We know how to integrate functions over intervals \([a,b]\). See also: convergence tests (in analogy with series). One can define \(\int_{a}^{\infty} f(x)\, dx\).
Definition (Convergence of an improper integral)
Let \(f:[a,+\infty] \to \mathbb{R}\) be Riemann integrable on \([a,b]\) for every \(b>a\). We say that \(\int_{a}^{\infty} f(x) \, dx\) converges if \(\lim_{ b \to \infty } \int_{a}^{b} f(x) \, dx\) exists.
Example
Consider \(\int_{1}^{\infty} \, \frac{1}{x^{p}}dx\) . Since
\[\int_{1}^{b} x^{-p} \, dx = \frac{x^{1-p}}{1-p} \biggr\rvert_{1}^{b } = \frac{b^{1-p}}{1-p} - \frac{1}{1-p}.\]Note that
\[-\frac{1}{1-p} \quad \text{if }p>1 \lim_{ b \to \infty } \frac{b^{1-p}}{1-p} - \frac{1}{1-p} = \begin{cases} -\frac{1}{1-p} \quad \text{if }p>1 \\ +\infty \quad \text{if }p <1 \end{cases}\]Moreover, if \(p=1\)
\[\int_{1}^{b} \frac{1}{x} \, dx = \ln b \to \infty.\]Lemma (The function must tend to zero)
Let \(f:[a,+\infty) \to \mathbb{R}\) be Riemann integrable on \([a,b]\) for every \(b>a\). If \(\lim_{ x \to \infty } f(x) = L \neq 0\), then \(\int_{a}^{\infty} f(x) \, dx\) diverges.
Proof: Assume first that \(L>0\) and take \(\varepsilon = \frac{L}{2} > 0\). By the definition of the limit, there exists \(M>a\) such that \(\lvert f(x)-L \rvert < \varepsilon\) whenever \(x>M\). In particular,
\[f(x) > L - \varepsilon = \frac{L}{2} > 0 \qquad \text{for every } x > M.\]Let \(b>M\). Splitting the integral at \(M\) and bounding \(f\) below by the constant \(L-\varepsilon\) on \([M,b]\),
\[\int_{a}^{b} f(x) \, dx = \int_{a}^{M} f(x) \, dx + \int_{M}^{b} f(x) \, dx \geq \int_{a}^{M} f(x) \, dx + (L-\varepsilon)(b-M).\]The term \(\int_{a}^{M} f(x) \, dx\) is a fixed real number and \(L-\varepsilon>0\), so the right-hand side tends to \(+\infty\) as \(b \to \infty\). Hence
\[\lim_{ b \to \infty } \int_{a}^{b} f(x) \, dx = +\infty,\]and therefore \(\int_{a}^{\infty} f(x) \, dx\) diverges.
If \(L<0\), apply the case already proved to \(-f\): since \(\lim_{ x \to \infty } (-f(x)) = -L > 0\), we obtain \(\lim_{ b \to \infty } \int_{a}^{b} (-f(x)) \, dx = +\infty\), that is, \(\lim_{ b \to \infty } \int_{a}^{b} f(x) \, dx = -\infty\), and the integral diverges as well.
Remark (a necessary Cauchy condition): Assume that \(\lim_{ b \to \infty } \int_{a}^{b} f(x) \, dx = I \in \mathbb{R}\). Given \(\varepsilon>0\) there exists \(M>0\) such that if \(b>M\), then
\[\left\lvert \int_{a}^{b} f(x) \, dx -I \right\rvert < \frac{\varepsilon}{2}.\]Take \(b_{2} > b_{1} > M\). Now,
\[\begin{aligned} \left\lvert \int_{b_{1}}^{b_{2}} f(x) \, dx \right\rvert &= \left\lvert \int_{a}^{b_{2}} f(x) \, dx - \int_{a}^{b_{1}} f(x) \, dx \right\rvert \\ &\leq \left\lvert \int_{a}^{b_{1}} f(x) \, dx - I \right\rvert + \left\lvert \int_{a}^{b_{2}} f(x) \, dx - I \right\rvert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{aligned}\]We conclude that \(\left\lvert \int_{b_{1}}^{b_{2}} f(x) \, dx \right\rvert < \varepsilon\) whenever \(b_{2}>b_{1}>M\); this is the converse of the Cauchy condition proved below.
