Example. We want to compute the area of the unit disc. The function \(f: [-1, 1] \to \mathbb{R}\), \(f(x) = \sqrt{1-x^2}\), describes the upper half of the disc, so we want to compute
\(2 \int_{-1}^1 \sqrt{1-x^2} dx.\)
We observe that \(1 - \cos^2(t) = \sin^2(t)\) and \(\cos(2t) = \cos^2(t) - \sin^2(t) = 1 - 2\sin^2(t)\), so we make the change of variables \(x = g(t) = \cos(t)\):
\(2 \int_{-1}^1 \sqrt{1-x^2} dx = 2 \int_{\pi}^0 \sqrt{1-\cos^2(t)} (-\sin(t)) dt\)
\(= 2 \int_0^\pi \sin^2(t) dt = \int_0^\pi (1 - \cos(2t)) dt\)
\(= \pi - \int_0^\pi \cos(2t) dt \xrightarrow{\text{Thm 48}} \pi - \frac{1}{2} \sin(2t) \big|_0^\pi = \pi.\)
Finally, we extend the definition of the Riemann integral to unbounded functions and functions with unbounded domains. Such integrals are called "improper integrals".
Definition. i) Let \(-\infty < a < b \le +\infty\). A function \(f: [a, b) \to \mathbb{R}\) is improperly Riemann-integrable on \([a, b)\) if, for every \(c \in (a, b)\), \(f \in \mathcal{R}([a, c])\)
and the limit \(\lim_{c \to b-} \int_a^c f(x) dx =: \int_a^{b-} f(x) dx\)
exists.
ii) Let \(-\infty \le a < b < +\infty\). A function \(f: (a, b] \to \mathbb{R}\) is improperly Riemann-integrable on \((a, b]\) if, for every \(c \in (a, b)\), \(f \in \mathcal{R}([c, b])\)
and the limit \(\lim_{c \to a+} \int_c^b f(x) dx =: \int_{a+}^b f(x) dx\)
exists.
iii) Let \(-\infty \le a < b \le +\infty\). A function \(f: (a, b) \to \mathbb{R}\) is improperly Riemann-integrable if there exists a \(c \in (a, b)\) such that f is improperly Riemann-integrable on \((a, c]\) and on \([c, b)\). In this case we set
\(\int_{a+}^{b-} f(x) dx := \int_{a+}^c f(x) dx + \int_c^{b-} f(x) dx.\)
Remarks. i) If \(f \in \mathcal{R}([a, b])\), then it is not difficult to show that f is improperly Riemann-integrable on \((a, b)\) with
\(\int_{a+}^{b-} f(x) dx = \int_a^b f(x) dx.\)
ii) When there is no ambiguity, we omit the plus and minus signs at the limits of the integral.
Examples. i) Let \(\alpha \in \mathbb{R}\), \(f: (0, 1] \to \mathbb{R}\), \(f(x) = \frac{1}{x^\alpha}\).
For every \(c \in (0, 1)\), we have
\(\int_c^1 \frac{1}{x^\alpha} dx \xrightarrow{\text{Thm 48}} \begin{cases} \log x \big|_c^1 = -\log c, & \alpha = 1 \\ \frac{1}{1-\alpha} x^{1-\alpha} \big|_c^1 = \frac{1}{1-\alpha} (1 - c^{1-\alpha}), & \alpha \neq 1 \end{cases}\)
\(\implies \lim_{c \to 0+} \int_c^1 \frac{1}{x^\alpha} dx = \begin{cases} +\infty, & \alpha \ge 1 \\ \frac{1}{1-\alpha}, & \alpha < 1 \end{cases}\)
\(\implies\) f is improperly Riemann-integrable on \((0, 1]\) if and only if \(\alpha < 1\). In this case we have
\(\int_0^1 \frac{1}{x^\alpha} dx = \frac{1}{1-\alpha}.\)
ii) Let \(\alpha \in \mathbb{R}\), \(f: [1, +\infty) \to \mathbb{R}\), \(f(x) = \frac{1}{x^\alpha}\).
Using the same arguments as in i), we obtain that f is improperly Riemann-integrable if and only if \(\alpha > 1\). In this case we have
\(\int_1^{+\infty} \frac{1}{x^\alpha} dx = \frac{1}{\alpha-1}.\)
As improper integrals do not necessarily converge, there are several convergence tests (similar to those for series).
Theorem 51. (Comparison Test for Integrals)
Let \(f, g: [a, b) \to \mathbb{R}\) be two functions such that \(f, g \in \mathcal{R}([a, c])\) for every \(c \in (a, b)\), and suppose that \(0 \le f(x) \le g(x)\) on \([a, b)\).
If \(\int_a^{b-} g(x) dx\) exists, then \(\int_a^{b-} f(x) dx\) exists, and in this case we have \(\int_a^{b-} f(x) dx \le \int_a^{b-} g(x) dx.\)
Remark An analogous result exists for integrals of the type \(\int_{a+}^b f(x) dx.\)
Proof. For \(c \in (a, b)\) we consider the functions
\(F(c) := \int_a^c f(x) dx, \quad G(c) := \int_a^c g(x) dx.\)
\(\implies (f, g \ge 0, \text{Thm 47, ii)})\) F and G are increasing on \([a, b)\) and
\(0 \le F(c) \le G(c) \to \int_a^{b-} g(x) dx, \ c \to b-.\)
\(\implies F(c)\) is increasing and bounded on \([a, b)\).
From this we can conclude that \(\lim_{c \to b-} F(c)\) exists (compare to Thm 6) and is \(\le \int_a^{b-} g(x) dx.\)
Example. For \(x > 0\), we consider \(\int_0^{+\infty} e^{-t} t^{x-1} dt\).
\(\alpha) \int_0^1 e^{-t} t^{x-1} dt\) exists: For \(x, t > 0\), we have
\(0 < e^{-t} t^{x-1} \le t^{x-1}=\frac{1}{t^{1-x}}\), and we know that
that \(\int_0^1 \frac{1}{t^{1-x}} dt\) exists (\(1-x < 1\)).
\(\beta) \int_1^{+\infty} e^{-t} t^{x-1} dt\) exists: For \(x > 0\), we have
\(\lim_{t \to +\infty} e^{-t} t^{x+1} = 0\), which implies that the function \(e^{-t} t^{x+1}\) is bounded on \([1, +\infty)\) (for fixed \(x > 0\)).
\(\implies \exists M > 0 \ \forall t \ge 1 : e^{-t} t^{x+1} \le M\)
\(\implies \forall t \ge 1 : e^{-t} t^{x-1} \le \frac{M}{t^2}.\)
We know that \(\int_1^{+\infty} \frac{M}{t^2} dt = M \int_1^{+\infty} \frac{1}{t^2} dt\) exists.
From \(\alpha)\) and \(\beta)\) it follows that \(\int_0^{+\infty} e^{-t} t^{x-1} dt\) exists for every \(x > 0\). The function \(\Gamma: (0, \infty) \to \mathbb{R}\), \(\Gamma(x) = \int_0^{+\infty} e^{-t} t^{x-1} dt\), is called "Gamma function" and plays a key role in many areas of mathematics.