6.5: Improper Integrals

Up to this point, we have covered integrals that featured finite bounds or were well defined on the interval of integration. But in this section, we discuss integrals that represent areas over infinite intervals or with infinite discontinuities. We call these integrals improper integrals, examples of which are \[\int_1^\infty \frac{1}{x^2} \di x \and \int_{-1}^0 \frac{1}{x^2} \di x \pd\] We discuss the following topics:

• Type I Improper Integrals
• \(p\)-Integrals
• Type II Improper Integrals
• Comparison Testing

Type I Improper Integrals

A Type I improper integral has one or more infinite limits of integration, such as \(\int_a^\infty f(x) \di x.\) In other words, we integrate \(f\) over the infinite interval \([a, \infty).\) But what does the upper bound of \(\infty\) mean? Let's first consider the proper integral \(\int_a^t f(x) \di x.\) For the case \(f(x) \geq 0\) and \(t \gt a,\) the integral represents the area under the graph \(y = f(x)\) from \(x = a\) to \(x = t.\) (See Figure 1.) As \(t\) grows, the line \(x = t\) shifts farther to the right, thus expanding region \(R.\) If \(t\) increases without bound, then the interval of integration becomes unbounded and so \(R\) represents an infinite region. Consequently, we write \[\int_a^\infty f(x) \di x = \lim_{t \to \infty} \int_a^t f(x) \di x \pd\] To evaluate this integral, we first evaluate \(\int_a^t f(x) \di x\) and then take the limit of the resulting expression as \(t \to \infty.\) If the limit exists, then we say \(\int_a^\infty f(x) \di x\) converges (or is convergent); otherwise, the improper integral diverges (or is divergent). For the special case \(f(x) \geq 0,\) \(\int_a^\infty f(x) \di x\) converges if the region \(R\) has a finite area. By similar logic, we define two other forms of Type I improper integrals using limits of integrals over finite intervals, as follows.

We replace the upper bound with a variable \(t\) that tends to infinity: \[\int_1^\infty \frac{1}{x^2} \di x = \lim_{t \to \infty} \int_1^t \frac{1}{x^2} \di x \pd\] We evaluate \(\int_1^t \dd x/x^2\) using Part II of the Fundamental Theorem of Calculus, and then apply the limit: \[ \ba \lim_{t \to \infty} \int_1^t \frac{1}{x^2} \di x &= \lim_{t \to \infty} \par{-\frac{1}{x}} \intEval_1^t \nl &= \lim_{t \to \infty} \par{1 - \frac{1}{t}} \nl &= 1 - 0 \nl &= \boxed 1 \ea \] Geometrically, the expression \[\int_1^t \frac{1}{x^2} \di x = 1 - \frac{1}{t}\] equals the area under the graph \(y = 1/x^2\) from \(x = 1\) to \(x = t.\) (See Figure 2A.) As \(t \to \infty,\) the line \(x = t\) shifts indefinitely to the right, and the unbounded region to the right of \(x = 1\) has an area of \(1.\) (See Figure 2B.)

We rewrite the integral as the limit \[\lim_{t \to \infty} \int_1^t \frac{1}{x} \di x \pd\] Applying Part II of the Fundamental Theorem of Calculus and taking the limit, we attain \[ \ba \lim_{t \to \infty} (\ln \abs x) \intEval_1^t &= \lim_{t \to \infty} (\ln t) - \ln 1 \nl &= \lim_{t \to \infty} \ln t \nl &= \infty \pd \ea \] The limit does not exist, so \[\int_1^\infty \frac{1}{x} \di x \divBoxed\] Accordingly, the unbounded region under \(y = 1/x\) to the right of \(x = 1\) has an infinite area (Figure 3).

