Improper Integrals
Solved Problems
Example 7.
Calculate the integral \[\int\limits_{ - \infty }^{ - 1} {\frac{{dx}}{{{x^2} + 4x + 5}}}.\]
Solution.
We complete the square in the denominator:
\[{x^2} + 4x + 5 = \left( {{x^2} + 4x + 4} \right) + 1 = {\left( {x + 2} \right)^2} + 1.\]
Then
\[\int\limits_{ - \infty }^{ - 1} {\frac{{dx}}{{{x^2} + 4x + 5}}} = \int\limits_{ - \infty }^{ - 1} {\frac{{dx}}{{{{\left( {x + 2} \right)}^2} + 1}}} = \lim\limits_{n \to - \infty } \int\limits_n^{ - 1} {\frac{{dx}}{{{{\left( {x + 2} \right)}^2} + 1}}} = \lim\limits_{n \to - \infty } \left[ {\arctan \left( {x + 2} \right)} \right]_n^{ - 1} = \lim\limits_{n \to - \infty } \left[ {\arctan 1 - \arctan \left( {n + 2} \right)} \right] = \frac{\pi }{4} - \left( { - \frac{\pi }{2}} \right) = \frac{{3\pi }}{4}.\]
Hence, the integral converges.
Example 8.
Calculate the integral \[\int\limits_1^\infty {\frac{{dx}}{{{x^2} - 6x + 13}}}.\]
Solution.
First we complete the square in the denominator:
\[{x^2} - 6x + 13 = \left( {{x^2} - 6x + 9} \right) + 4 = {\left( {x - 3} \right)^2} + {2^2}.\]
Taking the limit, we obtain:
\[\int\limits_1^\infty {\frac{{dx}}{{{x^2} - 6x + 13}}} = \int\limits_1^\infty {\frac{{dx}}{{{{\left( {x - 3} \right)}^2} + {2^2}}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{{{\left( {x - 3} \right)}^2} + {2^2}}}} = \lim\limits_{n \to \infty } \left[ {\frac{1}{2}\arctan \frac{{x - 3}}{2}} \right]_1^n = \frac{1}{2}\lim\limits_{n \to \infty } \Bigl[ {\arctan \frac{{n - 3}}{2} - \arctan \left( { - 1} \right)} \Bigr] = \frac{1}{2}\left( {\frac{\pi }{2} - \left( { - \frac{\pi }{4}} \right)} \right) = \frac{1}{2} \cdot \frac{{3\pi }}{4} = \frac{{3\pi }}{8}.\]
The integral converges.
Example 9.
Determine whether the improper integral \[{\int\limits_{ - \infty }^\infty} {\frac{{dx}}{{{x^2} + 2x + 8}}}\] converges or diverges?
Solution.
We can write this integral as
\[I = \int\limits_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 2x + 8}}} = \int\limits_{ - \infty }^0 {\frac{{dx}}{{{x^2} + 2x + 8}}} + \int\limits_0^\infty {\frac{{dx}}{{{x^2} + 2x + 8}}} .\]
By the definition of an improper integral, we have
\[I = \int\limits_{ - \infty }^\infty {\frac{{dx}}{{{x^2} + 2x + 8}}}
= \int\limits_{ - \infty }^0 {\frac{{dx}}{{{x^2} + 2x + 8}}}
+ \int\limits_0^\infty {\frac{{dx}}{{{x^2} + 2x + 8}}}
= \lim\limits_{M \to - \infty } \int\limits_M^0 {\frac{{dx}}{{{x^2} + 2x + 8}}}
+ \lim\limits_{N \to \infty } \int\limits_0^N {\frac{{dx}}{{{x^2} + 2x + 8}}}
= \lim\limits_{M \to - \infty } \int\limits_M^0 {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + 7}}}
+ \lim\limits_{N \to \infty } \int\limits_0^N {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + 7}}}
= \lim\limits_{M \to - \infty } \left. {\left( {\frac{1}{{\sqrt 7 }}\arctan \frac{{x + 1}}{{\sqrt 7 }}} \right)} \right|_M^0
+ \lim\limits_{N \to \infty } \left. {\left( {\frac{1}{{\sqrt 7 }}\arctan \frac{{x + 1}}{{\sqrt 7 }}} \right)} \right|_0^N
= \frac{1}{{\sqrt 7 }}\left( {\arctan \frac{1}{{\sqrt 7 }} - \lim\limits_{M \to - \infty } \arctan \frac{{M + 1}}{{\sqrt 7 }}} \right)
+ \frac{1}{{\sqrt 7 }}\left( {\lim\limits_{N \to \infty } \arctan \frac{{N + 1}}{{\sqrt 7 }} - \arctan \frac{1}{{\sqrt 7 }}} \right)
= \cancel{\frac{1}{{\sqrt 7 }}\arctan \frac{1}{{\sqrt 7 }}} - \frac{1}{{\sqrt 7 }} \cdot \left( { - \frac{\pi }{2}} \right)
+ \frac{1}{{\sqrt 7 }} \cdot \frac{\pi }{2} - \cancel{\frac{1}{{\sqrt 7 }}\arctan \frac{1}{{\sqrt 7 }}}
= \frac{1}{{\sqrt 7 }} \cdot \frac{\pi }{2} + \frac{1}{{\sqrt 7 }} \cdot \frac{\pi }{2}
= \frac{\pi }{{\sqrt 7 }}.\]
As both the limits exist and are finite, the given integral converges.
