Improper Integrals

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The goal of this lesson is to extend the concept of the definite integral \(\int\limits_a^b {f\left( x \right)dx} \) to improper integrals.

There are two types of improper integrals:

The limit \(a\) or \(b\) (or both the limits) are infinite;
The function \({f\left( x \right)}\) has one or more points of discontinuity in the interval \(\left[ {a,b} \right].\)

Type 1. Integration over an Infinite Domain

Let \({f\left( x \right)}\) be a continuous function on the interval \(\left[ {a,\infty} \right).\) We define the improper integral as

In order to integrate over the infinite domain \(\left[ {a,\infty } \right),\) we consider the limit of the form

\[\int\limits_a^\infty {f\left( x \right)dx} = \lim\limits_{n \to \infty } \int\limits_a^n {f\left( x \right)dx} .\]

Similarly, if a continuous function \(f\left(x\right)\) is given on the interval \(\left( {-\infty,b} \right],\) the improper integral of \(f\left(x\right)\) is defined as

\[\int\limits_{ - \infty }^b {f\left( x \right)dx} = \lim\limits_{n \to - \infty } \int\limits_n^b {f\left( x \right)dx} .\]

If these limits exist and are finite then we say that the improper integrals are convergent. Otherwise the integrals are divergent.

An improper integral might have two infinite limits. In this case, we can pick an arbitrary point \(c\) and break the integral up there. As a result, we obtain two improper integrals, each with one infinite limit:

\[\int\limits_{ - \infty }^\infty {f\left( x \right)dx} = \int\limits_{ - \infty }^c {f\left( x \right)dx} + \int\limits_c^\infty {f\left( x \right)dx} .\]

If, for some real number \(c,\) both of the integrals in the right-hand side are convergent, then we say that the integral \(\int\limits_{ - \infty }^\infty {f\left( x \right)dx} \) is also convergent; otherwise it is divergent.

Comparison Tests

Let \({f\left( x \right)}\) and \({g\left( x \right)}\) be continuous functions on the interval \(\left[ {a,\infty } \right).\) Suppose that \(0 \le g\left( x \right) \le f\left( x \right)\) for all \(x\) in the interval \(\left[ {a,\infty } \right).\)

If \(\int\limits_a^\infty {f\left( x \right)dx} \) is convergent, then \(\int\limits_a^\infty {g\left( x \right)dx} \) is also convergent;
If \(\int\limits_a^\infty {g\left( x \right)dx} \) is divergent, then \(\int\limits_a^\infty {f\left( x \right)dx} \) is also divergent;
If \(\int\limits_a^\infty {\left| {f\left( x \right)} \right|dx} \) is convergent, then \(\int\limits_a^\infty {f\left( x \right)dx} \) is also convergent. In this case, we say that the integral \(\int\limits_a^\infty {f\left( x \right)dx} \) is absolutely convergent.

It is often convenient to make comparisons with improper integrals of the form \(\int\limits_1^\infty {\frac{{dx}}{{{x^p}}},} \) where \(p \gt 0\) is a real number.

The integral \(\int\limits_1^\infty {\frac{{dx}}{{{x^p}}}} \) converges if \(p \gt 1,\) and diverges if \(p \lt 1.\) If \(p = 1,\) then the integral also diverges:

\[\lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{{x^p}}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{x}} = \lim\limits_{n \to \infty } \left. {\ln x} \right|_1^n = {\infty} .\]

Type 2. Improper Integrals with Infinite Discontinuities

This type of improper integrals refers to integrands that are undefined at one or more points of the domain of integration \(\left[ {a,b} \right].\)

Let \({f\left( x \right)}\) be a function which is continuous on the interval \(\left[ {a,b} \right),\) but is discontinuous at \(x = b.\) We define the improper integral as

\[\int\limits_a^b {f\left( x \right)dx} = \lim\limits_{\tau \to 0 + } \int\limits_a^{b - \tau } {f\left( x \right)dx} .\]

Similarly we can consider the case when the function \({f\left( x \right)}\) is continuous on the interval \(\left( {a,b} \right],\) but is discontinuous at \(x = a.\) Then

\[\int\limits_a^b {f\left( x \right)dx} = \lim\limits_{\tau \to 0 + } \int\limits_{a + \tau }^b {f\left( x \right)dx} .\]

If these limits exist and are finite then we say that the integrals are convergent; otherwise the integrals are divergent.

