5.4 Improper Integrals
We now extend the Riemann integral \(\int ^b_af(x)\,dx\) to unbounded intervals, and to functions that are unbounded near a point of a bounded interval.
Definition 5.4.1. If the interval of integration is unbounded (\(b = \infty \) or \(a = -\infty \), or both) and \(f\) is integrable on every bounded closed subinterval, then \(\int ^b_af(x)\,dx\) is called an improper integral of the first kind. For example, \[\int ^{\infty }_1\frac {1}{x^2}\, dx,\qquad \int ^0_{-\infty }e^x\,dx,\qquad \int ^{\infty }_0\sin x\, dx .\]
Definition 5.4.2. If \(a\) and \(b\) are real numbers and \(f\) is unbounded near some point of \([a,b]\), then \(\int ^b_af(x)\,dx\) is called an improper integral of the second kind. For example, \[\int ^2_0\frac {x}{1 - x}\, dx,\qquad \int ^1_0\frac {\sin x}{x^{3/2}}\,dx,\qquad \int ^1_0\ln x\, dx .\]
Definition 5.4.3. If the interval of integration is unbounded and \(f\) is also unbounded near some point of it, then \(\int ^b_af(x)\,dx\) is called an improper integral of the third kind. For example, \[\int ^{\infty }_0\frac {1}{x^2 + x^{1/2}}\, dx .\]
Definition 5.4.4. Let \(f\) be Riemann integrable on \([a,s]\) for every \(s\geq a\), and let \(F(s) = \int ^s_af(x)\,dx\). We say that \(\int ^{\infty }_af(x)\,dx\) converges to \(A\), and write \[\int ^{\infty }_af(x)\,dx = \lim _{s\rightarrow \infty } F(s) = A,\] if this limit exists and is finite. Otherwise \(\int ^{\infty }_af(x)\,dx\) is said to diverge. Integrals \(\int _{-\infty }^b f\) are treated in the same way. For \(\int _{-\infty }^\infty f\) we require both \(\int _{-\infty }^0 f\) and \(\int _0^\infty f\) to converge, and add them.
Similarly, if \(f\) is integrable on \([a,s]\) for every \(s<b\) but unbounded near \(b\), we define \(\int _a^bf = \lim _{s\rightarrow b^-}\int _a^sf\) when this limit exists, and likewise at the left end point. An integral of the third kind is split into pieces of the first and second kinds, each of which must converge.
Example 5.4.5. \(\int ^{\infty }_1\frac {1}{x^2}\, dx\) converges, since \[F(s) = \int ^s_1 \frac {1}{x^2}\, dx = 1 - \frac {1}{s} \longrightarrow 1 \quad \text {as } s\longrightarrow \infty .\] Hence \(\int ^{\infty }_1\frac {1}{x^2}\, dx = 1\).
Example 5.4.6. \(\int ^{\infty }_1\frac {1}{\sqrt {x}}\, dx\) diverges, since \[F(s) = \int ^s_1\frac {1}{\sqrt {x}}\,dx = 2(\sqrt {s} - 1) \longrightarrow \infty \quad \text {as } s \longrightarrow \infty .\]
Example 5.4.7. Consider \(\int ^{\infty }_0e^{-\alpha x}\, dx\), where \(\alpha \) is a real number. For \(\alpha \neq 0\), \[F(s) = \int ^s_0e^{-\alpha x}\, dx = \frac {1}{\alpha } (1 - e^{-\alpha s}) .\]
- (i)
- If \(\alpha > 0\), then \(e^{-\alpha s}\longrightarrow 0\) and \(F(s)\longrightarrow \frac {1}{\alpha }\).
- (ii)
- If \(\alpha < 0\), then \(e^{-\alpha s}\longrightarrow \infty \) and \(F(s)\longrightarrow \infty \).
- (iii)
- If \(\alpha = 0\), then \(F(s) = s\longrightarrow \infty \).
So the integral converges (to \(\frac {1}{\alpha }\)) if \(\alpha > 0\) and diverges if \(\alpha \leq 0\).
Example 5.4.8. Let \(a>0\). Show that \(\int _a^{\infty }\frac {1}{x^p}\,dx\) converges for \(p>1\) and diverges for \(p\leq 1\).
Solution. For \(p\neq 1\), \[F(s) = \int ^s_a\frac {dx}{x^p} = \frac {1}{1 - p}\left (s^{1-p} - a^{1-p}\right ).\] If \(p>1\) then \(s^{1-p}\longrightarrow 0\), and \(F(s)\longrightarrow \frac {a^{1-p}}{p-1}\). If \(p<1\) then \(s^{1-p}\longrightarrow \infty \) and \(F(s)\longrightarrow \infty \). When \(p = 1\), \(F(s) = \ln s - \ln a\longrightarrow \infty \). Hence the integral converges exactly when \(p>1\).
