1.3 Convergence and Divergence of Improper Integrals
We can test for convergence or divergence of an improper integral without solving the integral.
We have convergence theorems for improper integrals which mimic the convergence theorems for
series.
Theorem 1.3.1. An integral \[\int _a^{\infty } f(t)\, dt\] of the first kind with \(f(t)\geq 0\) for all \(t\) is convergent if there is a constant \(M>0\) such that \[\int _a^xf(t)\, dt \leq M\hspace {0.3cm}\text {when}\hspace {0.3cm} x > a.\] The value of the integral is then not more than \(M\).
Proof. Let \[F(x) = \int _a^xf(t)\, dt,\] then \[\int _a^{\infty } f(t)\, dt = \lim _{x\rightarrow \infty } F(x).\] Let \(f(t)\geq 0\) for \(t\geq a\). then \[F(x_2) - F(x_1) = \int _{x_1}^{x_2} f(t)\, dt\, \geq 0\] if \(a \leq x_1 \leq x_2.\)
This shows that \(F(x)\) is non decreasing for \(x\geq a\). There are two possibilities: either \(F(x)\) is bounded above or it is not. If it is bounded above, then there exist \(M > 0\) such that \(F(x) \leq M\) for all \(x\geq a\).
For this case, \(F(x)\) approaches a finite limit as \(x \rightarrow \infty \). Hence, \[\int _a^{\infty } f(t)\, dt\] converges.
If \(F(x)\) is not bounded above, for each \(M>0\) there is an \(x\) such that \(M<F(x)\) since \(F(x)\) is non decreasing as \(x\) increases, \(F(x) \rightarrow \infty \) as \(x \rightarrow \infty \). □
Theorem 1.3.2 (Comparison Test for Improper Integrals). Let \[\int _a^{\infty } f(x)\, dx\] and \[\int _b^{\infty } g(x)\, dx\] be two improper integrals of the first kind with non negative integrals, and suppose that \(f(x) \leq g(x)\) for all \(x\) beyond a certain value \(x = c\). Then if \[\int _b^{\infty } g(x)\, dx\] is convergent, so is \[\int _a^{\infty } f(x)\, dx.\] And if \[\int _a^{\infty } f(x)\, dx\] is divergent, so is \[\int _b^{\infty } g(x)\, dx.\]
Proof. The convergence or divergence of the integrals is not affected if we replace \(a\) or \(b\) with \(c\).
We then have that if \(x>c\) \[\int _c^xf(t)\, dt \leq \int _c^xg(t) \, dt.\] The result then follows from 1.3.1 □
Example 1.3.3. Use the comparison test to show that \[\int _0^{\infty } \frac {1}{\sqrt {1 + x^3}}\, dx\] is convergent.
Working. \[\frac {1}{\sqrt {1 + x^3}} \leq \frac {1}{\sqrt {x^3}} = \frac {1}{x^{\frac {3}{2}}}.\] Now \[\int _1^{\infty } \frac {1}{x^{\frac {3}{2}}}\, dx\] is a convergent \(p\)-integral with \(p = \frac {3}{2}.\) Hence, by comparison, \[\int _0^{\infty } \frac {1}{\sqrt {1 + x^3}}\, dx.\] □
Example 1.3.4. Show that \[\int _0^{\infty } \frac {1}{(1 + x^3)^{\frac {1}{3}}}\, dx\] is divergent, by comparison with the integral \[\int _0^{\infty } \frac {1}{1 + x}\, dx.\]
Working. \begin {align*} \int _0^{\infty }\frac {1}{x + 1}\, dx & = \lim _{t\rightarrow \infty }\int _0^t\frac {1}{x + 1}\, dx\\ & = \lim _{t\rightarrow \infty } \left [\ln |x + 1|\right ]_0^t\\ & = \lim _{t\rightarrow \infty } \left [\ln (t + 1) - \ln (0 + 1)\right ]\\ & = \infty , \end {align*}
diverges.
