W06 - Notes

Moments and center of mass

Videos

Review Videos

Videos, Math Dr. Bob:

• Moments and CoM 01: Points masses on a line
• Moments and CoM 02: Points masses in the plane
• Moments and CoM 03: Planar lamina of uniform density
• Moments and CoM 04: Integral formula for planar lamina
• Moments and CoM 05: Rod of non-uniform density

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03 Theory

Theory 1

Moment

The moment of a region to an axis is the total (integral) of mass times distance to that axis:

Moment to $x$-axis:

$$\begin{array}{cc}{M}_x& =\int \rho ydA\text{(general formula)}\\ & \\ M_x& ={\int }_c^d\rho y(g_2(y)-g_1(y))dy\text{(region between functions }{g}_2\text{ and }{g}_1\text{)}\end{array}$$

Moment to $y$-axis:

$$\begin{array}{cc}{M}_y& =\int \rho xdA\text{(general formula)}\\ & \\ M_y& ={\int }_a^b\rho x(f_2(x)-f_1(x))dx\text{(region between functions }{f}_2\text{ and }{f}_1\text{)}\end{array}$$

Notice the swap in letters

• $M_y$ integrand has $x$ factor
• $M_x$ integrand has $y$ factor

Notice the total mass

If you remove $x$ or $y$ factors from the integrands, the integrals give total mass $M$.

These formulas are obtained by slicing the region into rectangular strips that are parallel to the axis in question.

The area per strip is then:

• $f(x)dx$ — region under $y=f(x)$
• $(f_2-f_1)dx$ — region between $f_1$ and $f_2$
• $g(y)dy$ — region ‘under’ $x=g(y)$
• $(g_2-g_1)dy$ — region between $g_1$ and $g_2$

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The idea of moment is related to:

• Torque balance and angular inertia
• Center of mass

The center of mass (CoM) of a solid body is a single point with two important properties:

1. $\text{CoM}=$ “average position” of the body
   • The average position determines an effective center of dynamics. For example, gravity acting on every bit of mass of a rigid body acts the same as a force on the CoM alone.
2. $\text{CoM}=$ “balance point” of the body
   • The net torque (rotational force) about the CoM, generated by a force distributed over the body’s mass, equivalently a force on the CoM, is zero.

Centroid

When the body has uniform density, then the CoM is also called the centroid.

---

Center of mass from moments

Coordinates of the CoM:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Here $M$ is the total mass of the body.

Center of mass from moments - explanation

Notice how these formulas work. The total mass is always $M=\int \rho dA$. The moment to $y$ (for example) is $M_y=\int \rho xdA$. Dividing these two values:

$$\bar{x}=\frac{M_y}{m}\gg \gg \frac{\int \rho xdA}{\int \rho dA}\gg \gg \frac{\int xdA}{\int dA}\gg \gg \frac{\int xdA}{A}$$

where $A=\text{total area}$.

In other words, through the formula $\bar{x}=\frac{M_y}{M}$, we find that $\bar{x}$ is the average value of $x$ over the region with area $A$.

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04 Illustration

Example - CoM of a parabolic plate

CoM of a parabolic plate

Find the CoM of the region depicted:

Solution

(1) Compute the total mass:

Area under the curve with density factor $\rho$:

$$m={\int }_0^2\rho x^{2}dx\gg \gg \rho {\frac{x^3}{3}|}_0^2\gg \gg \frac{8\rho }{3}$$

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(2) Compute $M_y$:

Formula:

$$M_y={\int }_a^b\rho xdA$$

Interpret and calculate:

$$\begin{array}{c}{M}_y={\int }_0^2\rho xf(x)dx\gg \gg \rho {\int }_0^2{x}^{3}dx\\ \\ \gg \gg 4\rho =M_y\end{array}$$

---

(3) Compute $M_x$:

Formula:

$$M_x={\int }_c^d\rho ydA$$

Width of horizontal strips between the curves:

$$w(y)=2-\sqrt{y}$$

Interpret $dA$:

$$dA=(2-\sqrt{y})dy$$

Calculate integral:

$$\begin{array}{c}{M}_x={\int }_c^d\rho ydA\gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\\ \\ \gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\gg \gg {\int }_0^4\rho 2ydy-{\int }_0^4\rho y^{3/2}dy\\ \\ \gg \gg \frac{16\rho }{5}=M_x\end{array}$$

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(4) Compute CoM coordinates from moments:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Insert data:

$$\begin{array}{c}\bar{x}=\frac{4\rho }{8\rho /3}\gg \gg \frac{3}{2},\bar{y}=\frac{16\rho /5}{8\rho /3}\gg \gg \frac{6}{5}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{3}{2},\frac{6}{5})\end{array}$$ Link to original

05 Theory

Theory 2

A downside of the technique above is that to find $M_x$ we needed to convert the region into functions in $y$. This would be hard to do if the region was given as the area under a curve $y=f(x)$ but $f^{-1}(y)$ cannot be found analytically. An alternative formula can help in this situation.

