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Section 2.7 improper Integrals

Recall the the an essential Theorem of Calculus says that if (f) is a constant function top top the close up door interval ( ext,) then

where (F) is any antiderivative that (f ext.)

Both the continuity condition and also closed interval must host to usage the an essential Theorem that Calculus, and in this case, (dsint_a^b f(x),dx) to represent the network area under (f(x)) native (a) to (b ext:)

We start with an instance where blindly using the basic Theorem the Calculus can offer an incorrect result.

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Example 2.51. Using FTC.

Explain why (dsint_-1^1frac1x^2,dx) is not equal to (-2 ext.)

eginequation*eginsplitint_-1^1frac1x^2,dx amp= int_-1^1 x^-2,dx = -x^-1igg|_-1^1 = -frac1xigg|_-1^1\ amp=~ left(-frac11 ight) - left(-frac1(-1) ight) ~=~ -2endsplitendequation*

However, the above answer is WRONG! since (f(x)=1/x^2) is not continuous on (<-1,~1> ext,) we cannot directly use the fundamental Theorem of Calculus. Intuitively, we can see why (-2) is not the correct answer by looking in ~ the graph that (f(x)=1/x^2) on (<-1,~1> ext.) The shaded area shows up to grow without bound as watched in the number below.

Formalizing this example leads come the principle of an not correct integral. There space two means to expand the fundamental Theorem of Calculus. One is to usage an infinite interval, i.e., () or ((-infty,infty) ext.) The second is to enable the term () to contain an infinite discontinuity the (f(x) ext.) In either case, the integral is dubbed an improper integral. One of the most vital applications the this principle is probability distributions because determining quantities like the cumulative circulation or meant value typically require integrals on unlimited intervals.

Subsection 2.7.1 not correct Integrals: Infinite boundaries of Integration

To compute not correct integrals, we use the ide of limits in addition to the basic Theorem the Calculus.

Definition 2.52. Wrong Integrals — One infinite Limit of Integration.

If (f(x)) is consistent on (
eginequation*int_a^infty f(x),dx=lim_R oinftyint_a^R f(x),dx ext.endequation*

If (f(x)) is constant on ((-infty,b> ext,) then the improper integral of (f) over ((-infty,b>) is

eginequation*int_-infty^b f(x),dx=lim_R o -inftyint_R^b f(x),dx ext.endequation*

Since we are dealing with limits, we are interested in convergence and also divergence the the not correct integral. If the border exists and also is a finite number, we say the improper integral converges. Otherwise, us say the not correct integral diverges, i m sorry we capture in the complying with definition.

Definition 2.53. Convergence and also Divergence.

If the limit exists and also is a limited number, we say the wrong integral converges.

If the limit is (pminfty) or does no exist, we say the improper integral diverges.

To get an intuitive (though not completely correct) translate of not correct integrals, we attempt to analyze (dsint_a^infty f(x),dx) graphically. Below assume (f(x)) is consistent on (

us let (R) it is in a addressed number in (
eginequation*int_a^infty f(x),dx=lim_R oinftyint_a^R f(x),dx ext.endequation*

We can then use the an essential Theorem the Calculus to the critical integral as (f(x)) is constant on the closeup of the door interval ( ext.)

We next specify the improper integral for the interval ((-infty,~infty) ext.)

Definition 2.54. Not correct Integrals — two Infinite boundaries of Integration.

If both (dsint_-infty^a f(x),dx) and (dsint_a^infty f(x),dx) room convergent, climate the not correct integral the (f) end ((-infty,infty)) is

eginequation*int_-infty^infty f(x),dx=int_-infty^a f(x),dx+int_a^infty f(x),dxendequation*

The above an interpretation requires both the the integrals

to it is in convergent because that (dsint_-infty^infty f(x),dx) to additionally be convergent. If either of (dsint_-infty^a f(x),dx) or (dsint_a^infty f(x),dx) is divergent, climate so is (dsint_-infty^infty f(x),dx ext.)

Example 2.55. Wrong Integral—One boundless Limit the Integration.

Determine even if it is (dsint_1^inftyfrac1x,dx) is convergent or divergent.

Solution

Using the definition for not correct integrals we write this as:

eginequation*eginsplitint_1^infty frac1x,dxamp= lim_R oinfty int_1^Rfrac1x,dx = lim_R oinfty ln|x|igg|_1^R\ amp =lim_R oinfty ln|R| - ln|1| = lim_R oinfty ln|R| = +inftyendsplitendequation*

Therefore, the integral is divergent.