Finally, if the function \(f\left(x\right)\) is continuous on \(\left[ {a,c} \right) \cup \left( {c,b} \right]\) with an infinite discontinuity at \(x = c,\) then we define the improper integral as

\[\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^c {f\left( x \right)dx} + \int\limits_c^b {f\left( x \right)dx} ,\]

We say that the integral \(\int\limits_a^b {f\left( x \right)dx}\) is convergent if both of the integrals in the right side are also convergent. Otherwise the improper integral is divergent.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Determine for what values of \(k\) the integral \[{\int\limits_1^\infty} {\frac{{dx}}{{{x^k}}}}\;\left( {k \gt 0,k \ne 1} \right)\] converges.

Example 2

Calculate the integral \[\int\limits_1^\infty {\frac{{dx}}{{{x^2} + 1}}}.\]

Example 3

Calculate the integral \[\int\limits_0^\infty {\frac{{dx}}{{{x^2} + 16}}}.\]

Example 4

Calculate the integral \[\int\limits_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 4}}}.\]

Example 5

Determine whether the integral \[{\int\limits_1^\infty} {\frac{{dx}}{{{x^2}{e^x}}}}\] converges or diverges?

Example 6

Calculate the integral \[{\int\limits_{ - 2}^2} {\frac{{dx}}{{{x^3}}}}.\]

Example 1.

Determine for what values of \(k\) the integral \[{\int\limits_1^\infty} {\frac{{dx}}{{{x^k}}}}\;\left( {k \gt 0,k \ne 1} \right)\] converges.

Solution.

By the definition of an improper integral, we have

\[ \int\limits_1^\infty {\frac{{dx}}{{{x^k}}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{{x^k}}}} = \lim\limits_{n \to \infty } \int\limits_1^n {{x^{ - k}}dx} = \lim\limits_{n \to \infty } \left. {\left( {\frac{{{x^{ - k + 1}}}}{{ - k + 1}}} \right)} \right|_1^n = \frac{1}{{1 - k}} \cdot \lim\limits_{n \to \infty } \left. {\left( {{x^{ - k + 1}}} \right)} \right|_1^n = \frac{1}{{1 - k}} \cdot \lim\limits_{n \to \infty } \left( {{n^{ - k + 1}} - {1^{ - k + 1}}} \right) = \frac{1}{{k - 1}} \lim\limits_{n \to \infty } \left( {1 - {n^{1 - k}}} \right).\]

As you can see from the expression, there are \(2\) cases:

If \(0 \lt k \lt 1,\) then \({n^{1 - k}} \to \infty \) as \(n \to \infty\) and the integral diverges;
If \(k \gt 1,\) then
\[{n^{1 - k}} = {\frac{1}{{{n^{k - 1}}}}} \to 0\]

as \(n \to \infty\) and the integral converges.

Example 2.

Calculate the integral \[\int\limits_1^\infty {\frac{{dx}}{{{x^2} + 1}}}.\]

Solution.

\[\int\limits_1^\infty {\frac{{dx}}{{{x^2} + 1}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{{x^2} + 1}}} = \lim\limits_{n \to \infty } \left[ {\arctan x} \right]_1^n = \lim\limits_{n \to \infty } \left[ {\arctan n - \arctan 1} \right] = \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}.\]

Hence, the integral converges.

Example 3.

Calculate the integral \[\int\limits_0^\infty {\frac{{dx}}{{{x^2} + 16}}}.\]

Solution.

\[\int\limits_0^\infty {\frac{{dx}}{{{x^2} + 16}}} = \lim\limits_{n \to \infty } \int\limits_0^n {\frac{{dx}}{{{x^2} + 16}}} = \lim\limits_{n \to \infty } \int\limits_0^n {\frac{{dx}}{{{x^2} + {4^2}}}} = \lim\limits_{n \to \infty } \left. {\left( {\frac{1}{4}\arctan \frac{x}{4}} \right)} \right|_0^n = \frac{1}{4}\lim\limits_{n \to \infty } \left( {\arctan \frac{n}{4} - \arctan \frac{0}{4}} \right) = \frac{1}{4}\lim\limits_{n \to \infty } \left( {\arctan \frac{n}{4} - 0} \right) = \frac{1}{4} \cdot \frac{\pi }{2} = \frac{\pi }{8}.\]

The given integral converges.

Example 4.

Calculate the integral \[\int\limits_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 4}}}.\]

Solution.