Example 10.
Determine whether the integral \[\int\limits_0^\infty {\frac{{dx}}{{{x^2} + 2x + 2}}} \] converges or diverges?
Solution.
This improper integral has an infinite upper limit of integration. Hence, by definition,
\[\int\limits_0^\infty {\frac{{dx}}{{{x^2} + 2x + 2}}} = \lim\limits_{n \to \infty } \int\limits_0^n {\frac{{dx}}{{{x^2} + 2x + 2}}} = \lim\limits_{n \to \infty } \int\limits_0^n {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + 1}}} = \lim\limits_{n \to \infty } \left. {\arctan \left( {x + 1} \right)} \right|_0^n = \lim\limits_{n \to \infty } \left[ {\arctan \left( {n + 1} \right) - \arctan 1} \right] = \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}.\]
The integral converges and is equal to \(\frac{\pi }{4}.\)
Example 11.
Determine whether the integral \[{\int\limits_1^\infty} {{\frac{{\sin x}}{{\sqrt {{x^3}} }}} dx}\] converges or diverges?
Solution.
We can write the obvious inequality for the absolute values:
\[\left| {\frac{{\sin x}}{{\sqrt {{x^3}} }}} \right| \le \left| {\frac{1}{{\sqrt {{x^3}} }}} \right| = \left| {\frac{1}{{{x^{\frac{3}{2}}}}}} \right|.\]
It's easy to show that the integral \({\int\limits_1^\infty} {\left| {\frac{1}{{\sqrt {{x^3}} }}} \right|dx} \) converges (see also Example 1.) Indeed
\[
\int\limits_1^\infty {\left| {\frac{1}{{\sqrt {{x^3}} }}} \right|dx}
= \int\limits_1^\infty {\frac{{dx}}{{\sqrt {{x^3}} }}}
= \int\limits_1^\infty {\frac{{dx}}{{{x^{\frac{3}{2}}}}}}
= \int\limits_1^\infty {{x^{ - \frac{3}{2}}}dx}
= \lim\limits_{n \to \infty } \int\limits_1^n {{x^{ - \frac{3}{2}}}dx}
= \lim\limits_{n \to \infty } \left. {\left( {\frac{{{x^{ - \frac{3}{2} + 1}}}}{{ - \frac{3}{2} + 1}}} \right)} \right|_1^n
= - 2\lim\limits_{n \to \infty } \left. {\left( {\frac{1}{{\sqrt x }}} \right)} \right|_1^n
= - 2\lim\limits_{n \to \infty } \left( {\frac{1}{{\sqrt n }} - 1} \right)
= - 2\left( {0 - 1} \right) = 2.\]
Then we conclude that the integral \({\int\limits_1^\infty} {\left| {\frac{{\sin x}}{{\sqrt {{x^3}} }}} \right|dx} \) also converges by Comparison Test \(1.\) Hence, the given integral \({\int\limits_1^\infty} {{\frac{{\sin x}}{{\sqrt {{x^3}} }}} dx}\) converges (absolutely) by Comparison Test \(3.\)
Example 12.