Lemma (Reduction to sequences)
\(\int_{0}^{\infty} f(x)\, dx\) converges to \(L\) if, given \(\{ x_{n} \}_{n=1}^{\infty} \subseteq [a,+\infty]\) with \(x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \infty\), we have that
\[\lim_{ n \to \infty } \int_{a}^{x_{n}}f(x) \, dx = F(x_{n})\text{ exists and converges to } L.\]Lemma (Cauchy condition for improper integrals)
If, given \(\varepsilon>0\), there exists \(M>0\) such that \(\left\lvert \int_{c}^{d} f(x)\, dx \right\rvert < \varepsilon\) whenever \(c,d > M\), then \(\int_{0}^{\infty} f(x) \, dx\) converges.
Proof: Let \(\{ x_{n} \}_{n=1}^{\infty} \subseteq [a,+\infty]\) be such that \(x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \infty\). Take \(z_{n} = \int_{a}^{x_{n}} f(x)\, dx\). Let \(\varepsilon>0\); then there exists \(M \in \mathbb{N}\) such that \(\left\lvert \int_{c}^{d} f(x)\, dx \right\rvert < \varepsilon\) for all \(c>d\geq M\). Now, if we take \(x_{n} < x_{m}\), then
\[\lvert z_{n}-z_{m} \rvert = \left\lvert \int_{x_{n}}^{x_{m}} f(x) \, dx \right\rvert < \varepsilon.\]Let \(N \in \mathbb{N}\) be such that \(x_{k}\geq M\) for every \(k\geq N\), which we know exists because \(x_{n} \underset{n \rightarrow \infty}{\longrightarrow} \infty\). Hence, if \(n,m \geq N\), \(\left\lvert \int_{x_{n}}^{x_{m}} f(x) \, dx \right\rvert < \varepsilon\). Then \(z_{n}\) is Cauchy and converges. We will now prove that they all converge to the same limit. Let \(\{ y_{n} \}_{n=1}^{\infty} \subseteq [a, +\infty]\) be such that \(y_{n} \underset{n \rightarrow \infty}{\longrightarrow} \infty\). We must show that
\[\int_{a}^{x_{n}} f(x) \, dx - \int_{a}^{y_{n}} f(x) \, dx = \int_{\min\{ x_{n}, y_{n}\}}^{\max \{ x_{n},y_{n} \}} f(x) \, dx \underset{n \rightarrow \infty}{\longrightarrow} 0.\]Take \(N_{1}\) such that \(x_{n}, y_{n} \geq M\); then for \(n\geq N_{1}\),
\[\left\lvert \int_{\min\{ x_{n},y_{n} \}}^{\max\{ x_{n},y_{n} \}} f(x)\, dx \right\rvert < \varepsilon.\]Conclude that any two sequences converge to the same value.
Theorem (Absolute convergence)
If \(\int_{a}^{\infty} \lvert f(x) \rvert \, dx\) converges, then \(\int_{a}^{\infty} f(x) \, dx\) converges.
Proof: \(\int_{0}^{\infty} f(x) \, dx\) exists if and only if, given \(\varepsilon>0\), there exists \(M>0\) such that \(\left\lvert\int_{c}^{d} f(x) \, dx \right\rvert < \varepsilon\) whenever \(c,d \geq M\). Note that
\[\left\lvert \int_{c}^{d} f(x) \, dx \right\rvert < \int_{c}^{d} \lvert f(x) \rvert \, dx.\]Hence, if \(\int_{c}^{\infty} \lvert f(x) \rvert \, dx\) converges, then given \(\varepsilon>0\) there exists \(M > 0\) such that \(\int_{c}^{d} \lvert f(x) \rvert \, d < \varepsilon\) whenever \(c,d \geq M\). Therefore
\[\left\lvert \int_{c}^{d} f(x) \, dx \right\rvert \leq \int_{c}^{d} \lvert f(x) \rvert \, dx < \varepsilon.\]So \(\int_{a}^{\infty} f(x) \, dx\) converges.
Theorem (Comparison)
If there exists \(c \in [a, \infty)\) such that \(0 \leq f(x) \leq g(x)\) for every \(x \geq c\), then:
- \(\int_{a}^{\infty} g(x) \, dx\) converges if \(\int_{a}^{\infty} f(x) \, dx\) converges.
- \(\int_{a}^{\infty} f(x) \, dx\) diverges if \(\int_{a}^{\infty} g(x) \, dx\) diverges.