The original integral has two infinite limits. Therefore we split it into two integrals and evaluate each as a one-sided improper integral:

\[I = \int\limits_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 4}}} = \int\limits_{ - \infty }^0 {\frac{{dx}}{{{x^2} + {2^2}}}} + \int\limits_0^\infty {\frac{{dx}}{{{x^2} + {2^2}}}} = {I_1} + {I_2}.\]

Calculate each integral:

\[{I_1} = \int\limits_{ - \infty }^0 {\frac{{dx}}{{{x^2} + {2^2}}}} = \lim\limits_{n \to - \infty } \int\limits_n^0 {\frac{{dx}}{{{x^2} + {2^2}}}} = \lim\limits_{n \to - \infty } \left[ {\frac{1}{2}\arctan \frac{x}{2}} \right]_n^0 = \frac{1}{2}\lim\limits_{n \to - \infty } \left[ {\arctan 0 - \arctan \frac{n}{2}} \right] = \frac{1}{2}\left( {0 - \left( { - \frac{\pi }{2}} \right)} \right) = \frac{\pi }{4};\]

\[{I_2} = \int\limits_0^\infty {\frac{{dx}}{{{x^2} + {2^2}}}} = \lim\limits_{n \to \infty } \int\limits_0^n {\frac{{dx}}{{{x^2} + {2^2}}}} = \lim\limits_{n \to \infty } \left[ {\frac{1}{2}\arctan \frac{x}{2}} \right]_0^n = \frac{1}{2}\lim\limits_{n \to - \infty } \left[ {\arctan \frac{n}{2} - \arctan 0} \right] = \frac{1}{2}\left( {\frac{\pi }{2} - 0} \right) = \frac{\pi }{4}.\]

Hence,

\[I = {I_1} + {I_2} = \frac{\pi }{4} + \frac{\pi }{4} = \frac{\pi }{2}.\]

We see that the integral converges.

Example 5.

Determine whether the integral \[{\int\limits_1^\infty} {\frac{{dx}}{{{x^2}{e^x}}}}\] converges or diverges?

Solution.

Note, that

\[{\frac{1}{{{x^2}{e^x}}}} \le {\frac{1}{{{x^2}}}}\]

for all values \(x \ge 1.\) Since the improper integral \({\int\limits_1^\infty} {\frac{{dx}}{{{x^2}}}}\) is convergent according to the results in Example \(1,\) the given integral \({\int\limits_1^\infty} {\frac{{dx}}{{{x^2}{e^x}}}}\) is also convergent by Comparison Test \(1.\)

Example 6.

Calculate the integral \[{\int\limits_{ - 2}^2} {\frac{{dx}}{{{x^3}}}}.\]

Solution.

There is a discontinuity at \(x = 0,\) so that we must consider two improper integrals:

\[\int\limits_{ - 2}^2 {\frac{{dx}}{{{x^3}}}} = \int\limits_{ - 2}^0 {\frac{{dx}}{{{x^3}}}} + \int\limits_0^2 {\frac{{dx}}{{{x^3}}}} .\]

Using the definition of improper integral, we obtain

\[\int\limits_{ - 2}^2 {\frac{{dx}}{{{x^3}}}} = \int\limits_{ - 2}^0 {\frac{{dx}}{{{x^3}}}} + \int\limits_0^2 {\frac{{dx}}{{{x^3}}}} = \lim\limits_{\tau \to 0 + } \int\limits_{ - 2}^{ - \tau } {\frac{{dx}}{{{x^3}}}} + \lim\limits_{\tau \to 0 + } \int\limits_\tau ^2 {\frac{{dx}}{{{x^3}}}} .\]

For the first integral,

\[\lim\limits_{\tau \to 0 + } \int\limits_{ - 2}^{ - \tau } {\frac{{dx}}{{{x^3}}}} = \lim\limits_{\tau \to 0 + } \left. {\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right)} \right|_{ - 2}^{ - \tau } = - \frac{1}{2}\lim\limits_{\tau \to 0 + } \left. {\left( {\frac{1}{{{x^2}}}} \right)} \right|_{ - 2}^{ - \tau } = - \frac{1}{2}\lim\limits_{\tau \to 0 + } \left[ {\frac{1}{{{{\left( { - \tau } \right)}^2}}} - \frac{1}{{{{\left( { - 2} \right)}^2}}}} \right] = - \frac{1}{2}\lim\limits_{\tau \to 0 + } \left( {\frac{1}{{{\tau ^2}}} + \frac{1}{8}} \right) = \infty .\]

Since it is divergent, the initial integral also diverges.