Determine whether the integral \[{\int\limits_0^4} {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}}\] converges or diverges?
Solution.
There is a discontinuity in the integrand at \(x = 2,\) so that we must consider two improper integrals:
\[\int\limits_0^4 {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}} = \int\limits_0^2 {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}} + \int\limits_2^4 {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}} .\]
Using the definition of an improper integral, we obtain
\[
\int\limits_0^2 {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}} + \int\limits_2^4 {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}}
= \lim\limits_{\tau \to 0 + } \int\limits_0^{2 - \tau } {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}}
+ \lim\limits_{\tau \to 0 + } \int\limits_{2 + \tau }^4 {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}} .\]
For the first integral,
\[
\lim\limits_{\tau \to 0 + } \int\limits_0^{2 - \tau } {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}}
= \lim\limits_{\tau \to 0 + } \int\limits_0^{2 - \tau } {{{\left( {x - 2} \right)}^{ - 3}}d\left( {x - 2} \right)}
= \lim\limits_{\tau \to 0 + } \left[ {\frac{{{{\left( {x - 2} \right)}^{ - 3 + 1}}}}{{ - 3 + 1}}} \right]_0^{2 - \tau }
= - \frac{1}{2}\lim\limits_{\tau \to 0 + } \left. {\left[ {\frac{1}{{{{\left( {x - 2} \right)}^2}}}} \right]} \right|_0^{2 - \tau }
= - \frac{1}{2}\lim\limits_{\tau \to 0 + } \left[ {\frac{1}{{{{\left( {2 - \tau - 2} \right)}^2}}} - \frac{1}{{{{\left( {0 - 2} \right)}^2}}}} \right]
= - \frac{1}{2}\lim\limits_{\tau \to 0 + } \left( {\frac{1}{{{\tau ^2}}} - \frac{1}{4}} \right) = - \infty .\]
As it is divergent, the given integral \({\int\limits_0^4} {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}}\) is also divergent.
Example 13.
Determine for what values of \(k\) the integral \[{\int\limits_0^1} {\frac{{dx}}{{{x^k}}}}\;\left( {k \gt 0,k \ne 1} \right)\] converges?
Solution.
The integrand has discontinuity at the point \(x = 0,\) so that we can write the integral as
\[
\int\limits_0^1 {\frac{{dx}}{{{x^k}}}}
= \lim\limits_{\tau \to 0 + } \int\limits_\tau ^1 {\frac{{dx}}{{{x^k}}}}
= \lim\limits_{\tau \to 0 + } \int\limits_\tau ^1 {{x^{ - k}}dx}
= \lim\limits_{\tau \to 0 + } \left. {\left( {\frac{{{x^{ - k + 1}}}}{{ - k + 1}}} \right)} \right|_\tau ^1
= \frac{1}{{1 - k}} \cdot \lim\limits_{\tau \to 0 + } \left. {\left( {{x^{ - k + 1}}} \right)} \right|_\tau ^1
= \frac{1}{{1 - k}} \cdot \lim\limits_{\tau \to 0 + } \left( {{1^{ - k + 1}} - {\tau ^{ - k + 1}}} \right)
= \frac{1}{{1 - k}} \lim\limits_{\tau \to 0 + } \left( {1 - {\tau ^{1 - k}}} \right).\]
As you can see from this expression, there are \(2\) cases:
If \(0 \lt k \lt 1,\) then
\[\lim\limits_{\tau \to 0 +} {\tau ^{1 - k}} = 0\]
and the integral converges;
If \(k \gt 1,\) then
\[\lim\limits_{\tau \to 0 +} {\tau ^{1 - k}} = \lim\limits_{\tau \to 0 +} {\frac{1}{{{\tau ^{k - 1}}}}} = \infty\]
and the integral diverges.
Example 14.
Find the area above the curve \[y = \ln x\] in the lower half-plane between \(x = 0\) and \(x = 1.\)
Solution.