Proof of (1): Assume that \(0 \leq f(x) \leq g(x)\) for every \(x \geq c\), with \(a < c\). Then
\[0\leq \int_{c}^{d} f(x) \, dx \leq \int_{c}^{d} g(x) \, dx.\]Suppose \(\int_{a}^{\infty} g(x) \, dx\) converges. Then there exists \(M\geq 0\) such that for all \(\ell,q \geq M\),
\[0 < \int_{\ell}^{q} g(x) \, dx < \varepsilon\]Hence \(0<\int_{\ell}^{q} f(x) \, dx < \varepsilon\) and therefore it is Cauchy and converges. The proof of 2 is left as an exercise. One hint is to note that \(F(x) = \int_{a}^{u} f(u) \, du\) is increasing if \(f(x)\geq_{0}\).
Theorem (Limit comparison)
Let \(f:[a,\infty) \to \mathbb{R}\) and \(g:[a,+\infty] \to \mathbb{R}\) be such that \(0\leq f(x)\leq g(x)\) for \(x\geq c\), where \(c>a\).
- If \(\lim_{ n \to \infty } \frac{f(x)}{g(x)} = \ell \neq 0\), then \(\int_{a}^{\infty} f(x) \, dx\) converges if and only if \(\int_{a}^{\infty} g(x)\, dx\) converges.
- If \(\lim_{ n \to \infty } \frac{f(x)}{g(x)} = \infty\), we have that
- \(\int_{a}^{\infty} g(x) \, dx\) converges if \(\int_{a}^{\infty} f(x) \, dx\) converges.
- \(\int_{a}^{\infty} f(x) \, dx\) diverges if \(\int_{a}^{\infty} g(x) \, dx\) diverges.
Proof: Assume that \(\lim_{ n \to \infty } \frac{f(x)}{g(x)} = \ell \neq 0\). Given \(\varepsilon>0\), there exists \(M>0\) such that for every \(x \geq M\),
\[\ell - \varepsilon < \frac{f(x)}{g(x)} < \ell + \varepsilon.\]Hence, if we take \(\varepsilon < \ell\), then
\[0 \leq (l-\varepsilon) g(x) \leq f(x) \leq (l+\varepsilon) g(x).\]The result follows from the comparison theorem. On the other hand, if \(\lim_{ n \to \infty } \frac{f(x)}{g(x)} = +\infty\): given \(M>0\), there exists \(N>0\) such that \(\frac{f(x)}{g(x)} \geq M\) if \(x>N\). Then \(f(x) \geq Mg(x)\) if \(x\geq N\).
For integrals of the other kinds, compute the limit towards the problematic point.
Example
Let \(\Gamma(\alpha) = \int_{0}^{\infty} e^{-x} x^{\alpha-1} \, dx\) for \(\alpha \geq 1\). Note that
\[\lim_{ x \to \infty } \frac{e^{-x}x^{\alpha-1}}{e^{-x/2}} = \lim_{ x \to \infty } e^{-x/2} x^{\alpha-1} = 0.\]Since
\[\int_{0}^{a} e^{x/2} \, dx = -\frac{1}{2} e^{-x/2} \biggr\rvert_{a}^{0} = \frac{1}{2} - \frac{1}{2} e^{-a/2} \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{2},\]that is, \(\int_{0}^{\infty} e^{-x/2} \, dx\) converges, we conclude by limit comparison that \(\int_{0}^{\infty} e^{-x} x^{\alpha-1} \, dx\) converges.
Exercise
Show that \(\Gamma(1) = 1\) and that \(\Gamma(\alpha) = \alpha\Gamma(\alpha-1)\) for \(\alpha \geq 2\).
Theorem (Dirichlet)
Let \(f:[a,+\infty) \to \mathbb{R}\) and \(g:[a, +\infty) \to \mathbb{R}\) be functions such that
- There exists \(M>0\) such that \(\int_{a}^{b} f(x) \, dx \leq M\) for all \(a<b \in \mathbb{R}\).
- \(g\) is decreasing and \(\lim_{ x \to \infty } g(x) = 0\). Then \(\int_{a}^{\infty} f(x)g(x) \, dx\) converges.
