5.4 Improper Integrals
We want now to extend the definition of the Riemann integral \(\hspace {0.1cm}\displaystyle {\int ^b_af(x)\,dx}\hspace {0.1cm}\) to integrals over unbounded intervals
as well as for functions which are unbounded at a point in the finite interval.
Definition 5.4.1. Let \(f\) be bounded on the interval \([a,b]\). If either \(b = \infty \) or \(a = -\infty \), then the integral \(\hspace {0.1cm}\displaystyle {\int ^b_af(x)\,dx}\hspace {0.1cm}\) is called an improper integral of the first kind.
\[\text {e.g}\hspace {0.5cm}\int ^{\infty }_0\dfrac {1}{x^2}\, dx\hspace {1cm},\hspace {1cm} \int ^0_{-\infty }e^x\,dx\hspace {1cm}, \hspace {1cm} \int ^{\infty }_0\sin x\, dx\]
Definition 5.4.2. If \(a\) and \(b\) are real numbers and \(f\) is unbounded at some point in \([a,b]\), then the
integral \(\, \displaystyle {\int ^b_af(x)\,dx}\,\) is called an improper integral of the second kind
\[\text {e.g}\hspace {0.5cm} \int ^2_0\dfrac {x}{1 - x}\, \, dx\hspace {0.5cm},\hspace {1cm}\int ^1_0\dfrac {\sin x}{x^{3/2}}\,dx\hspace {0.5cm},\hspace {1cm}\int ^0_{-2}\dfrac {x^{1/2}}{1 + x}\, dx\]
Definition 5.4.3. If the integral \(\, \displaystyle {\int ^b_af(x)\,dx}\,\) is defined over an unbounded interval and \(f(x)\) is also unbounded
at some point in the unbounded interval, then \(\, \displaystyle {\int ^b_af(x)\,dx}\,\) is said to be an improper integral of the third
kind
\[\text {e.g}\hspace {1cm}\int ^{\infty }_0\dfrac {1}{x^2 + x^{1/2}}\, dx\]
Definition 5.4.4. Let \(f\) be Riemann integrable for every \(s\geq a\) and let \(\, F(s) = \displaystyle {\int ^s_af(x)\,dx}\,\) be its indefinite integral. Then
\(\,\displaystyle {\int ^{\infty }_af(x)\,dx}\,\) is said to be convergent to \(A\) if \(\, F(s) \longrightarrow A\). Thus
\[\int ^{\infty }_af(x)\,dx = \lim _{s\rightarrow \infty }\, F(s) = A\]
If \(\, \displaystyle {\int ^{\infty }_af(x)\,dx}\,\) does not converge then \(\, \displaystyle {\int ^{\infty }_af(x)\,dx}\,\) is said to be divergent.
\(\displaystyle {\int ^{\infty }_1\dfrac {1}{x^2}\, dx}\,\) is convergent since \(\, F(s) = \displaystyle {\int ^s_1 \dfrac {1}{x^2}\, dx = 1 - \dfrac {1}{s}}\,\) so that \(\, \displaystyle {\lim _{s\rightarrow \infty }\Bigg (1 - \dfrac {1}{s}\Bigg )} = 1\)
Hence \(\, \displaystyle {\int ^{\infty }_1\dfrac {1}{x^2}\,\, dx} = 1\)
\(\displaystyle {\int ^{\infty }_0\dfrac {1}{\sqrt {x}}\, dx}\,\) is divergent since \(\, F(s) = \displaystyle {\int ^s_0\frac {1}{\sqrt {x}}\,\,dx = 2\Big (\sqrt {s} - 1\Big )}\,\) so that \(\, \lim \limits _{s\rightarrow \infty }2\Big (\sqrt {s} - 1\Big )\,\) is infinity.
Example 5.4.7. consider the integral \(\displaystyle {\int ^{\infty }_0e^{-\alpha \, x}\, dx}\,\) where \(\alpha \) is real number. Then \[F(s) = \int ^s_0e^{-\alpha \, x}\, dx = \dfrac {1}{\alpha }\, \Big (1 - e^{-\alpha \, s}\Big )\]
- (i).