Example 2.56. Wrong Integral — two Infinite limits of Integration.

Determine even if it is (dsint_-infty^infty xsin(x^2),dx) is convergent or divergent.

Solution

We should compute both (dsint_0^infty xsin(x^2),dx) and also (dsint_-infty^0 xsin(x^2),dx ext.) note that we don"t have to separation the integral up in ~ (0 ext,) any kind of finite worth (a) will certainly work. Very first we compute the unknown integral. Allow (u=x^2 ext,) then (du=2x,dx) and also hence,

eginequation*int xsin(x^2),dx=frac12 int sin(u),du=-frac12cos(x^2)+Cendequation*

Using the definition of wrong integral gives:

eginequation*eginsplitint_0^infty xsin(x^2),dx amp = lim_R oinfty int_0^R xsin(x^2),dx =lim_R oinfty left<-frac12cos(x^2) ight> igg|_0^R\ amp = -frac12 lim_R oinfty cos(R^2) +frac12endsplitendequation*

This limit does not exist since (cos x) oscillates between (-1) and also (+1 ext.) In particular, (cos x) go not technique any particular value together (x) it s okay larger and larger. Thus, (dsint_0^infty xsin(x^2),dx) diverges, and hence, the integral (dsint_-infty^infty xsin(x^2),dx) diverges.

Subsection 2.7.2 not correct Integrals: Discontinuities

When there is a discontinuity in () or in ~ an endpoint, then the not correct integral is as follows.

Definition 2.57.

Improper Integrals — Discontinuities top top Integration Bounds If (f(x)) is consistent on ((a,b> ext,) then the improper integral that (f) end ((a,b>) is

eginequation*int_a^b f(x),dx=lim_R o a^+int_R^b f(x),dx ext.endequation*

If (f(x)) is continuous on (
eginequation*int_a^b f(x),dx=lim_R o b^-int_a^R f(x),dx ext.endequation*

Definition 2.53 ~ above convergence and divergence the an wrong integral holds here as well: If the limit above exists and also is a limited number, us say the improper integral converges. Otherwise, we say the improper integral diverges.

When there is a discontinuity in the inner of ( ext,) we use the complying with definition.

Definition 2.58. Not correct Integrals—Discontinuities in ~ Integration Interval.

If (f) has actually a discontinuity in ~ (x=c) wherein (cin ext,) and both (dsint_a^c f(x),dx) and also (dsint_c^b f(x),dx) are convergent, then (f) end () is

eginequation*int_a^b f(x),dx=int_a^c f(x),dx+int_c^b f(x),dx ext.endequation*

Again, us can get an intuitive feeling of this concept by analyzing (dsint_a^b f(x),dx) graphically. Here assume (f(x)) is continuous on ((a,b>) but discontinuous at (x=a ext:)

us let (R) it is in a addressed number in ((a,b) ext.) climate by acquisition the limit as (R) viewpoints (a) from the right, we obtain the wrong integral:

eginequation*int_a^b f(x),dx=lim_R o a^+int_R^b f(x),dx ext.endequation*

Now we can use FTC to the last integral as (f(x)) is constant on ( ext.)

Example 2.59. A Divergent Integral.

Determine if (dsint_-1^1frac1x^2,dx) is convergent or divergent.

Solution

The function (f(x)=1/x^2) has a discontinuity at (x=0 ext,) which lies in (<-1,1> ext.) We should compute (dsint_-1^0 frac1x^2,dx) and (dsint_0^1 frac1x^2,dx ext.) Let"s begin with (dsint_0^1 frac1x^2,dx ext:)

eginequation*int_0^1 frac1x^2,dx = lim_R o 0^+ int_R^1 frac1x^2 ,dx = lim_R o 0^+ -frac1xigg|_R^1 = -1 + lim_R o 0^+ frac1Rendequation*

which quarter to (+infty ext.) Therefore, (dsint_-1^1frac1x^2,dx) is divergent due to the fact that one that (dsint_-1^0 frac1x^2,dx) and also (dsint_0^1 frac1x^2,dx) is divergent.

Example 2.60. Integral the the Logarithm.

Determine if (dsint_0^1 ln x ,dx) is convergent or divergent. Evaluate it if that is convergent.