Midpoint of strips for opposite variables

When the region lies between $f_1(x)$ and $f_2(x)$, we can find $M_x$ with an $x$-integral:

$$M_x={\int }_a^b\rho \frac{1}{2}(f_2^2-f_1^2)dx\text{(region between }{f}_1\text{ and }{f}_2\text{)}$$

When the region lies between $g_1(y)$ and $g_2(y)$, we can find $M_y$ with a $y$-integral:

$$M_y={\int }_c^d\rho \frac{1}{2}(g_2^2-g_1^2)dy\text{(region between }{g}_1\text{ and }{g}_2\text{)}$$

Region under a curve

For the region “under the curve” $y=f(x)$, just set:

$$f_1=0,f_2=f$$

For the region “under the curve” $x=g(x)$, set:

$$g_1=0,g_2=g$$

The idea for these formulas is to treat each vertical strip as a point concentrated at the CoM of the vertical strip itself.

The height to this midpoint is $\frac{1}{2}f(x)$, and the area of the strip is $f(x)dx$, so the integral becomes $\int \rho \frac{1}{2}f{(x)}^{2}dx$.

Midpoint of strips formula - full explanation

• If the strip is located at some $x$, with $y$ values from $0$ up to $f(x)$, then:

$$\text{CoM of strip}=(x,\frac{1}{2}f(x))$$

• The area of the strip is $dA=f(x)dx$. So the integral formula for $M_x$ can be recast:

$$M_x=\int ydA\gg \gg {\int }_a^b\frac{1}{2}f(x)\cdot f(x)dx\gg \gg {\int }_a^b\frac{1}{2}{f}^{2}dx$$

• If the vertical strips are between $f_1(x)$ and $f_2(x)$, then the midpoints of the strips are given by the ‘average’ function:

$$\frac{1}{2}(f_1(x)+f_2(x))$$

• The height of each strip is $f_2(x)-f_1(x)$, so $dA=(f_2-f_1)dx$.
• Putting this together:

$$\begin{array}{c}{M}_x=\int ydA\gg \gg {\int }_a^b\frac{1}{2}(f_1+f_2)\cdot (f_2-f_1)dx\\ \\ \gg \gg {\int }_a^b\frac{1}{2}(f_2^2-f_1^2)dx\end{array}$$

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06 Illustration

Example - Computing CoM using only vertical strips

Computing CoM using only vertical strips

Find the CoM of the region:

Solution

(1) Compute the total mass $m$:

Area under the curve times density $\rho$:

$${\int }_0^{\pi /2}\rho \cos xdx=\rho \sin x{|}_0^{\pi /2}=\rho$$

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(2) Compute $M_x$, using vertical strips and the ‘midpoints’ method:

Plug $f_2(x)=f(x)=\cos x$ and $f_1(x)=0$ into formula:

$$M_x={\int }_0^{\pi /2}\rho \frac{1}{2}{f}_2^{2}dx\gg \gg {\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx$$

Integration by ‘power to frequency conversion’:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx\gg \gg \frac{\rho }{4}{\int }_0^{\pi /2}(1+\cos 2x)dx\\ \\ \gg \gg \frac{\rho }{4}x{|}_0^{\pi /2}+{\frac{\rho \sin 2x}{8}|}_0^{\pi /2}=\frac{\pi \rho }{8}\end{array}$$

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(3) Compute $M_y$ using vertical strips and the regular method:

Plug $f(x)=\cos x$ into formula:

$$M_y={\int }_a^b\rho xf(x)dx\gg \gg {\int }_0^{\pi /2}\rho x\cos xdx$$

Integration by parts. Set $u=x$, $v^{\prime }=\cos x$ and so $u^{\prime }=1$, $v=\sin x$:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho x\cos xdx\gg \gg \rho x\sin x{|}_0^{\pi /2}-\rho {\int }_0^{\pi /2}\sin xdx\\ \\ \gg \gg \frac{\pi \rho }{2}\cdot 1-\rho (-\cos \frac{\pi }{2}--\cos 0)=\rho (\frac{\pi }{2}-1)\end{array}$$

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(4) Compute CoM:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Plug in data:

$$\begin{array}{c}\bar{x}=\frac{\rho (\pi /2-1)}{\rho }\gg \gg \frac{\pi }{2}-1\\ \\ \bar{y}=\frac{\pi \rho /8}{\rho }\gg \gg \frac{\pi }{8}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{\pi }{2}-1,\frac{\pi }{8})\end{array}$$ Link to original

07 Theory

Theory 4

CoM of a triangle: “ $h/3$ trick”

Given a triangle with edge $b$ and altitude $h$ (measured along a perpendicular to $b$), the CoM of the triangle lies on the line parallel to $b$ and shifted upwards by $h/3$.

For example, if the triangle is positioned with base $b$ on the $x$-axis, then its CoM will lie on the line $y=h/3$.

Proof of “ $h/3$ trick”

Quick Linear Interpolation Function:

$$\begin{array}{c}w(y)=b+\frac{0-b}{h}(y-0)\\ \\ \gg \gg w(y)=b-\frac{b}{h}y\end{array}$$

Thus:

$$\begin{array}{c}{M}_x=\rho {\int }_0^{h}yw(y)dy\\ \\ \gg \gg \rho {\int }_0^{h}y(b-\frac{b}{h}y)dy\\ \\ \gg \gg \rho {(\frac{b{y}^2}{2}-\frac{b{y}^3}{3h})|}_0^h\gg \gg \rho \frac{b{h}^2}{6}\end{array}$$

Total mass:

$$m=\rho \cdot (\text{area})=\rho \cdot \frac{1}{2}bh$$

Therefore:

$$\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho b{h}^2/6}{\rho bh/2}\gg \gg \frac{h}{3}$$

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07 Theory

Theory 3

Two useful techniques for calculating moments and (thereby) CoMs:

• Additivity principle
• Symmetry

Additivity says that you can add moments of parts of a region to get the total moment of the region (to a given axis).

A symmetry principle is that if a region is mirror symmetric across some line, then the CoM must lie on that line.

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08 Illustration

Example - Center of mass using moments and symmetry

Center of mass using moments and symmetry

Find the center of mass of the two-part region:

Solution

(1) Symmetry: CoM on $y$-axis

Because the region is symmetric in the $y$-axis, the CoM must lie on that axis. Therefore $\bar{x}=0$.

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(2) Additivity of moments:

Write $M_x$ for the total $x$-moment (distance measured to the $x$-axis from above).

Write $M_x^{\mathrm{tri}}$ and $M_x^{\mathrm{circ}}$ for the $x$-moments of the triangle and circle.

Additivity of moments equation:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}$$

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(3) Find moment of the circle $M_x^{\mathrm{circ}}$ using Symmetry:

By symmetry we know ${\bar{x}}^{\mathrm{circ}}=0$.

By symmetry we know ${\bar{y}}^{\mathrm{circ}}=5$.

Area of circle with $r=2$ is $A=4\pi$, so total mass is $M=4\pi \rho$.

Centroid-from-moments equation:

$${\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}$$

Solve the equation for $M_x^{\mathrm{circ}}$:

$$\begin{array}{c}{\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}\gg \gg 5=\frac{M_x^{\mathrm{circ}}}{4\pi \rho }\\ \\ \gg \gg M_x^{\mathrm{circ}}=20\pi \rho \end{array}$$

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(4) Find moment of the triangle $M_x^{\mathrm{tri}}$ using Symmetry and the “$h/3$ trick”:

By Symmetry, the CoM of the triangle must lie on the vertical line $x=0$.

By the “$h/3$ trick,” the CoM of the triangle must lie on the horizontal line at $y=h/3$. In this case $h=3$, so it lies on the line $y=1$.