- if \(\alpha > 0\) then \(\hspace {0.3cm}\displaystyle {\lim _{s \rightarrow \infty }\,\frac {1}{\alpha }\,\Big (1 - e^{-\alpha \, s}\Big ) = \dfrac {1}{\alpha }}\)
- (ii).
- if \(\alpha < 0\) then \(\hspace {0.3cm}\displaystyle {\lim _{s \rightarrow \infty }\,\frac {1}{\alpha }\,\Big (1 - e^{-\alpha \, s}\Big ) = \infty }\)
Therefore, the integral \(\, \displaystyle {\int _0^{\infty }e^{-\alpha \,x}\, dx}\,\) converges if \(\alpha > 0\) and diverges in \(\alpha < 0\).
Example 5.4.8. Show that the integral \(\, \displaystyle {\int _a^{\infty }\frac {1}{x^p}\,dx}\hspace {0.3cm}x>a > 0\)
- (i).
- convergent for \(p>1\)
- (ii).
- divergent for \(p\leq 1\)
Solution. The function \(\, f(x) = \dfrac {1}{x^p}\,\) is continuous for any \(x> a\) and \[F(s) = \int ^s_a\dfrac {1}{x^p}\, \, dx = \dfrac {1}{1 - p}\Bigg [\dfrac {1}{s^{p - 1}} \, -\, \dfrac {1}{a^{p-1}}\Bigg ]\]
Hence \(\hspace {0.2cm}\displaystyle {\lim _{s\rightarrow \infty } F(s) = \dfrac {-1}{\big (1 - p\big )\, a^{p - 1}}}\,\) if \(p>1\) and \(\, \lim \limits _{s\rightarrow \infty } F(s) = \infty \, \) if \(p<1\).
When \(p = 1\) we have \(\, \displaystyle {F(s) = \int ^s_a\dfrac {1}{x}\, dx = \ln (s) - \ln (a) = \ln \big (\dfrac {s}{a}\big )}\,\) so that \(\, \lim \limits _{s\rightarrow \infty } F(s) = \infty \).
Hence the integral converges for \(P>1\) and diverges for \(p\leq 1\).
Theorem 5.4.9 (Change of Variables). Let \(f\in R[a,b]\). Let \(\Phi :\, [\alpha , \beta ]\longrightarrow [a,b]\,\) be differentiable, strictly monotone function
on \([\alpha , \beta ]\ni \, \Phi '\in R[a,b]\). Then
\[\int ^b_af(x)\, dx = \int ^{\beta }_{\alpha }f\big (\Phi (t)\big )\, \Phi '(t)\, dt\]
Proof. We assume that \(\Phi \) is increasing function on \([\alpha ,\beta ]\). Let \(\Phi = \big \{\alpha = t_0,\, t_1,\, \cdots \cdots ,\, t_n = \beta \big \}\,\) be the partition of \([\alpha ,\beta ]\) and \(\,P = \big \{ a = x_0,\, x_1,\, \cdots \cdots ,\, x_n = b\big \}\,\) be the
corresponding partition of \([a,b]\,\ni \, x_i= \Phi (t_i)\, , \, i = 0,\, 1,\, 2\, \cdots \cdots ,\, n\). Then the Mean Value Theorem implies that
\[\Delta x_r = x_r - x_{r - 1} = \Phi (t_r) - \Phi (t_{r - 1}) = \Phi (\eta _r)\, \Delta t_r\]
where \(\, \eta _r \in [t_{r-1},t_r]\, \, r = 1,\, 2, \, \cdots \cdots , n\).
