1.12: Improper Integrals

• Joel Feldman, Andrew Rechnitzer and Elyse Yeager
• University of British Columbia

Definitions

• finite limits of integration \(a\) and \(b\text\) and
• a bounded integrand \(f(x)\) (and in fact continuous except possibly for finitely many jump discontinuities).

Not all integrals we need to study are quite so nice.

Definition 1.12.1

An integral having either an infinite limit of integration or an unbounded integrand is called an improper integral.

Two examples are

The first has an infinite domain of integration and the integrand of the second tends to \(\infty\) as \(x\) approaches the left end of the domain of integration. We'll start with an example that illustrates the traps that you can fall into if you treat such integrals sloppily. Then we'll see how to treat them carefully.

Example 1.12.2 \(\int_^1 \frac\, d\)

Consider the integral

If we “do” this integral completely naively then we get

which is wrong 1 . In fact, the answer is ridiculous. The integrand \(\frac \gt 0\text\) so the integral has to be positive.

The flaw in the argument is that the fundamental theorem of calculus, which says that

is applicable only when \(F'(x)\) exists and equals \(f(x)\) for all \(a\le x\le b\text<.>\) In this case \(F'(x)=\frac\) does not exist for \(x=0\text<.>\) The given integral is improper. We'll see later that the correct answer is \(+\infty\text<.>\)

Let us put this example to one side for a moment and turn to the integral \(\int_a^\infty\frac>\text<.>\) In this case, the integrand is bounded but the domain of integration extends to \(+\infty\text<.>\) We can evaluate this integral by sneaking up on it. We compute it on a bounded domain of integration, like \(\int_a^R\frac>\text\) and then take the limit \(R\rightarrow\infty\text<.>\)

Let us put this into practice:

Example 1.12.3 \(\int_a^\infty\frac>\)

Solution:

• Since the domain extends to \(+\infty\) we first integrate on a finite domain \begin \int_a^R\frac>&= \arctan x\bigg|_a^R\\ &= \arctan R - \arctan a \end
• We then take the limit as \(R \to +\infty\text\) \begin \int_a^\infty \frac>&= \lim_ \int_a^R\frac>\\ &= \lim_ \big[ \arctan R - \arctan a\big]\\ &= \frac<\pi>- \arctan a. \end

To be more precise, we actually formally define an integral with an infinite domain as the limit of the integral with a finite domain as we take one or more of the limits of integration to infinity.

Definition 1.12.4 Improper integral with infinite domain of integration

1. If the integral \(\int_a^R f(x)\, d\) exists for all \(R \gt a\text\) then \[ \int_a^\infty f(x)\, d=\lim_\int_a^R f(x)\, d \nonumber \] when the limit exists (and is finite).
2. If the integral \(\int_r^b f(x)\, d\) exists for all \(r \lt b\text\) then \[ \int_<-\infty>^b f(x)\, d=\lim_\int_r^b f(x)\, d \nonumber \] when the limit exists (and is finite).
3. If the integral \(\int_r^R f(x)\, d\) exists for all \(r \lt R\text\) then \[ \int_<-\infty>^\infty f(x)\, d=\lim_\int_r^c f(x)\, d +\lim_\int_c^R f(x)\, d \nonumber \] when both limits exist (and are finite). Any \(c\) can be used.

When the limit(s) exist, the integral is said to be convergent. Otherwise it is said to be divergent.

We must also be able to treat an integral like \(\int_0^1\frac>\) that has a finite domain of integration but whose integrand is unbounded near one limit of integration 2 Our approach is similar — we sneak up on the problem. We compute the integral on a smaller domain, such as \(\int_t^1\frac>\text\) with \(t \gt 0\text\) and then take the limit \(t\rightarrow 0+\text<.>\)

Example 1.12.5 \(\int_0^1 \frac\, d\)

Solution:

• Since the integrand is unbounded near \(x=0\text\) we integrate on the smaller domain \(t\leq x \leq 1\) with \(t \gt 0\text\) \begin \int_t^1 \frac\, d &= \log|x| \bigg|_t^1 = -\log|t| \end
• We then take the limit as \(t \to 0^+\) to obtain \begin \int_0^1 \frac\, d &= \lim_\int_t^1 \frac\, d = \lim_ -\log|t| = +\infty \end Thus this integral diverges to \(+\infty\text\)

Indeed, we define integrals with unbounded integrands via this process:

Definition 1.12.6 Improper integral with unbounded integrand

1. If the integral \(\int_t^b f(x)\, d\) exists for all \(a \lt t \lt b\text\) then \[ \int_a^b f(x)\, d=\lim_\int_t^b f(x)\, d \nonumber \] when the limit exists (and is finite).
2. If the integral \(\int_a^T f(x)\, d\) exists for all \(a \lt T \lt b\text\) then \[ \int_a^b f(x)\, d=\lim_\int_a^T f(x)\, d \nonumber \] when the limit exists (and is finite).
3. Let \(a \lt c \lt b\text\) If the integrals \(\int_a^T f(x)\, d\) and \(\int_t^b f(x)\, d\) exist for all \(a \lt T \lt c\) and \(c \lt t \lt b\text\) then \[ \int_a^b f(x)\, d=\lim_\int_a^T f(x)\, d +\lim_\int_t^b f(x)\, d \nonumber \] when both limit exist (and are finite).

When the limit(s) exist, the integral is said to be convergent. Otherwise it is said to be divergent.

Notice that (c) is used when the integrand is unbounded at some point in the middle of the domain of integration, such as was the case in our original example

A quick computation shows that this integral diverges to \(+\infty\)

More generally, if an integral has more than one “source of impropriety” (for example an infinite domain of integration and an integrand with an unbounded integrand or multiple infinite discontinuities) then you split it up into a sum of integrals with a single “source of impropriety” in each. For the integral, as a whole, to converge every term in that sum has to converge.

Example 1.12.7 \(\int_<-\infty>^\infty\frac>\)

Consider the integral

• The domain of integration that extends to both \(+\infty\) and \(-\infty\text\)
• The integrand is singular (i.e. becomes infinite) at \(x=2\) and at \(x=0\text\)
• So we would write the integral as \begin \int_<-\infty>^\infty\frac>&=\int_<-\infty>^ \frac>+\int_^0 \frac>+\int_0^b \frac>\\ &+\int_b^2 \frac>+\int_2^c \frac>+\int_c^\infty \frac>\end where
  • \(a\) is any number strictly less than \(0\text\)
  • \(b\) is any number strictly between \(0\) and \(2\text\) and
  • \(c\) is any number strictly bigger than \(2\text\)

  So, for example, take \(a=-1, b=1, c=3\text<.>\)

  • the first integral has domain of integration extending to \(-\infty\)
  • the second integral has an integrand that becomes unbounded as \(x\rightarrow 0-\text\)
  • the third integral has an integrand that becomes unbounded as \(x\rightarrow 0+\text\)
  • the fourth integral has an integrand that becomes unbounded as \(x\rightarrow 2-\text\)
  • the fifth integral has an integrand that becomes unbounded as \(x\rightarrow 2+\text\) and
  • the last integral has domain of integration extending to \(+\infty\text\)

  Examples

  With the more formal definitions out of the way, we are now ready for some (important) examples.

  Example 1.12.8 \(\int_1^\infty\frac>\) with \(p \gt 0\)

  Solution:

  • Fix any \(p \gt 0\text\)
  • The domain of the integral \(\int_1^\infty\frac>\) extends to \(+\infty\) and the integrand \(\frac\) is continuous and bounded on the whole domain.
  • So we write this integral as the limit \begin \int_1^\infty\frac>&=\lim_ \int_1^R\frac>\end
  • The antiderivative of \(1/x^p\) changes when \(p=1\text\) so we will split the problem into three cases, \(p \gt 1\text\) \(p=1\) and \(p \lt 1\text\)
  • When \(p \gt 1\text\) \begin \int_1^R \frac>&= \fracx^ \bigg|_1^R\\ &= \frac\end Taking the limit as \(R \to \infty\) gives \begin \int_1^\infty\frac>&= \lim_ \int_1^R\frac>\\ &= \lim_ \frac\\ &= \frac= \frac\end since \(1-p \lt 0\text\)
  • Similarly when \(p \lt 1\) we have \begin \int_1^\infty\frac>&= \lim_ \int_1^R \frac>&= \lim_ \frac\\ &= +\infty \end because \(1-p \gt 0\) and the term \(R^\) diverges to \(+\infty\text\)
  • Finally when \(p=1\) \begin \int_1^R\frac>&= \log|R|-\log 1 = \log R \end Then taking the limit as \(R \to \infty\) gives us \begin \int_1^\infty\frac>&= \lim_ \log|R| = +\infty. \end
  • So summarising, we have \begin \int_1^\infty\frac>&=\begin \text & \text p\le 1 \\ \frac& \text p \gt 1 \end \end

  Example 1.12.9 \(\int_0^1\frac>\) with \(p \gt 0\)

  Solution:

  • Again fix any \(p \gt 0\text\)
  • The domain of integration of the integral \(\int_0^1\frac>\) is finite, but the integrand \(\frac\) becomes unbounded as \(x\) approaches the left end, \(0\text\) of the domain of integration.
  • So we write this integral as \begin \int_0^1\frac>&=\lim_ \int_t^1\frac>\end
  • Again, the antiderivative changes at \(p=1\text\) so we split the problem into three cases.
  • When \(p \gt 1\) we have \begin \int_t^1\frac>&= \fracx^\bigg|_t^1\\ &= \frac<1-t^>\end Since \(1-p \lt 0\) when we take the limit as \(t\to 0\) the term \(t^\) diverges to \(+\infty\) and we obtain \begin \int_0^1\frac>&= \lim_ \frac<1-t^>= +\infty \end
  • When \(p=1\) we similarly obtain \begin \int_0^1\frac>&= \lim_ \int_t^1\frac>\\ &= \lim_ \big( -\log|t| \big)\\ &= +\infty \end
  • Finally, when \(p \lt 1\) we have \begin \int_0^1\frac>&= \lim_\int_t^1\frac>\\ &= \lim_ \frac<1-t^>= \frac\end since \(1-p \gt 0\text\)
  • In summary \begin \int_0^1\frac>&=\begin \frac& \text p \lt 1 \\ \text & \text p\ge 1 \end \end

  Example 1.12.10 \(\int_0^\infty\frac>\) with \(p \gt 0\)

  Solution:

  • Yet again fix \(p \gt 0\text\)
  • This time the domain of integration of the integral \(\int_0^\infty\frac>\) extends to \(+\infty\text\) and in addition the integrand \(\frac\) becomes unbounded as \(x\) approaches the left end, \(0\text\) of the domain of integration.
  • So we split the domain in two — given our last two examples, the obvious place to cut is at \(x=1\text\) \[ \int_0^\infty\frac>=\int_0^1\frac>+ \int_1^\infty\frac>\nonumber \]
  • We saw, in Example 1.12.9, that the first integral diverged whenever \(p\ge 1\text\) and we also saw, in Example 1.12.8, that the second integral diverged whenever \(p\le 1\text\)
  • So the integral \(\int_0^\infty\frac>\) diverges for all values of \(p\text\)

  Example 1.12.11 \(\int_^1\frac>\)

  This is a pretty subtle example. Look at the sketch below:

  This suggests that the signed area to the left of the \(y\)-axis should exactly cancel the area to the right of the \(y\)-axis making the value of the integral \(\int_^1\frac>\) exactly zero.

  But both of the integrals

  diverge so \(\int_^1\frac>\) diverges. Don't make the mistake of thinking that \(\infty-\infty=0\text<.>\) It is undefined. And it is undefined for good reason.