Proof: Given \(\varepsilon>0\), there exists \(K\) such that \(0\leq g(x) < \frac{\varepsilon}{M}\) for \(x\geq K\). Moreover, given \(c,d\geq K\), there exists \(c_{1} \in \mathbb{R}\) such that
\[\int_{c}^{d} f(x)g(c) \, dx = g(c) \int_{c}^{c_{1}} f(x) \, dx,\]where \(c\leq c_{1} \leq d\). Then, for all \(d\geq c\geq k\)
\[\left\lvert \int_{c}^{d} f(x)g(x) \, dx \right\rvert = g(c) \left\lvert \int_{c}^{d} f(x) \, dx \right\rvert < \frac{\varepsilon}{M} \cdot M = \varepsilon,\]and therefore the integral converges.
Example
Consider \(\int_{0}^{\infty} \frac{\sin x}{x^{p}}\, dx\) for \(p>0\). Note that \(\lim_{ n \to \infty } \frac{1}{x^{p}} = 0\) and \(\frac{1}{x^{p}}\) is decreasing. Moreover,
\[\left\lvert \int_{a}^{b} \sin x \, dx \right\rvert = \left\lvert -\cos x \biggr\rvert_{a}^{b } \right\rvert = \lvert \cos b-\cos a \rvert \leq 2.\]Therefore it converges.
Example
Recall that \(\int_{1}^{\infty} \frac{\sin x}{x} \, dx\) converges by Dirichlet’s test. What happens with \(\int_{\pi}^{\infty} \left\lvert \frac{\sin x}{x} \right\rvert \, dx\)? Consider
\[\int_{\pi}^{k \pi} \left\lvert \frac{\sin x}{x} \right\rvert \, dx = \sum_{j=1}^{k-1} \int_{j \pi}^{(j+1)\pi} \left\lvert \frac{\sin x}{x} \right\rvert \, dx.\]Now,
\[\begin{aligned} \frac{2}{\pi(j+1)} =\frac{1}{(j+1)\pi} \int_{j \pi}^{(j+1) \pi} \left\lvert \sin x \right\rvert \, dx &\leq \int_{j \pi}^{(j+1)\pi} \left\lvert \frac{\sin x}{x} \right\rvert \, dx \\ &\leq \frac{1}{j\pi} \int_{j \pi}^{(j+1) \pi} \left\lvert \sin x \right\rvert \, dx = \frac{2}{j \pi}. \end{aligned}\]Then
\[\sum_{j=1}^{k-1} \frac{2}{\pi} \frac{1}{j+1} \leq \int_{\pi}^{k \pi} \frac{1}{(j+1)\pi} \int_{j \pi}^{(j+1) \pi} \left\lvert \frac{\sin x}{x} \right\rvert \, dx \leq \sum_{j=1}^{k-1} \frac{2}{\pi} \frac{1}{j}.\]Since \(\sum_{k=1}^\infty \frac{1}{n} = +\infty\), we get \(\lim_{ k \to \infty } \int_{0}^{k \pi} \left\lvert \frac{\sin x}{x} \right\rvert \, dx = \infty\).
Lemma (Abel’s test)
Let \(f:[a,+\infty) \to \mathbb{R}\) be such that
- \(\int_{a}^{\infty} f(x) \, dx\) converges.
Assume moreover that \(g:[a,+\infty) \to \mathbb{R}\)
- is monotone,
- \(\lim_{ x \to \infty } g(x) = \ell\).
Then \(\int_{0}^{\infty} f(x)g(x) \, dx\) converges.
Proof: Let \(\varepsilon>0\). Let \(c,d \geq M\). Then there exists \(c_{1} \in [c,d]\) such that
\[\int_{c}^{d} f(x)g(x) \, dx = g(c) \int_{c}^{c_{1}} f(x) \, dx + g(d) \int_{c_{1}}^{d} f(x) \, dx,\]where \(M\) is such that \(\ell-\varepsilon < g(x) < \ell+\varepsilon\) if \(x\geq M\). Then there exists \(K\) such that \(\lvert g(x) \rvert \leq K\) for \(x \geq M\). Let \(M_{1}\) be such that
\[\left\lvert \int_{p}^{q} f(x) \, dx \right\rvert < \frac{\varepsilon}{2K}\]for \(p,q \geq M_{1}\geq M\). Then
\[\left\lvert \int_{c}^{d} f(x)g(x) \, dx \right\rvert \leq \lvert g(c) \rvert \left\lvert \int_{c}^{c_{1}} f(x) \, dx \right\rvert + \lvert g(d) \rvert \left\lvert \int_{c_{1}}^{d} f(x) \, dx \right\rvert \leq K \frac{\varepsilon}{2K} + K \frac{\varepsilon}{2K} = \varepsilon.\]Conclude that it is Cauchy and therefore converges.