Therefore the triangle CoM is:

$$({\bar{x}}^{\text{tri}},{\bar{y}}^{\text{tri}})=(\mathrm{0,1})$$

To get moments, multiply by mass $m=\rho \cdot \frac{1}{2}(4)(3)=6\rho$:

$$M_y^{\text{tri}}=0,M_x^{\text{tri}}=(6\rho )(1)=6\rho$$

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(5) Apply additivity:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}\gg \gg \rho (20\pi +6)$$

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(6) Total mass of region:

Area of circle is $4\pi$. Area of triangle is $\frac{1}{2}\cdot 4\cdot 3=6$. Thus $m=\rho A=\rho (4\pi +6)$.

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(7) Compute center of mass $\bar{y}$ from total $M_x$ and total $M$:

We have $M_x=\rho (20\pi +6)$ and $M=\rho (4\pi +6)$. Plug into formula:

$$\begin{array}{c}\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho (20\pi +6)}{\rho (4\pi +6)}\approx 3.71\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\mathrm{0,3.71})\end{array}$$ Link to original

Example - Center of mass - two part region

Center of mass - two part region

Find the center of mass of the region which combines a rectangle and triangle (as in the figure) by computing separate moments. What are those separate moments? Assume the mass density is $\rho =1$.

Solution

(1) Apply symmetry to rectangle:

By symmetry, the center of mass of the rectangle is located at $(-\mathrm{1,2})$.

Thus ${\bar{x}}^{\text{rect}}=-1$ and ${\bar{y}}^{\text{rect}}=2$.

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(2) Find moments of the rectangle:

Total mass of rectangle $=m^{\text{rect}}=\rho \times \text{area}=1\cdot 8=8$. Thus:

$$\begin{array}{cc}{\bar{x}}^{\text{rect}}=\frac{M_y^{\text{rect}}}{m^{\text{rect}}}& \gg \gg M_y^{\text{rect}}=-8\\ & \\ {\bar{y}}^{\text{rect}}=\frac{M_x^{\text{rect}}}{m^{\text{rect}}}& \gg \gg M_x^{\text{rect}}=16\end{array}$$

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(3) Find moments of the triangle:

CoM of triangle using “$h/3$ trick:”

From the base on the $x$-axis, we have $h=4$ so the CoM lies on the line $y=4/3$.

From the base on the $y$-axis, we have $h=4$ so the CoM lies on the line $x=4/3$.

Multiply by the triangle’s mass to get its moment:

$$m^{\text{tri}}=\rho \cdot \frac{1}{2}(4)(4)\gg \gg 8$$

Therefore:

$$\begin{array}{c}{M}_x^{\text{tri}}=(8\rho )(4/3)\gg \gg \frac{32}{3}\\ \\ M_y^{\text{tri}}=(8\rho )(4/3)\gg \gg \frac{32}{3}\end{array}$$

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(4) Add up total moments:

General formulas: $M_x=M_x^{\text{tri}}+M_x^{\text{rect}}$ and $M_y=M_y^{\text{tri}}+M_y^{\text{rect}}$

Plug in data: $M_x=\frac{32}{3}+16=\frac{80}{3}$ and $M_y=\frac{32}{3}-8=\frac{8}{3}$

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(5) Find center of mass from moments:

Total mass of triangle $=M^{\text{tri}}=\rho \times \text{area}=1\cdot \frac{1}{2}\cdot 4\cdot 4=8$.

Total combined mass $=m=m^{\text{tri}}+m^{\text{rect}}=8+8=16$.

Apply moment relation:

$$\begin{array}{c}\bar{x}=\frac{M_y}{m}\gg \gg \frac{8/3}{16}\gg \gg \frac{1}{6}\\ \\ \bar{y}=\frac{M_x}{m}\gg \gg \frac{80/3}{16}\gg \gg \frac{5}{3}\\ \\ \text{CoM}=(\bar{x},\bar{y})=(\frac{1}{6},\frac{5}{3})\end{array}$$Link to original

Improper integrals

Videos

Review Videos

Videos, Math Dr. Bob:

• Improper integrals: Infinite limits
• Improper integrals: Vertical asymptote
• Improper integrals: $\int \frac{1}{x^p}dx$
• Improper integrals: $\int \frac{1}{x^2}{e}^{-1/x}dx$

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03 Theory

Theory 1

Improper integrals are those for which either a bound or the integrand itself become infinite somewhere on the interval of integration.