Solution

Note the (f(x)=ln x) is discontinuous at the endpoint (x=0 ext.) We an initial use Integration by components to compute (dsintln x,dx ext.) us let (u=ln x) and (dv=dx ext.) then (du=(1/x)dx ext,) (v=x ext,) giving:

eginalign*int ln x,dx amp = ds xln x-int xcdotfrac1x,dx\amp = xln x-int 1,dx\amp = xln x-x+Cendalign*

Now utilizing the meaning of wrong integral because that (dsint_0^1 ln x ,dx ext:)

eginequation*eginsplitint_0^1 ln x ,dx amp= lim_R o 0^+ int_R^1 ln x,dx = lim_R o 0^+ (xln x-x)igg|_R^1 \ amp= -1 - lim_R o 0^+(Rln R) + lim_R o 0^+Rendsplitendequation*

Note the (dslim_R o 0^+R=0 ext.) We following compute (dslim_R o 0^+(Rln R) ext.) First, we rewrite the expression as follows:

eginequation*lim_x o0^+(Rln R)=lim_R o0^+fracln R1/R ext.endequation*

Now the limit is that the indeterminate kind ((-infty)/(infty)) and also l"Hôpital"s preeminence can it is in applied.

eginequation*lim_R o0^+(Rln R) =lim_R o0^+fracln R1/R =lim_R o0^+frac1/R-1/R^2 =lim_R o0^+-fracR^2R =lim_R o0^+(-R) =0endequation*

Thus, (dslim_R o 0^+(Rln R)=0 ext.) Thus

eginequation*int_0^1 ln x ,dx = -1 ext,endequation*

and the integral is convergent to (-1 ext.)

Graphically, one might interpret this to typical that the net area under (ln x) on (<0,1>) is (-1) (the area in this situation lies below the (x)-axis).
Example 2.61. Integral the a Square Root.

Determine if (dsint_0^4fracdxsqrt4-x) is convergent or divergent. Advice it if that is convergent.

Solution

Note the (frac1sqrt4-x) is discontinuous in ~ the endpoint (x=4 ext.) We usage a (u)-substitution to compute (int fracdxsqrt4-x ext.) we let (u=4-x ext,) then (du=-dx ext,) giving:

eginalign*dsintfracdxsqrt4-xamp =int-fracduu^1/2\amp =int -u^-1/2,du\amp =-2(u)^1/2+C\amp =-2sqrt4-x+Cendalign*

Now utilizing the definition of improper integrals for (dsint_0^4fracdxsqrt4-x ext:)

eginequation*dsint_0^4fracdxsqrt4-x=lim_R o4^-(-2sqrt4-x)igg|_0^R=lim_R o4^--2sqrt4-R+2sqrt4=4endequation*
Example 2.62. Improper Integral.

Determine if (dsint_1^2dfracdxleft( x-1 ight) ^1/3) is convergent or divergent. Advice it if it is convergent.

Solution

Note that (fleft( x ight) =dfrac1left( x-1 ight) ^1/3) is discontinuous at the endpoint (x=1 ext.) We first use substitution to uncover (dsint dfracdxleft( x-1 ight) ^1/3 ext.) we let (u=x-1 ext.) then (du=dx ext,) giving

eginequation*int dfracdxleft( x-1 ight) ^1/3=int fracduu^1/3=int u^-1/3du=frac32u^2/3+C=frac32left( x-1 ight) ^2/3+C ext.endequation*

Now making use of the meaning of wrong integral because that (dsint_1^2dfracdx left( x-1 ight) ^1/3:)

eginequation*eginsplitint_1^2dfracdxleft( x-1 ight) ^1/3amp=lim_R ightarrow 1^+int_R^2dfracdxleft( x-1 ight) ^1/3=left. lim_R ightarrow 1^+frac32left( x-1 ight) ^2/3 ightvert _R^2 \ amp=frac32-lim_R ightarrow 1^+frac32left( R-1 ight) ^2/3=frac32endsplit ext,endequation*

and the integral is convergent come (frac32 ext.) Graphically, one could interpret this to typical that the net area under (dfrac1left( x-1 ight)^1/3) top top (left< 1,2 ight>) is (frac32 ext.)

Subsection 2.7.3 (p)-Integrals

Integrals of the type (ds frac1x^p) come up again in the research of series. These integrals deserve to be one of two people classified together an not correct integral with an unlimited limit of integration, (dsint_a^infty dfrac1x^p,dx ext,) or together an improper integral through discontinuity in ~ (x=0 ext,) (dsint_0^a dfrac1x^p,dx ext.) In asymptotic analysis, that is valuable to recognize when one of two people of this intervals converge or diverge.

