ${\displaystyle {\int }_2^{\infty }\frac{x^3}{x^4-1}dx}$

(b) ${\displaystyle {\int }_1^{\infty }\frac{1}{x^2+x+1}dx}$

Solution

(a) (1) Observe large $x$ tendency:

Consider large values. Notice the integrand tends toward $1/x$ for large $x$.

$$\frac{x^3}{x^4-1}⟶\frac{x^3}{x^4}\text{for}x\to \infty ,\text{and}\frac{x^3}{x^4}=\frac{1}{x}$$

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(2) Try comparison to $1/x$:

$$\frac{x^3}{x^4-1}\stackrel{?}{>}\frac{1}{x}$$

Validate. Notice $x^4-1>0$ and $x>0$ when $x\ge 2$.

$$\begin{array}{c}\frac{x^3}{x^4-1}\stackrel{?}{>}\frac{1}{x}\\ \\ \gg \gg x^3\cdot x\stackrel{?}{>}1\cdot (x^4-1)\gg \gg x^4\check{>}{x}^4-1\end{array}$$

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(3) Apply comparison test:

We know:

$$\frac{x^3}{x^4-1}>\frac{1}{x},{\int }_2^{\infty }\frac{1}{x}dx\text{diverges}$$

We conclude:

$${\int }_2^{\infty }\frac{x^3}{x^4-1}dx\text{diverges}$$

(b) (1) Observe large $x$ tendency:

Consider large values. Notice the integrand tends toward $1/x^2$ for large $x$.

$$\frac{1}{x^2+x+1}⟶\frac{1}{x^2}\text{for}x\to \infty$$

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(2) Try comparison to $1/x^2$:

$$\frac{1}{x^2+x+1}\stackrel{?}{<}\frac{1}{x^2}$$

Validate. Notice $x^2+x+1>0$ and $x^2>0$ when $x\ge 1$.

$$\begin{array}{c}\frac{1}{x^2+x+1}\stackrel{?}{<}\frac{1}{x^2}\\ \\ \gg \gg 1\cdot x^2\stackrel{?}{<}1\cdot (x^2+x+1)\gg \gg x^2\check{<}{x}^2+x+1\end{array}$$

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(3) Apply comparison test:

We know:

$$\frac{1}{x^2+x+1}<\frac{1}{x^2},{\int }_1^{\infty }\frac{1}{x^2}dx\text{converges}$$

We conclude:

$${\int }_1^{\infty }\frac{1}{x^2+x+1}dx\text{converges}$$