  For example, we have just seen that the area to the right of the \(y\)-axis is

  and that the area to the left of the \(y\)-axis is (substitute \(-7t\) for \(T\) above)

  If \(\infty-\infty=0\text\) the following limit should be \(0\text<.>\)

  This appears to give \(\infty-\infty=\log 7\text<.>\) Of course the number \(7\) was picked at random. You can make \(\infty-\infty\) be any number at all, by making a suitable replacement for \(7\text<.>\)

  Example 1.12.12 Example 1.12.2 revisited

  The careful computation of the integral of Example 1.12.2 is

  Hence the integral diverges to \(+\infty\text<.>\)

  Example 1.12.13 \(\int_<-\infty>^\infty\frac>\)

  The integral \(\int_<-\infty>^\infty\frac>\) converges and takes the value \(\pi\text<.>\)

  Example 1.12.14 When does \(\int_e^\infty\frac>\) converge?

  For what values of \(p\) does \(\int_e^\infty\frac>\) converge?

  Solution:

  • For \(x\ge e\text\) the denominator \(x(\log x)^p\) is never zero. So the integrand is bounded on the entire domain of integration and this integral is improper only because the domain of integration extends to \(+\infty\) and we proceed as usual.
  • We have \begin \int_e^\infty\frac>&=\lim_ \int_e^R\frac>\qquad\qquad\qquad \text\\ &=\lim_ \int_1^<\log R>\frac>\qquad\qquad\textu=\log x,\, d=\frac>\\ &=\lim_ \begin \frac\Big[(\log R)^-1\Big] & \text p\ne 1\\ \log(\log R) & \text p=1 \end\\ &=\begin \text & \text p\le 1\\ \frac& \text p \gt 1 \end \end In this last step we have used similar logic that that used in Example 1.12.8, but with \(R\) replaced by \(\log R\text\)

  Example 1.12.15 The gamma function

  The gamma function \(\Gamma(x)\) is defined by the improper integral

  \[ \Gamma(t) = \int_0^\infty x^e^\, d \nonumber \]

  We shall now compute \(\Gamma(n)\) for all natural numbers \(n\text<.>\)

  • To get started, we'll compute \begin \Gamma(1) &= \int_0^\infty e^\, d = \lim_\int_0^R e^\, d = \lim_\Big[-e^\Big]_0^R = 1 \end
  • Then compute \[\begin \Gamma(2) &= \int_0^\infty x e^\, d\\ &=\lim_ \int_0^R x e^\, d\\ \end\] Use integration by parts with \(u=x, \, d=e^\, d,\) so \(v=-e^, \, d=\, d\) \begin & = \lim_\bigg[- xe^ \Big|_0^R + \int_0^R e^\, d\bigg]\\ & = \lim_\Big[- xe^ - e^\Big]_0^R\\ & = 1 \end For the last equality, we used that \(\lim\limits_x e^=0\text\)
  • Now we move on to general \(n\text\) using the same type of computation as we just used to evaluate \(\Gamma(2)\text\) For any natural number \(n\text\) \[\begin \Gamma(n+1) &= \int_0^\infty x^n e^\, d\\ &=\lim_ \int_0^R x^n e^\, d\\ \end\] Again integrate by parts with \(u=x^n,\, d= e^\, d\text\) so \(v=-e^, \, d=nx^\, d\) \begin & = \lim_\bigg[- x^ne^\Big|_0^R + \int_0^R nx^e^\, d\bigg]\\ & = \lim_ n\int_0^R x^ e^\, d\\ & = n\Gamma(n) \end To get to the third row, we used that \(\lim\limits_x^n e^=0\text\)
  • Now that we know \(\Gamma(2)=1\) and \(\Gamma(n+1)= n\Gamma(n)\text\) for all \(n\in\mathbb\text\) we can compute all of the \(\Gamma(n)\)'s. \begin\Gamma(2)&=1\\ \Gamma(3)&=\Gamma(2+1)=2\Gamma(2)=2\cdot 1\\ \Gamma(4)&=\Gamma(3+1)=3\Gamma(3)=3\cdot2\cdot 1\\ \Gamma(5)&=\Gamma(4+1)=4\Gamma(4)=4\cdot3\cdot 2\cdot 1\\ &\vdots\\ \Gamma(n)&=(n-1)\cdot(n-2)\cdots 4\cdot 3\cdot 2\cdot 1 = (n-1)! \end That is, the factorial is just 3 the Gamma function shifted by one.