Now we can confirm that \[\frac {1}{x + 1} \leq \frac {1}{\left (1 + x^3\right )^{\frac {1}{3}}}\] for \(x>0\) since \[(x + 1)^3 = 1 + 3x + 3x^2 + x^3 \geq 1 + x^3.\] Thus by comparison test \[\int _0^{\infty } \frac {1}{(1 + x^3)^{\frac {1}{3}}}\] diverges. □
Working. \[\int _0^{\infty } e^{-x^2}\, dx = \int _0^1e^{-x^2}\, dx + \int _1^{\infty }e^{-x^2}\, dx.\] Now if \(x \geq 1, \, x^2 \geq x, \, \, -x^2 \leq -x,\) and so \(\, e^{-x^2} \leq e^{-x}\). Now \[\int ^{\infty }_0e^{-x} \, dx = \lim _{t\rightarrow \infty } \int _0^te^{-x}\, dx = \lim _{t \rightarrow \infty }\left [-e^{-x}\right ]_0^t = 1.\] Hence, by comparison test \[\int _0^{\infty } e^{-x^2}\, dx \] is also convergent. □
Exercise 1.3.6. Use the comparison test for integrals to determine whether the integral converges or diverges.
- (a).
- \(\displaystyle {\int _1^{\infty } \frac {\cos ^2 x}{1 + x^2}}\, dx\hspace {0.3cm}\) Hint: \(\cos ^2 x \leq 1\)
- (b).
- \(\displaystyle {\int _1^{\infty } \frac {1}{x + e^{2x}}\, dx}\)
- (c).
- \(\displaystyle {\int _1^{\infty }\frac {\sqrt {1 + \sqrt {x}}}{\sqrt {x}}}\, dx\)
- (d).
- \(\displaystyle {\int _0^{\frac {\pi }{2}} \frac {1}{x\,\sin x}}\, dx\)
- (e).
- \(\displaystyle {\int _0^1\frac {e^{-x}}{\sqrt {x}}}\, dx\)
- (f).
- \(\displaystyle {\int _1^{\infty }\frac {1}{(1 + x)\sqrt {x}}}\, dx\)
- (g).
- \(\displaystyle {\int _2^{\infty } \frac {1}{x\, \sqrt {1 + x^2}}}\, dx\)
- (h).
- \(\displaystyle {\int _4^{\infty } \frac {\sqrt {x + 1}}{x + \sqrt {x}}}\, dx\)
- (i).
- \(\displaystyle {\int _2^{\infty } \frac {x^2 + 4x + 4}{(\sqrt {x} - 1)^3\, \sqrt {x^3 - 1}}}\, dx\)
- (j).
- \(\displaystyle {\int _1^{\infty } \frac {1 + e^{-x}}{x}}\, dx\)
Exercise 1.3.7. Evaluate the integral \[\int _0^{\infty } \frac {1}{\sqrt {x}\, \sqrt {1 + x}}\, dx\] by expressing it as a sum of a type I and a type II integral.
Theorem 1.3.8. Suppose \[\int _a^{\infty } f(x)\, dx\] and \[\int _b^{\infty } g(x)\, dx\] are type I(one) with positive integrals, and suppose that \[\lim _{x\rightarrow \infty } \frac {f(x)}{g(x)} = L\] exists and is non-zero. Then either both integrals are convergent or both are divergent.
Proof. Suppose \[\lim _{x\rightarrow \infty } \frac {f(x)}{g(x)} = L \neq 0.\] For \(x\) large enough, we have that \[\frac {1}{2} L < \frac {f(x)}{g(x)} < \frac {3}{2}L.\] From this we get that \(g(x) < \frac {2}{L}\) and also \(f(x) < \frac {3}{2}L\, g(x)\). □
- (a).
- Show that
\[\int _1^{\infty } \frac {x^2}{2x^4 - x + 1}\, dx\]
is convergent.