We replace the lower bound \(-\infty\) with a variable \(t\) that tends to \(-\infty,\) attaining \[ \ba \lim_{t \to -\infty} \int_t^{-1} \frac{1}{\sqrt[3]{x}} \di x &= \lim_{t \to -\infty} \int_t^{-1} x^{-1/3} \di x \nl &= \lim_{t \to -\infty} \par{\tfrac{3}{2} x^{2/3}} \intEval_t^{-1} \nl &= \tfrac{3}{2} - \lim_{t \to -\infty} \par{\tfrac{3}{2} t^{2/3}} \nl &= -\infty \pd \ea \] Because the limit does not exist, \[\int_{-\infty}^{-1} \frac{1}{\sqrt[3]{x}} \di x \divBoxed\] Figure 4 shows the unbounded region between the graph \(y = 1/\sqrt[3] x,\) the \(x\)-axis, and the line \(x = -1.\)

The integral is equivalent to \[ \ba \lim_{t \to \infty} \int_0^t \cos x \di x &= \lim_{t \to \infty} \sin x \intEval_0^t \nl &= \lim_{t \to \infty} \sin t - \sin 0 \nl &= \lim_{t \to \infty} \sin t \pd \ea \] But because \(\lim_{t \to \infty} \sin t\) does not exist, \[\int_0^{\infty} \cos x \di x \divBoxed\] The signed area bounded by the \(x\)-axis and the graph of \(y = \cos x\) does not approach any final value as \(t \to \infty,\) as shown by Figure 5.

\(p\)-Integrals

We have discovered a paradoxical fact: \(\int_1^\infty \dd x/x^2\) converges (Example 1), while the similar \(\int_1^\infty \dd x/x\) diverges (Example 2). How can we generalize the convergence or divergence of the family of integrals \(\int_1^\infty \dd x/x^p,\) called a \(\bf p\)-integral? As previously established, the integral converges for \(p = 2\) and diverges for \(p = 1.\) Generally, for \(p \ne 1\) we see \begin{align} \int_1^\infty \frac{1}{x^p} \di x &= \lim_{t \to \infty} \par{\frac{x^{1 - p}}{1 - p}} \intEval_1^t \nonum \nl &= \lim_{t \to \infty} \par{\frac{t^{1 - p}}{1 - p}} - \frac{1}{1 - p} \pd \label{eq:p-int-lim} \end{align} If \(p \lt 1,\) then \(1 - p\) is positive and so \(t^{1 - p} \to \infty\) as \(t \to \infty.\) Then the expression in \(\eqref{eq:p-int-lim}\) becomes \(\infty,\) so \(\int_1^\infty \dd x/x^p\) diverges for \(p \lt 1.\) Conversely, if \(p \gt 1\) then \(1 - p\) is negative, so \(t^{1 - p} \to 0\) as \(t \to \infty.\) As a result, \(\eqref{eq:p-int-lim}\) becomes \[ 0 -\frac{1}{1 - p} = \frac{1}{p - 1} \cma \] a finite number, so \(\int_1^\infty \dd x/x^p\) converges. Consequently, the \(p\)-integral \(\int_1^\infty \dd x/x^p\) diverges for \(p \leq 1\) and converges for \(p \gt 1.\)

Type II Improper Integrals

In Figure 6 the graph of \(y = f(x)\) has a vertical asymptote at \(x = b.\) The region \(R\) is infinite, so the integral \(\int_a^b f(x) \di x\) is improper. Using a similar procedure from Type I integrals, we imagine moving a vertical line \(x = t,\) \(a \lt t \lt b,\) closer and closer to the vertical asymptote at \(x = b.\) During this procedure, the area \(\int_a^t f(x) \di x\) better approximates the area of region \(R.\) We therefore define \[\int_a^b f(x) \di x = \lim_{t \to b^-} \int_a^t f(x) \di x \pd\] This definition applies to any function \(f\) that is continuous on \([a, b),\) not just positive functions. This integral is an example of a Type II improper integral, in which the integrand has an infinite discontinuity between the limits of integration. We define two other forms of Type II improper integrals, as shown below.