Example
For \(\int_{1}^{\infty} \frac{\sin x \arctan x}{x} \, dx\), take \(g(x) = \arctan x\) and \(f(x)=\frac{\sin x}{x}\). Since \(g(x)\) is monotone, \(\lim_{ x \to \infty } \arctan x = \frac{\pi}{2}\) and \(\int_{1}^{\infty} \frac{\sin x}{x} \, dx\) converges, the given integral converges.
Other types of improper integrals
From -\(\infty\) to \(b\)
Let \(f:(-\infty,b] \to \mathbb{R}\) be such that \(f:[a,b] \to \mathbb{R}\) is Riemann integrable for every \(a<b\). If \(\lim_{ a \to -\infty } \int_{a}^{b} f(x) \, dx\) exists, we say that \(\int_{-\infty}^{b} f(x) \, dx\) converges \(\int_{0}^{1} \, dx\)
Note that \(\int_{a}^{b} f(x) \, dx = \int_{-b}^{-a} f(u) \, du\), so \(\int_{-\infty}^{b}f(x) \, dx\) converges if and only if \(\int_{-b}^{\infty} f(-x) \, dx\) converges.
From \(-\infty\) to \(+\infty\)
Let \(f:(-\infty,+\infty) \to \mathbb{R}\). Then \(\int_{-\infty}^{+\infty} f(x) \, dx\) converges if and only if \(\int_{0}^{+\infty} f(x)\, dx\) and \(\int_{-\infty}^{0} f(x) \, dx\) converge.
Example
Consider the following improper integral: \(\int_{-\infty}^{+\infty} x^{2}e^{-\lvert x \rvert} \, dx\). Note that
\[\int_{0}^{+\infty} x^{2}e^{-\lvert x \rvert } \, dx = \int_{0}^{+\infty} x^{2}e^{-x}\, dx.\]On the other hand,
\[\int_{-\infty}^{0} x^{2} e^{-\lvert x \rvert } \, dx = \int_{-\infty}^{0} x^{2} e^{x} \, dx = \int_{+\infty}^{0} (-u)^{2}e^{-u} - \, du = \int_{0}^{+\infty} (-u)^{2} e^{-u} \, du.\]Now,
\[\lim_{ x \to \infty } \frac{ x^{2}e^{-x}}{e^{-x/2}} = 0.\]Since \(\int_{0}^{+\infty} e^{-x/2} \, dx = -\frac{1}{2} e^{-x/2} \biggr\rvert_{0}^{+\infty } = \frac{1}{2}\), we have that \(\int_{0}^{\infty} x^{2} e^{-x} \, dx\) converges.
Example
Consider \(\int_{\alpha}^{1} \frac{1}{x^{p}} \, dx\) with \(\alpha>0\). This function is not Riemann integrable on \([0,1]\) if \(p>0\), but does \(\lim_{ \alpha \to 0^{+} } \int_{\alpha}^{1} \frac{1}{x^{p}} \, dx\) exist?
Note that
\[\int_{\alpha}^{1} \frac{1}{x^{p}} \, dx = \begin{cases} \frac{1}{1-p} x^{1-p} \biggr\rvert_{\alpha}^{1} \quad\text{if } p\neq 1, \\ \ln x \biggr\rvert_{\alpha}^{1} = \ln(1) - \ln(\alpha) \quad \text{if }p=1. \end{cases}\]Then
\[\lim_{ \alpha \to 0^{+} } \int_{\alpha}^{1} \frac{1}{x^{p}} \, dx = \begin{cases} \frac{1}{1-p} \quad\text{if } 0<p<1, \\ +\infty \quad \text{otherwise}. \end{cases}\]Making the change of variable \(x = \frac{1}{u} \implies dx = -\frac{du}{u^{2}}\), we have that
\[\int_{\alpha}^{1} \frac{1}{x^{p}} \, dx = \int_{1}^{1/\alpha} \frac{1}{\left( \frac{1}{u} \right)^{p}}\, \frac{du}{u^{2}} = \int_{1}^{1/\alpha} \frac{1}{u^{2-p}} \, du .\]So \(\int_{\alpha}^{1} \frac{1}{x^{p}} \, dx\) converges if and only if \(\int_{1}^{+\infty} \frac{1}{u^{2-p}} \, dx\) converges.