Working. Let \begin {align*} f(x) & = \frac {x^2}{2x^4 - x + 1}\\ & = \frac {\frac {x^2}{x^4}}{\frac {2x^4}{x^4} - \frac {x}{x^4} + \frac {1}{x^4}}\\ & = \frac {\frac {1}{x^2}}{2 - \frac {1}{x^3} + \frac {1}{x^4}}\, \, , \hspace {0.3cm} \text {set}\, \hspace {0.2cm}g(x) = \frac {1}{x^2} \end {align*}
Then \[\frac {f(x)}{g(x)} = \frac {1}{2 - \frac {1}{x^3} + \frac {1}{x^4}}\] so \[\lim _{x\rightarrow \infty } \frac {f(x)}{g(x)} = \frac {1}{2}.\] Now \[\int _1^{\infty } \frac {1}{x^2}\, dx\] is convergent.
Therefore, \[\int _1^{\infty } f(x)\, dx \] must also converge. □
- (b).
- The integral
\[\int _0^{\infty } \frac {x^3}{16 + x^4}\, dx\]
diverges .
Working. \[f(x) = \frac {x^3}{16 + x^4} = \frac {\frac {1}{x}}{\frac {16}{x^4} + 1}.\] Then set \[g(x) = \frac {1}{x}.\] □
Exercise 1.3.10. Check for convergence or divergence.
- (a).
- \(\displaystyle {\int _0^{\infty } \frac {x}{(1 + x)^3}}\, dx\)
- (b).
- \(\displaystyle {\int _1^{\infty } \frac {\sqrt {x}}{(1 + x)^2}}\, dx\)
Corollary 1.3.11. If \(L=0\) in 1.3.8 and if \[\int _b^{\infty }g(x)\, dx\] is convergent, so is \[\int _a^{\infty } f(x)\, dx.\] Also, if \(L = +\infty \), then if \[\int _b^{\infty } g(x) \, dx \] is divergent, so is \[\int _a^{\infty }f(x)\, dx.\]
Proof. \[L = \lim _{x\rightarrow \infty } \frac {f(x)}{g(x)},\] if \(L = 0\), then \[\frac {f(x)}{g(x)} < 1\] eventually, so \(f(x) \leq g(x)\).
If \(L = \infty \), then \[\frac {f(x)}{g(x)} > 1\] eventually, and so \(g(x) < f(x)\). □
- (a).
- The integral
\[\int _1^{\infty } x^{\alpha }\, e^{-x}\, dx\]
is convergent for all \(\alpha \in \mathbb {R}\).
Working. We apply the limit test using the convergent integral \[\int _1^{\infty } \frac {1}{x^2}\, dx.\] So \(f(x) = x^{\alpha }\, e^{-x}\, ,, \hspace {0.3cm} g(x) = \frac {1}{x^2}\), so \[\frac {f(x)}{g(x)} = \frac {x^{\alpha }\, e^{-x}}{x^{-2}} = \frac {x^{\alpha + 2}}{e^{x}}\, \longrightarrow \, 0\] as \(x \rightarrow \infty \) by repeated application of L’Hopital’s rule. Hence, the integral converges. □
- (b).
- The integral
\[\int _2^{\infty } \frac {dx}{(\log x)^p},\]
\(p>0\) is divergent.
Working. Let \(f(x) = \frac {1}{(\log x)^p}\) and \(g(x) = \frac {1}{x}\). Then \begin {align*} \lim _{x\rightarrow \infty } \frac {f(x)}{g(x)} & = \lim _{x\rightarrow \infty } \frac {x}{(\log x)^p}\\ & = \lim _{x\rightarrow \infty } \frac {1}{p(\log x)^{p-1}\, \cdot \frac {1}{x}}\hspace {0.6cm} \text {L'Hospital's rule}\\ & = \lim _{x \rightarrow \infty } \frac {x}{p(\log x)^{p-1}}\\ & = \lim \frac {x}{p\, (p-1)\, (p-2)\, \cdots \, (p-p+1)\, (\log x)^{p-p+1}}\\ & = \infty . \end {align*}
When \(p-p + 1 < 0\) since \[\int _2^{\infty }\frac {1}{x}\, dx\] diverges so is \[\int _2^{\infty } \frac {1}{(\log x)^p}\, dx \hspace {0.4cm} p > 0.\] □
- (a).