The graph of \(y = \ln x\) has a vertical asymptote at \(x = 0,\) so we write \[\int_0^1 \ln x \di x = \lim_{t \to 0^+} \int_t^1 \ln x \di x \pd\] By the result of Example 6.1-3, the improper integral is \[ \ba \lim_{t \to 0^+} \par{x \ln x - x} \intEval_t^1 &= -1 - \lim_{t \to 0^+} \par{t \ln t - t} \nl &= -1 - \lim_{t \to 0^+} t \ln t + 0 \pd \ea \] To evaluate \(\lim_{t \to 0^+} t \ln t,\) we force the product into a quotient and then use L'Hôpital's Rule, as follows: \[ \ba \lim_{t \to 0^+} t \ln t &= \lim_{t \to 0^+} \frac{\ln t}{1/t} \nl &= \lim_{t \to 0^+} \frac{1/t}{-1/t^2} \nl &= \lim_{t \to 0^+} (-t) \nl &= 0 \pd \ea \] Consequently, \[ \int_0^1 \ln x \di x = -1 - 0 = \boxed{-1} \] Thus, the region below the \(x\)-axis bounded by \(y = \ln x\) has area \(1\) (Figure 7).

The graph of \(y = 1/x^2\) has a vertical asymptote at \(x = 0\) (Figure 8), so we split the integral into a sum of limits: \[ \ba \int_{-1}^1 \frac{1}{x^2} \di x &= \int_{-1}^0 \frac{1}{x^2} \di x + \int_0^1 \frac{1}{x^2} \di x \nl &= \lim_{P \to 0^-} \int_{-1}^P \frac{1}{x^2} \di x + \lim_{Q \to 0^+} \int_Q^1 \frac{1}{x^2} \di x \pd \ea \] The improper integral \(\int_{-1}^1 \dd x/x^2\) converges if both limits exist; otherwise, it diverges. The first limit is \[\lim_{P \to 0^-} \par{-\frac{1}{x}} \intEval_{-1}^P = \lim_{P \to 0^-} \par{-\frac{1}{P}} - 1 \pd \] Since the limit is nonexistent, \[\int_{-1}^1 \frac{1}{x^2} \di x \divBoxed\]

Example 7 demonstrates a common trap—Type II improper integrals are not obvious to identify. From this point on, you must determine whether a definite integral is improper. For example, it is tempting to write \[\int_{-1}^1 \frac{1}{x^2} \di x = - \frac{1}{x} \intEval_{-1}^1 = -2 \cma\] but this result is nonsensical because \(1/x^2\) is strictly positive. It turns out that Part II of the Fundamental Theorem of Calculus requires the integrand to be continuous over \([-1, 1],\) so it cannot be applied to \(\int_{-1}^1 \dd x/x^2.\)

Comparison Testing

The improper integral \(\int_1^\infty \dd x/x^4\) converges because it is a \(p\)-integral with \(p = 4 \gt 1.\) But does the similar-looking integral \(\int_1^\infty \dd x/(x^4 + 1)\) converge or diverge? Evaluating the integral is very difficult, so another viable option is needed to attain a conclusion—a comparison to another integral.

To introduce this idea, let \(f\) and \(g\) be continuous functions on \([a, \infty).\) If \[0 \leq f(x) \leq g(x)\] and \(\int_a^\infty g(x) \di x\) converges, then \(\int_a^\infty f(x) \di x\) also converges because \(\int_a^\infty f(x) \di x\) is bounded between two finite numbers. For the special case where \(f\) and \(g\) are positive, the area under \(y = f(x)\) is less than the area under \(y = g(x),\) so \(\int_a^\infty f(x) \di x\) is finite. (See Figure 9.) Informally, if the larger integral converges, then the smaller integral also converges. By contrast, if \[0 \leq g(x) \leq f(x)\] and \(\int_a^\infty g(x) \di x\) diverges, then \(\int_a^\infty f(x) \di x\) also diverges. Informally, if the smaller integral diverges, then the larger integral also diverges. We summarize these facts below for Type I improper integrals.