Theorem 5.4.9 (Change of Variables). Let \(\Phi : [\alpha , \beta ]\longrightarrow \mathbb {R}\) have a continuous derivative on \([\alpha ,\beta ]\), and let \(f\) be continuous on an interval containing \(\Phi ([\alpha ,\beta ])\). Then \[\int ^{\Phi (\beta )}_{\Phi (\alpha )}f(x)\, dx = \int ^{\beta }_{\alpha }f(\Phi (t))\, \Phi '(t)\, dt .\]
Proof. Let \(J\) be an interval containing \(\Phi ([\alpha ,\beta ])\) on which \(f\) is continuous, fix \(u_0\in J\), and let \(G(u) = \int _{u_0}^uf(x)\,dx\) for \(u\in J\). By the First Fundamental Theorem, \(G' = f\) on \(J\). By the Chain Rule, \(H = G\circ \Phi \) is differentiable on \([\alpha ,\beta ]\) with \[H'(t) = G'(\Phi (t))\,\Phi '(t) = f(\Phi (t))\,\Phi '(t),\] which is continuous. By the Second Fundamental Theorem, \[\int ^{\beta }_{\alpha }f(\Phi (t))\, \Phi '(t)\, dt = H(\beta ) - H(\alpha ) = G(\Phi (\beta )) - G(\Phi (\alpha )) = \int ^{\Phi (\beta )}_{\Phi (\alpha )}f(x)\, dx .\] Note that \(\Phi \) need not be monotone. □
Theorem 5.4.10. Suppose that \(\int ^{\infty }_af(x)\, dx\) and \(\int ^{\infty }_ag(x)\, dx\) both converge. Then \(\int ^{\infty }_a(f(x)\pm g(x))\, dx\) converges and \[\int ^{\infty }_a\bigl (f(x) \pm g(x)\bigr )\, dx = \int ^{\infty }_af(x)\, dx \pm \int ^{\infty }_ag(x)\, dx .\]
Proof. Let \(F(s) = \int _a^s f\longrightarrow A\) and \(G(s) = \int _a^sg\longrightarrow B\) as \(s\longrightarrow \infty \). By the linearity of the integral on \([a,s]\), \[\int ^s_a(f\pm g) = F(s)\pm G(s)\longrightarrow A\pm B .\] Hence \(\int ^{\infty }_a(f\pm g)\) converges, to \(A\pm B\). □
Example 5.4.11. Decide whether the integrals \(\int _0^1\frac {dx}{\sqrt {x}}\), \(\int _0^1\frac {dx}{x}\) and \(\int _0^1\ln x\,dx\) converge, and evaluate those that do.
Solution. Each integrand is unbounded near \(0\), so these are improper integrals of the second kind, computed as limits as \(s\longrightarrow 0^+\).
\(\int _s^1x^{-1/2}dx = 2 - 2\sqrt {s}\longrightarrow 2\), so the first integral converges to \(2\).
\(\int _s^1x^{-1}dx = -\ln s\longrightarrow \infty \), so the second diverges.
Since \((x\ln x - x)' = \ln x\), \(\int _s^1\ln x\,dx = -1 - s\ln s + s\longrightarrow -1\), because \(s\ln s\longrightarrow 0\). So \(\int _0^1\ln x\,dx = -1\).
In general \(\int _0^1x^{-p}\,dx\) converges exactly when \(p<1\): the opposite of the condition at \(\infty \).
Example 5.4.12 (Comparison test). Let \(0\leq f\leq g\) on \([a,\infty )\), with both integrable on every \([a,s]\). Show that if \(\int _a^\infty g\) converges then so does \(\int _a^\infty f\). Deduce that \(\int _1^\infty e^{-x^2}dx\) and \(\int _1^\infty \frac {\sin x}{x^2}\,dx\) converge.
Solution. \(F(s) = \int _a^sf\) is increasing in \(s\), because \(f\geq 0\), and bounded above by \(\int _a^\infty g\), because \(f\leq g\). An increasing function bounded above has a finite limit as \(s\longrightarrow \infty \) (the proof is that of the monotone convergence of sequences), so \(\int _a^\infty f\) converges.
For \(x\geq 1\), \(0<e^{-x^2}\leq e^{-x}\), and \(\int _1^\infty e^{-x}dx = e^{-1}\) converges. So \(\int _1^\infty e^{-x^2}dx\) converges.
For the second integral, \(\left |\frac {\sin x}{x^2}\right |\leq \frac {1}{x^2}\) and \(\int _1^\infty \frac {dx}{x^2}\) converges. Apply the test to \(|f|\) and to \(|f| - f\) (both lie between \(0\) and \(\frac {2}{x^2}\)); since \(f = |f| - (|f| - f)\), the integral of \(f\) converges as a difference of two convergent integrals.
Example 5.4.13. Show that \(\int _0^\infty \sin x\,dx\) diverges, although \(\int _0^s\sin x\,dx\) stays bounded.
Solution. \(F(s) = \int _0^s\sin x\,dx = 1 - \cos s\), which takes every value in \([0,2]\) as \(s\) runs through each interval of length \(2\pi \). So \(F(s)\) has no limit as \(s\longrightarrow \infty \), and the integral diverges. Bounded partial integrals are not enough; the limit must exist.
Example 5.4.14. Use the Change of Variables Theorem to evaluate \(\int _0^1x\sqrt {1-x^2}\,dx\) and \(\int _0^{\pi /2}\sin ^3t\cos t\,dt\).
Solution. For the first, take \(\Phi (x) = 1 - x^2\) on \([0,1]\), with \(\Phi ' = -2x\) continuous. Then \(x\sqrt {1-x^2} = -\frac {1}{2}\sqrt {\Phi (x)}\,\Phi '(x)\), and \[\int _0^1x\sqrt {1-x^2}\,dx = -\frac {1}{2}\int _{\Phi (0)}^{\Phi (1)}\sqrt {u}\,du = -\frac {1}{2}\int _1^0\sqrt {u}\,du = \frac {1}{2}\cdot \frac {2}{3} = \frac {1}{3} .\] For the second, take \(\Phi (t) = \sin t\), so \(\Phi ' = \cos t\), \(\Phi (0) = 0\) and \(\Phi \left (\frac {\pi }{2}\right ) = 1\): \[\int _0^{\pi /2}\sin ^3t\cos t\,dt = \int _0^1u^3\,du = \frac {1}{4} .\]
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