The given region is sketched in Figure \(7.\)
Figure 7. Since it is infinite, we calculate the improper integral to find the area:
\[\int\limits_0^1 {\ln xdx} = \lim\limits_{\tau \to 0 + } \int\limits_\tau ^1 {\ln xdx} .\]
Use integration by parts. Let \(u = \ln x,\) \(dv = dx.\) Then \(du = {\frac{{dx}}{x}},\) \(v = x.\)
Thus
\[
\lim\limits_{\tau \to 0 + } \int\limits_\tau ^1 {\ln xdx}
= \lim\limits_{\tau \to 0 + } \left[ {\left. {\left( {x\ln x} \right)} \right|_\tau ^1 - \int\limits_\tau ^1 {x\frac{{dx}}{x}} } \right]
= \lim\limits_{\tau \to 0 + } \left. {\left[ {x\ln x - x} \right]} \right|_\tau ^1
= \lim\limits_{\tau \to 0 + } \left[ {\left( {\ln 1 - 1} \right) - \left( {\tau \ln \tau - \tau } \right)} \right]
= \left( {0 - 1} \right) - \lim\limits_{\tau \to 0 + } \left[ {\tau \left( {\ln \tau - 1} \right)} \right]
= - 1 - \lim\limits_{\tau \to 0 + } \frac{{\ln \tau - 1}}{{\frac{1}{\tau }}}.\]
We can apply L'Hopital's rule to find the limit:
\[\lim\limits_{\tau \to 0 + } \frac{{\ln \tau - 1}}{{\frac{1}{\tau }}} = \lim\limits_{\tau \to 0 + } \frac{{\frac{1}{\tau }}}{{ - \frac{1}{{{\tau ^2}}}}} = - \lim\limits_{\tau \to 0 + } \frac{{{\tau ^2}}}{\tau } = - \lim\limits_{\tau \to 0 + } \tau = 0.\]
Hence, the improper integral is
\[\lim\limits_{\tau \to 0 + } \int\limits_\tau ^1 {\ln xdx} = - 1 - \lim\limits_{\tau \to 0 + } \frac{{\ln \tau - 1}}{{\frac{1}{\tau }}} = - 1 - 0 = - 1.\]
As you can see from the figure above, the required area is
\[S = \left| {\lim\limits_{\tau \to 0 + } \int\limits_\tau ^1 {\ln xdx} } \right| = 1.\]
Example 15.
Find the circumference of the unit circle.
Solution.
We calculate the length of the arc of the circle in the first quadrant between \(x = 0\) and \(x = 1\) and then multiply the result by \(4.\)
The equation of the circle centered at the origin is
\[{x^2} + {y^2} = 1.\]
Then the arc of the circle in the first quadrant (Figure \(8\)) is described by the function
\[y = \sqrt {1 - {x^2}} ,\;\;\; 0 \le x \le 1.\]
Figure 8. Find the derivative of the function:
\[y' = \frac{d}{{dx}}\sqrt {1 - {x^2}} = \frac{{ - \cancel{2}x}}{{\cancel{2}\sqrt {1 - {x^2}} }} = \frac{{ - x}}{{\sqrt {1 - {x^2}} }}.\]
Since the length of an arc is given by \(\int\limits_{x = \alpha }^{x = \beta } {\sqrt {1 + {{\left( {y'} \right)}^2}} dx},\) we obtain
\[\int\limits_0^1 {\sqrt {1 + {{\left( {\frac{{ - x}}{{\sqrt {1 - {x^2}} }}} \right)}^2}} dx} = \int\limits_0^1 {\sqrt {1 + \frac{{{x^2}}}{{1 - {x^2}}}} dx} = \int\limits_0^1 {\sqrt {\frac{{1 - \cancel{x^2} + \cancel{x^2}}}{{1 - {x^2}}}} dx} = \int\limits_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} .\]
Now we calculate the improper integral \(\int\limits_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}:\)
\[
\int\limits_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}
= \lim\limits_{\tau \to 0 + } \int\limits_0^{1 - \tau } {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}
= \lim\limits_{\tau \to 0 + } \left. {\left( {\arcsin x} \right)} \right|_0^{1 - \tau }
= \lim\limits_{\tau \to 0 + } \left[ {\arcsin \left( {1 - \tau } \right) - \arcsin 0} \right]
= \arcsin 1 - 0 = \frac{\pi }{2}.\]
Thus, the circumference of the unit circle is \(\frac{\pi }{2} \cdot 4 = 2\pi .\)