- For what values of \(p\) is the integral
\[\int _a^{\infty }\frac {dx}{x\, (\log x)^p}\]
converges, \(a > 1\).
Working. Let \(u = \log x\), \(\, du = \frac {1}{x}\, dx\) then \begin {align*} \int _a^{\infty } \frac {dx}{x\,(\log x)^p} & = \int ^{\infty }_{\log a} \frac {1}{u^{p}}\, du\\ & = \lim _{t\rightarrow \infty } \left [\frac {u^{-p + 1}}{1 - p}\right ]^t_{\log a}\\ & = \lim _{t\rightarrow \infty } \frac {t^{1 - p}\, (\log a)^{1-p}}{1 - p}\\ & = \frac {(\log a)^{1-p}}{p - 1}\hspace {0.5cm} \text {when}\,\,\, 1 - p < 0 \,\, \text {or}\,\,\,p>1. \end {align*}
So \[\int _a^{\infty } \frac {1}{x\, (\log x)^p}\, dx\] converges for \(p >1\). □
- (b).
- Determine whether the following converge or diverge:
- (i).
- \(\displaystyle {\int _1^{\infty } \frac {1}{\sqrt {x^2 + 1}\, [\log (1 + x)]^2}}\, dx\)
- (ii).
- \(\displaystyle {\int _6^{\infty } \frac {x + 1}{(x^2 - 2)\left (\log \frac {x}{2} - 1\right )}}\, dx\)
Working.
- (i).
-
\[ \frac {1}{\sqrt {x^2 + 1}\, [\log (1 + x)]^2} \leq \frac {1}{\sqrt {x^2}\, [\log (x)]^2} = \frac {1}{x[\log x]^2},\]
whose integral converges with \(p=2\).
So the integral converges.
- (ii).
- \[ \frac {x + 1}{(x^2 - 2)\left (\log \frac {x}{2} - 1\right )} \geq \frac {x + 1}{x^2 \log x} = \frac {1}{x\log x} + \frac {1}{x^2\log x}.\] Now \[\int _6^{\infty } \frac {1}{x\log x}\, dx\] diverges, since \(p = 1\). Hence, the integral also diverges.
Exercise 1.3.14. Determine whether each of the following integrals converges or diverges.
- (a).
- \(\displaystyle {\int _0^{\infty } \frac {x}{(1 + x)^3}\, dx}\) (C)
- (b).
- \(\displaystyle {\int _0^{\infty }e^{-x}\, x^2 (\log x)^3}\, dx\) (C)
- (c).
- \(\displaystyle {\int _0^{\infty } \left [\frac {\pi }{2} - \tan ^{-1}x\right ]}\, dx\) (D)
- (d).
- \(\displaystyle {\int _1^{\infty } \frac {\frac {\pi }{2} - \tan ^{-1} x}{x}}\, dx\) (D)
- (e).
- \(\displaystyle {\int _0^{\infty } \frac {1}{\sqrt {x}\, (\log x)^3}}\, dx\) (D)
- (f).
- \(\displaystyle {\int _4^{\infty } \frac {1}{(x - 1)\, \log (x - 2)\, \log x}}\, dx \hspace {0.5cm} \text {(D)}\)
- (g).
- \(\displaystyle {\int _3^{\infty } \frac {1}{\sqrt {x + 1}\, \log x\, [\log (\log x)]^2}}\, dx\hspace {0.5cm}\text {(C)}\)
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.