  Convergence Tests for Improper Integrals

  It is very common to encounter integrals that are too complicated to evaluate explicitly. Numerical approximation schemes, evaluated by computer, are often used instead (see Section 1.11). You want to be sure that at least the integral converges before feeding it into a computer 4 . Fortunately it is usually possible to determine whether or not an improper integral converges even when you cannot evaluate it explicitly.

  Remark 1.12.16

  For pedagogical purposes, we are going to concentrate on the problem of determining whether or not an integral \(\int_a^\infty f(x)\, d\) converges, when \(f(x)\) has no singularities for \(x\ge a\text<.>\) Recall that the first step in analyzing any improper integral is to write it as a sum of integrals each of has only a single “source of impropriety” — either a domain of integration that extends to \(+\infty\text\) or a domain of integration that extends to \(-\infty\text\) or an integrand which is singular at one end of the domain of integration. So we are now going to consider only the first of these three possibilities. But the techniques that we are about to see have obvious analogues for the other two possibilities.

  Now let's start. Imagine that we have an improper integral \(\int_a^\infty f(x)\, d\text\) that \(f(x)\) has no singularities for \(x\ge a\) and that \(f(x)\) is complicated enough that we cannot evaluate the integral explicitly 5 . The idea is find another improper integral \(\int_a^\infty g(x)\, d\)

  • with \(g(x)\) simple enough that we can evaluate the integral \(\int_a^\infty g(x)\, d\) explicitly, or at least determine easily whether or not \(\int_a^\infty g(x)\, d\) converges, and
  • with \(g(x)\) behaving enough like \(f(x)\) for large \(x\) that the integral \(\int_a^\infty f(x)\, d\) converges if and only if \(\int_a^\infty g(x)\, d\) converges.

  So far, this is a pretty vague strategy. Here is a theorem which starts to make it more precise.

  Theorem 1.12.17 Comparison

  Let \(a\) be a real number. Let \(f\) and \(g\) be functions that are defined and continuous for all \(x\ge a\) and assume that \(g(x)\ge 0\) for all \(x\ge a\text<.>\)

  1. If \(|f(x)|\le g(x)\) for all \(x\ge a\) and if \(\int_a^\infty g(x)\, d\) converges then \(\int_a^\infty f(x)\, d\) also converges.
  2. If \(f(x)\ge g(x)\) for all \(x\ge a\) and if \(\int_a^\infty g(x)\, d\) diverges then \(\int_a^\infty f(x)\, d\) also diverges.

  We will not prove this theorem, but, hopefully, the following supporting arguments should at least appear reasonable to you. Consider the figure below:

  • If \(\int_a^\infty g(x)\, d\) converges, then the area of \begin \big\ \text < is finite.>\end When \(|f(x)|\le g(x)\text\) the region \begin \big\ \text < is contained inside >\big\ \end and so must also have finite area. Consequently the areas of both the regions \begin \big\ \text < and >\big\ \end are finite too 6 .
  • If \(\int_a^\infty g(x)\, d\) diverges, then the area of \begin \big\ \text < is infinite.>\end When \(f(x)\ge g(x)\text\) the region \begin \big\ \text < contains the region >\big\ \end and so also has infinite area.

  Example 1.12.18 \(\int_1^\infty e^\, d\)

  We cannot evaluate the integral \(\int_1^\infty e^\, d\) explicitly 7 , however we would still like to understand if it is finite or not — does it converge or diverge?

  Solution: We will use Theorem 1.12.17 to answer the question.