Examples:

$$(a){\int }_1^{\infty }\frac{1}{x^2}dx,(b){\int }_0^2\frac{1}{x}dx,(c){\int }_{-1}^{+1}\frac{1}{x^2}dx$$

(a) the upper bound is $\infty$ (b) the integrand goes to $\infty$ as $x\to 0^{+}$ (c) the integrand is $\infty$ at the point $0\in [-\mathrm{1,1}]$

The limit interpretation of (a) is this:

$${\int }_1^{\infty }\frac{1}{x^2}dx=\underset{R\to \infty }{\lim }{\int }_1^R\frac{1}{x^2}dx$$

The limit interpretation of (b) is this:

$${\int }_0^2\frac{1}{x}dx=\underset{R\to 0^{+}}{\lim }{\int }_R^2\frac{1}{x}dx$$

The limit interpretation of (c) is this:

$$\begin{array}{c}{\int }_{-1}^{+1}\frac{1}{x^2}dx={\int }_{-1}^0\frac{1}{x^2}dx+{\int }_0^{+1}\frac{1}{x^2}dx\\ \\ =\underset{R\to 0^{-}}{\lim }{\int }_{-1}^R\frac{1}{x^2}dx+\underset{R\to 0^{+}}{\lim }{\int }_R^{+1}\frac{1}{x^2}dx\end{array}$$

These limits are evaluated using the usual methods.

An improper integral is said to be convergent or divergent according to whether it may be assigned a finite value through the appropriate limit interpretation.

For example, $(a)$ converges while $(b)$ diverges.

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04 Illustration

Example - Improper integral - infinite bound

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 $R\to \infty$:

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

Improper integral - infinite integrand

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 $R\to 0^{+}$:

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

Example - Improper integral - infinity inside the interval

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:

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

---

(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.

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05 Theory

Theory 2

Two tools allow us to determine convergence of a large variety of integrals. They are the comparison test and the $p$-integral cases.

Comparison test - integrals

The comparison test says:

• When an improper integral converges, every smaller integral converges.
• When an improper integral diverges, every bigger integral diverges.

Here, smaller and bigger are comparisons of the integrand at all values (accounting properly for signs), and the bounds are assumed to be the same.

For example, ${\int }_2^{\infty }\frac{dx}{x^3}$ converges, and $x^4>x^3$ implies $\frac{1}{x^4}<\frac{1}{x^3}$ (when $x>1$), therefore the comparison test implies that ${\int }_2^{\infty }\frac{dx}{x^4}$ converges.

$p$-integral cases

Assume $p>0$ and $a>0$. We have:

$$\begin{array}{cccc}p>1:& {\int }_a^{\infty }\frac{dx}{x^p}\text{converges}& \text{and}& {\int }_0^a\frac{dx}{x^p}\text{diverges}\\ & & & \\ p<1:& {\int }_a^{\infty }\frac{dx}{x^p}\text{diverges}& \text{and}& {\int }_0^a\frac{dx}{x^p}\text{converges}\\ & & & \\ p=1:& {\int }_a^{\infty }\frac{dx}{x}\text{diverges}& \text{and}& {\int }_0^a\frac{dx}{x}\text{diverges}\end{array}$$

Proving the $p$-integral cases

It is easy to prove the convergence / divergence of each $p$-integral case using the limit interpretation and the power rule for integrals. (Or for $p=1$, using $\int \frac{1}{x}dx=\ln x+C$.)

Additional improper integral types

The improper integral ${\int }_{-\infty }^{a}f(x)dx$ also has a limit interpretation:

$${\int }_{-\infty }^{a}f(x)dx=\underset{R\to -\infty }{\lim }{\int }_R^{a}f(x)dx$$

The double improper integral ${\int }_{-\infty }^{\infty }f(x)dx$ has this limit interpretation:

$${\int }_{-\infty }^{\infty }f(x)dx=\underset{R\to -\infty }{\lim }{\int }_R^{a}f(x)dx+\underset{R\to \infty }{\lim }{\int }_a^{R}f(x)dx$$

Where $a$ is any finite number. This double integral does not exist if either limit does not exist for any value of $a$.

Double improper is not simultaneous!

Watch out! This may happen:

$${\int }_{-\infty }^{\infty }f(x)dx\ne \underset{R\to \infty }{\lim }{\int }_{-R}^{R}f(x)dx$$

This simultaneous limit might exist only because of internal cancellation in a case where the separate individual limits do not exist! We do not say ‘convergent’ in these cases!

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06 Illustration

Example - Comparison to $p$-integrals

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 $x\ge 2$.

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

---

(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:

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