Theorem 2.63. (p)-Test for infinite Limit.

For (a>0 ext:)

If (p>1 ext,) then (dsint_a^infty frac1x^p,dx) converges.

If (pleq 1 ext,) climate (dsint_a^infty frac1x^p~dx) diverges.

Proof.

If (p>1 ext,) we have (dsint_a^infty frac1x^p,dx=lim_R ightarrow infty left. fracx^1-p1-p ight|^R_a = lim_R ightarrow infty fracR^1-p1-p - fraca^1-p1-p=fraca^1-pp-1 ext.)

If (pleq 1 ext,) the above tells us that the resulting border is infinite.

Theorem 2.64. (p)-Test for Discontinuity.

For (a>0 ext:)

If (plt 1 ext,) climate (dsint_0^afrac1x^p,dx) converges.

If (pgeq 1 ext,) climate (dsint_0^afrac1x^p,dx) diverges.

Proof.

If (plt 1 ext,) we need to (dsint_0^a frac1x^p, dx=lim_R ightarrow 0^+ left. fracx^1-p1-p ight|^a_R = lim_R ightarrow 0^+ fraca^1-p1-p - fracR^1-p1-p=fraca^1-p1-p ext.)

If (pgeq 1 ext,) the over tells us that the resulting limit is infinite.

With Example 2.55 and Example 2.59, girlfriend have currently seen exactly how the (p)-Test is applied. For an excellent measure, here is one more example.

Example 2.65. (p)-Test.

Determine if the adhering to integrals room convergent or divergent.

(dsint_1^infty frac1x^3,dx)

(dsint_0^5 frac1x^4,dx)

Solution

This is a (p)-integral v an limitless upper border of integration and also (p=3 > 1 ext.) Therefore, through the (p)-Test for unlimited Limit, (dsint_1^infty frac1x^3,dx) converges.

We classify (dsint_0^5 frac1x^4,dx) together a (p)-integral v a discontinuity at (x=0) and (p=4 geq 1 ext.) Thus, by the (p)-Test for Discontinuity, the integral diverges.

Subsection 2.7.4 comparison Test

The complying with test allows us to recognize convergence/divergence information about improper integrals that are hard to compute by comparing them to much easier ones. Us state the test because that (Theorem 2.66. To compare Test for Improper Integrals.

Assume that (f(x)geq g(x)geq 0) because that (xgeq a ext.)

If (dsint_a^infty f(x),dx) hmfontconverges, climate (dsint_a^infty g(x),dx) additionally hmfontconverges.

If (dsint_a^infty g(x),dx) hmfontdiverges, climate (dsint_a^infty f(x),dx) also hmfontdiverges.

Informally, (i) claims that if (f(x)) is larger than (g(x) ext,) and also the area under (f(x)) is limited (converges), climate the area under (g(x)) must also be finite (converges). Informally, (ii) says that if (f(x)) is larger than (g(x) ext,) and the area under (g(x)) is boundless (diverges), climate the area under (f(x)) must also be unlimited (diverges).

Example 2.67. To compare Test.

Show the (dsint_2^infty fraccos^2xx^2 ,dx) converges.

Solution

We usage the Comparison check to display that it converges. Note that (0leq cos^2xleq 1) and hence

eginequation*0 leqfraccos^2xx^2leqfrac1x^2 ext.endequation*

Thus, acquisition (f(x)=1/x^2) and also (g(x)=cos^2x / x^2) we have (f(x)geq g(x)geq 0 ext.) One can quickly see that (dsint_2^infty frac1x^2,dx) converges. Therefore, (dsint_2^infty fraccos^2xx^2 ,dx) additionally converges.

Exercises because that Section 2.7.Exercise 2.7.1.

Determine whether the complying with improper integrals space convergent or divergent. Advice those that are convergent.

Exercise 2.7.2.

Prove that the integral (dsint_1^inftyfrac1x^p,dx) is convergent if (p>1) and also divergent if (0lt pleq 1 ext.)

Solution

Since

eginequation*int_1^inftyfrac1x^p,dx = lim_R oinfty fracx^1-p1-p igglvert_1^R = lim_R oinfty fracR^1-p1-p - frac11-p ext.endequation*

So if (p>1 ext,) we view that the integral converges. Yet for (0 lt ns leq 1 ext,) the integral diverges.

Exercise 2.7.3.

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Suppose that (p>0 ext.) uncover all worths of (p) because that which (dsint_0^1dfrac1x^p,dx) converges.