  • So we want to find another integral that we can compute and that we can compare to \(\int_1^\infty e^\, d\text\) To do so we pick an integrand that looks like \(e^\text\) but whose indefinite integral we know — such as \(e^\text\)
  • When \(x\ge 1\text\) we have \(x^2\ge x\) and hence \(e^\le e^\text\) Thus we can use Theorem 1.12.17 to compare \begin \int_1^\infty e^\, d \text < with >\int_1^\infty e^\, d \end
  • The integral \begin \int_1^\infty e^\, d &=\lim_\int_1^R e^\, d\\ &=\lim_\Big[-e^\Big]_1^\\ &=\lim_\Big[e^-e^\Big] =e^ \end converges.
  • So, by Theorem 1.12.17, with \(a=1\text\) \(f(x)=e^\) and \(g(x)=e^\text\) the integral \(\int_1^\infty e^\, d\) converges too (it is approximately equal to \(0.1394\)).

  Example 1.12.19 \(\int_^\infty e^\, d\)

  Solution:

  • The integral \(\int_^\infty e^\, d\) is quite similar to the integral \(\int_1^\infty e^\, d\) of Example 1.12.18. But we cannot just repeat the argument of Example 1.12.18 because it is not true that \(e^\le e^\) when \(0 \lt x \lt 1\text\)
  • In fact, for \(0 \lt x \lt 1\text\) \(x^2 \lt x\) so that \(e^ \gt e^\text\)
  • However the difference between the current example and Example 1.12.18 is \begin \int_^\infty e^\, d-\int_1^\infty e^\, d &= \int_^1 e^\, d \end which is clearly a well defined finite number (its actually about \(0.286\)). It is important to note that we are being a little sloppy by taking the difference of two integrals like this — we are assuming that both integrals converge. More on this below.
  • So we would expect that \(\int_^\infty e^\, d\) should be the sum of the proper integral integral \(\int_^1 e^\, d\) and the convergent integral \(\int_1^\infty e^\, d\) and so should be a convergent integral. This is indeed the case. The Theorem below provides the justification.

  Theorem 1.12.20

  Let \(a\) and \(c\) be real numbers with \(a \lt c\) and let the function \(f(x)\) be continuous for all \(x\ge a\text<.>\) Then the improper integral \(\int_a^\infty f(x)\ \, d\) converges if and only if the improper integral \(\int_c^\infty f(x)\ \, d\) converges.

  By definition the improper integral \(\int_a^\infty f(x)\, d\) converges if and only if the limit

  exists and is finite. (Remember that, in computing the limit, \(\int_a^c f(x)\, d\) is a finite constant independent of \(R\) and so can be pulled out of the limit.) But that is the case if and only if the limit \(\lim_\int_c^R f(x)\, d\) exists and is finite, which in turn is the case if and only if the integral \(\int_c^\infty f(x)\, d\) converges.

  Example 1.12.21 Does \(\int_1^\infty\frac>\, d\) converge?

  Does the integral \(\int_1^\infty\frac>\, d\) converge or diverge?

  Solution:

  • Our first task is to identify the potential sources of impropriety for this integral.
  • The domain of integration extends to \(+\infty\text\) but we must also check to see if the integrand contains any singularities. On the domain of integration \(x\ge 1\) so the denominator is never zero and the integrand is continuous. So the only problem is at \(+\infty\text\)
  • Our second task is to develop some intuition 8 . As the only problem is that the domain of integration extends to infinity, whether or not the integral converges will be determined by the behavior of the integrand for very large \(x\text\)
  • When \(x\) is very large, \(x^2\) is much much larger than \(x\) (which we can write as \(x^2\gg x\)) so that the denominator \(x^2+x\approx x^2\) and the integrand \begin \frac>& \approx \frac>=\frac> \end
  • By Example 1.12.8, with \(p=\frac\text\) the integral \(\int_1^\infty \frac>>\) converges. So we would expect that \(\int_1^\infty\frac>\, d\) converges too.
  • Our final task is to verify that our intuition is correct. To do so, we want to apply part (a) of Theorem 1.12.17 with \(f(x)= \frac>\) and \(g(x)\) being \(\frac>\text\) or possibly some constant times \(\frac>\text\) That is, we need to show that for all \(x \geq 1\) (i.e. on the domain of integration) \begin \frac>\leq \frac> \end for some constant \(A\text\) Let's try this.
  • Since \(x\geq 1\) we know that \[\begin x^2+x & \gt x^2\\ \end\] Now take the reciprocal of both sides: \begin \frac& \lt \frac\\ \end Multiply both sides by \(\sqrt\) (which is always positive, so the sign of the inequality does not change) \begin \frac>& \lt \frac>= \frac> \end
  • So Theorem 1.12.17(a) and Example 1.12.8, with \(p=\frac\) do indeed show that the integral \(\int_1^\infty\frac>\, d\) converges.

