Riemann Integrals of Real Functions#

Definition: partition

A partition of a compact interval \([a,b]\) is a set

\[P = \{x_0, x_1, \dotsc, x_n\}\]

with \(a = x_0 \lt x_1 \lt \cdots \lt x_n = b\).

Definition: Darboux Sums

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function which is bounded on some compact interval \([a,b] \subseteq \mathcal{D}\) and let \(P = \{x_0, x_1, \dotsc, x_n\}\) be a partition of \([a,b]\).

The lower Darboux sum of \(f\) with respect to \(P\) is defined using the infima of \(f\), on each interval \([x_{i-1}; x_i]\):

\[L_{f, P} \overset{\text{def}}{=} \sum_{i=1}^n (x_i - x_{i-1}) \inf_{x \in [x_{i-1}; x_i]} f(x)\]

The upper Darboux sum of \(f\) with respect to \(P\) is defined using the suprema of \(f\), on each interval \([x_{i-1}; x_i]\):

\[U_{f, P} \overset{\text{def}}{=} \sum_{i=1}^n (x_i - x_{i-1}) \sup_{x \in [x_{i-1}; x_i]} f(x)\]

Definition: Darboux Integrals

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function which is bounded on a compact interval \([a,b] \subseteq \mathcal{D}\).

The lower Darboux integral of \(f\) on \([a,b]\) is the supremum of all its lower Darboux sums on \([a,b]\):

\[L_f \overset{\text{def}}{=} \sup \{L_{f, P} \mid P \text{ is a partition of } [a,b]\}\]

The upper Darboux integral on \([a,b]\) of \(f\) is the infimum of all its upper Darboux sums on \([a,b]\):

\[U_f \overset{\text{def}}{=} \inf \{U_{f, P} \mid P \text{ is a partition of } [a,b]\}\]

Definition: Riemann Integral

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function which is bounded on a compact interval \([a,b] \subseteq \mathcal{D}\).

We say that \(f\) is Riemann-integrable on \([a,b]\) if its lower Darboux integral and upper Darboux integral on \([a,b]\) are equal. This common value is the Riemann integral of \(f\) on \([a,b]\).

Notation

The Riemann integral of \(f\) on \([a,b]\) is denoted by

\[\int_a^b f(x) \mathop{\mathrm{d}x},\]

where \(x\) is just a label.

We additionally use the following notation to simplify the wording of some theorems:

\[\int_b^a f(x) \mathop{\mathrm{d}x} = -\int_a^b f(x) \mathop{\mathrm{d}x}\]

\[\int_c^c f(x) \mathop{\mathrm{d}x} = 0 \qquad \forall c \in [a,b]\]

Definition: Local Riemann Integrability

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) is locally Riemann-integrable on an arbitrary interval \(I \subseteq \mathcal{D}\) if it is Riemann-integrable on every compact interval contained in \(I\).

Theorem: Linearity of the Riemann Integral

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions.

If \(f\) and \(g\) are Riemann-integrable on a compact interval \([a,b] \subseteq \mathcal{D}_f \cap \mathcal{D}_g\), then so is \(\lambda f + \mu g\) for all \(\lambda,\mu \in \mathbb{R}\) with

\[\int_a^b \lambda f(x) + \mu g(x) \mathop{\mathrm{d}x} = \lambda \int_a^b f(x) \mathop{\mathrm{d}x} + \mu \int_a^b g(x) \mathop{\mathrm{d}x}.\]

Proof

TODO

Theorem: Domain Additivity for Riemann Integrals

Let \([a, b] \subset \mathbb{R}\) be a compact interval and let \(c \in (a,b)\).

A real function \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) (where \(\mathcal{D} \supset [a,b]\)) is Riemann-integrable on \([a,b]\) if and only if it is Riemann-integrable on \([a,c]\) and \([c, b]\). Moreover:

\[\int_a^b f(x) \mathop{\mathrm{d}x} = \int_a^c f(x) \mathop{\mathrm{d}x} + \int_c^b f(x) \mathop{\mathrm{d}x}\]

Proof

TODO

Theorem: Riemann Integral under Alterations

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions, let \([a,b] \subseteq \mathcal{D}_f \cap \mathcal{D}_g\) be a compact interval and \(X = \{x \in [a, b] \mid f(x) \ne g(x)\}\).

If \(f\) is Riemann-integrable on \([a,b]\) and \(X\) is finite, then \(g\) is also Riemann-integrable on \([a,b]\) with

\[\int_a^b f(x) \mathop{\mathrm{d}x} = \int_a^b g(x) \mathop{\mathrm{d}x}.\]

Proof

TODO

Theorem: Inequality \(\implies\) Riemann Integral Inequality

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions which are Riemann-integrable on a compact interval \([a,b] \subseteq \mathcal{D}_f \cap \mathcal{D}_g\).