Let \(\, \zeta _r = \Phi (\eta _r)\, ,\hspace {0.2cm} r = 1,\, 2,\, \cdots \cdots , n\).
\[\sum f(\zeta _r) \, \Delta x_r = \sum f\big (\Phi (\eta _r)\big )\, \Phi '(\eta _r)\, \Delta t_r\hspace {0.3cm}\cdots \cdots \cdots \,\hspace {0.3cm} (1)\]
Since \(\Phi \) is differentiable on \([\alpha ,\beta ]\) it is uniformly continuous on \([\alpha ,\beta ]\) and consequently \(\, \begin {vmatrix} \Phi \\ \end {vmatrix}\longrightarrow 0\,\) as \(\, \begin {vmatrix} P\\ \end {vmatrix}\longrightarrow 0\).
Now let \(\, \begin {vmatrix} P\\ \end {vmatrix}\longrightarrow 0\,\) so that \(\, \displaystyle {\sum f(\zeta _r)\, \Delta x_r \, \longrightarrow \, \, \int ^b_af(x)\, dx}\,\) and
\(\displaystyle {\sum f\Big (\Phi (\eta _r)\Big )\, \Phi '(\eta _r)\, \Delta t_r \longrightarrow \int ^{\beta }_{\alpha } f\big [\Phi (t)\big ]\, \Phi '(t)\, dt}\)
Express (1) then gives \(\, \displaystyle {\int ^b_a f(x)\, dx \, = \, \int ^{\beta }_{\alpha } f\big (\Phi (t)\big )\, \Phi '(t)\, dt}\)
Theorem also holds for monotone decreasing function with minor adjustment in the above
argument.
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If \(\hspace {0.2cm}\displaystyle {\int ^{\infty }_af(x)\, dx}\hspace {0.2cm}\) and \(\hspace {0.2cm}\displaystyle {\int ^{\infty }_ag(x)\, dx} \hspace {0.2cm}\) are both convergent, then \(\hspace {0.2cm}\displaystyle {\int ^{\infty }_a\big [f(x)\pm g(x)\big ]\, dx}\hspace {0.2cm}\) is convergent and
\[\int ^{\infty }_a\big [f(x) \pm g(x)\big ]\, dx = \int ^{\infty }_af(x)\, dx + \int ^{\infty }_ag(x)\, dx\]
Proof. Since \(\hspace {0.1cm}\displaystyle {\int ^{\infty }_af(x)\, dx}\hspace {0.1cm}\) and \(\hspace {0.1cm}\displaystyle {\int ^{\infty }_ag(x)\, dx}\hspace {0.1cm}\) are both convergent, we have \(\lim \limits _{s\rightarrow \infty } F(s)\hspace {0.1cm}\) and \(\hspace {0.1cm}\lim \limits _{s\rightarrow \infty } G(s) \hspace {0.1cm}\) exists. Let \(\lim \limits _{s\rightarrow \infty } F(s) = A\hspace {0.1cm}\) and \(\lim \limits _{s\rightarrow \infty } G(s) = B.\hspace {0.1cm}\) Then \begin {align*} \int ^{\infty }_a\big [f(x)\pm g(x)\big ]\, dx & = \lim _{s\rightarrow \infty }\, \int ^s_a\big [f(x)\pm g(x)\big ]\, dx\\\\ & =\lim _{s\rightarrow \infty }\, \Bigg [\int ^s_af(x)\, dx \, \pm \, \int ^s_ag(x)\, dx\Bigg ]\\\\ & =\lim _{s\rightarrow \infty }\, \int ^s_af(x)\, dx \, \pm \,\lim _{s\rightarrow \infty }\, \int ^s_ag(x)\, dx\\\\ & = A\, \pm \, B \end {align*}
This proves that \(\hspace {0.2cm}\displaystyle {\int ^{\infty }_a\big [f(x)\,\pm \,g(x)\big ]\, dx}\hspace {0.1cm}\) is convergent and
\[\hspace {0.2cm}\displaystyle {\int ^{\infty }_a\big [f(x)\,\pm \,g(x)\big ]\, dx = \int ^{\infty }_af(x)\, dx + \int ^{\infty }_ag(x)\, dx}\]
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