  Notice that in this last example we managed to show that the integral exists by finding an integrand that behaved the same way for large \(x\text<.>\) Our intuition then had to be bolstered with some careful inequalities to apply the comparison Theorem 1.12.17. It would be nice to avoid this last step and be able jump from the intuition to the conclusion without messing around with inequalities. Thankfully there is a variant of Theorem 1.12.17 that is often easier to apply and that also fits well with the sort of intuition that we developed to solve Example 1.12.21.

  A key phrase in the previous paragraph is “behaves the same way for large \(x\)”. A good way to formalise this expression — “\(f(x)\) behaves like \(g(x)\) for large \(x\)” — is to require that the limit

  Suppose that this is the case and call the limit \(L\ne 0\text<.>\) Then

  • the ratio \(\frac\) must approach \(L\) as \(x\) tends to \(+\infty\text\)
  • So when \(x\) is very large — say \(x \gt B\text\) for some big number \(B\) — we must have that \begin \fracL \leq \frac\leq 2L && \text \end Equivalently, \(f(x)\) lies between \(\fracg(x)\) and \(2Lg(x)\text\) for all \(x\ge B\text\)
  • Consequently, the integral of \(f(x)\) converges if and only if the integral of \(g(x)\) converges, by Theorems 1.12.17 and 1.12.20.

  These considerations lead to the following variant of Theorem 1.12.17.

  Theorem 1.12.22 Limiting comparison

  Let \(-\infty \lt a \lt \infty\text<.>\) Let \(f\) and \(g\) be functions that are defined and continuous for all \(x\ge a\) and assume that \(g(x)\ge 0\) for all \(x\ge a\text<.>\)

  1. If \(\int_a^\infty g(x)\, d\) converges and the limit \begin \lim_\frac\end exists, then \(\int_a^\infty f(x)\, d\) converges.
  2. If \(\int_a^\infty g(x)\, d\) diverges and the limit \begin \lim_\frac\end exists and is nonzero, then \(\int_a^\infty f(x)\) diverges.

  Note that in (b) the limit must exist and be nonzero, while in (a) we only require that the limit exists (it can be zero).

  Here is an example of how Theorem 1.12.22 is used.

  Example 1.12.23 \(\int_1^\infty\frac+x^2>\, d\)

  Does the integral \(\displaystyle \int_1^\infty\frac+x^2>\, d\) converge or diverge?

  Solution:

  • Our first task is to identify the potential sources of impropriety for this integral.
  • The domain of integration extends to \(+\infty\text\) On the domain of integration the denominator is never zero so the integrand is continuous. Thus the only problem is at \(+\infty\text\)
  • Our second task is to develop some intuition about the behavior of the integrand for very large \(x\text\) A good way to start is to think about the size of each term when \(x\) becomes big.
  • When \(x\) is very large:
    • \(e^ \ll x^2\text\) so that the denominator \(e^+x^2\approx x^2\text\) and
    • \(|\sin x|\le 1 \ll x\text\) so that the numerator \(x+\sin x\approx x\text\) and
    • the integrand \(\frac \approx \frac=\frac\text\)

    Notice that we are using \(A \ll B\) to mean that “\(A\) is much much smaller than \(B\)”. Similarly \(A\gg B\) means “\(A\) is much much bigger than \(B\)”. We don't really need to be too precise about its meaning beyond this in the present context.