If \(f(x) \le g(x)\) for all \(x \in [a,b]\), then

\[\int_a^b f(x) \mathop{\mathrm{d}x} \le \int_a^b g(x) \mathop{\mathrm{d}x}.\]

Proof

TODO

The Mean Value Theorem for Riemann Integrals

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions and let \([a,b] \subseteq \mathcal{D}_f \cap \mathcal{D}_g\) be a compact interval.

If \(f\) is continuous on \([a,b]\) and \(g\) is Riemann-integrable on \([a,b]\) and does not change sign on \([a,b]\), then the product \(fg\) is Riemann-integrable on \([a,b]\) and there exists some \(\xi \in [a,b]\) such that

\[\int_a^b f(x) g(x) \,\mathrm{d}x = f(\xi)\int_a^b g(x) \,\mathrm{d}x\]

Tip: \(g(x) = 1\)

In the case of \(g(x) = 1\) for all \(x\), we get

\[\int_a^b f = f(\xi)(b - a).\]

Proof

Since \(f\) is continuous on \([a,b]\), the extreme value theorem tells us that \(f\) attains a minimum \(m\) and a maximum \(M\) on \([a,b]\):

\[m \le f(x) \le M \qquad \forall x \in [a,b]\]

Since \(g\) does not change sign on \([a,b]\), there are two cases which we need to consider:

(I) \(g(x) \ge 0\) for all \(x \in [a,b]\);
(II) \(g(x) \le 0\) for all \(x \in [a,b]\).

Case (I):

Multiplying the inequality for \(f\) by \(g\), we obtain the following:

\[mg(x) \le f(x)g(x) \le Mg(x) \qquad \forall x \in [a,b]\]

Since \(f\) is continuous on \([a,b]\) and \(g\) is Riemann-integrable on \([a,b]\), we know that \(fg\) is Riemann-integrable on \([a,b]\). Combining this with the above inequality, we get:

\[\int_a^b mg(x)\,\mathrm{d}x \le \int_a^b f(x)g(x) \,\mathrm{d}x \le \int_a^b Mg(x) \,\mathrm{d}x\]

\[m\int_a^b g(x)\,\mathrm{d}x \le \int_a^b f(x)g(x) \,\mathrm{d}x \le M\int_a^b g(x) \,\mathrm{d}x\]

If \(\int_a^b g(x)\,\mathrm{d}x = 0\), we have \(0 \le \int_a^b f(x)g(x)\,\mathrm{d}x \le 0\) and so \(\int_a^b f(x)g(x) \,\mathrm{d}x = 0\). Therefore, \(\int_a^b f(x)g(x)\,\mathrm{d}x = f(\xi) \cdot \int_a^b g(x)\,\mathrm{d}x\) for any \(\xi \in [a,b]\), since \(0 \cdot f(\xi) = 0\).

If \(\int_a^b g(x)\,\mathrm{d}x \gt 0\), we can divide the inequality by it:

\[m \le \frac{\int_a^b f(x)g(x)\,\mathrm{d}x}{\int_a^b g(x)\,\mathrm{d}x} \le M\]

Since \(f\) is continuous on \([a,b]\), the intermediate value theorem tells us that \(f\) attains every value between its minimum \(m\) and maximum \(M\), i.e. there exists some \(\xi \in [a,b]\) such that

\[f(\xi) = \frac{\int_a^b f(x)g(x)\,\mathrm{d}x}{\int_a^b g(x)\,\mathrm{d}x}\]

Multiplying by \(\int_a^b g(x)\,\mathrm{d}x\), we obtain:

\[\int_a^b f(x) g(x) \,\mathrm{d}x = f(\xi)\int_a^b g(x) \,\mathrm{d}x\]

Case (II):

Multiplying the inequality for \(f\) by \(g\), we obtain the following:

\[mg(x) \ge f(x)g(x) \ge Mg(x) \qquad \forall x \in [a,b]\]

Since \(f\) is continuous on \([a,b]\) and \(g\) is Riemann-integrable on \([a,b]\), we know that \(fg\) is Riemann-integrable on \([a,b]\). Combining this with the above inequality, we get:

\[\int_a^b mg(x)\,\mathrm{d}x \ge \int_a^b f(x)g(x) \,\mathrm{d}x \ge \int_a^b Mg(x) \,\mathrm{d}x\]

\[m\int_a^b g(x)\,\mathrm{d}x \ge \int_a^b f(x)g(x) \,\mathrm{d}x \ge M\int_a^b g(x) \,\mathrm{d}x\]

If \(\int_a^b g(x)\,\mathrm{d}x = 0\), we have \(0 \ge \int_a^b f(x)g(x)\,\mathrm{d}x \ge 0\) and so \(\int_a^b f(x)g(x) \,\mathrm{d}x = 0\). Therefore, \(\int_a^b f(x)g(x) \,\mathrm{d}x = f(\xi) \cdot \int_a^b g(x)\,\mathrm{d}x\) for any \(\xi \in [a,b]\), since \(0 \cdot f(\xi) = 0\).

If \(\int_a^b g(x)\,\mathrm{d}x \lt 0\), we can divide the inequality by it:

\[m \le \frac{\int_a^b f(x)g(x)\,\mathrm{d}x}{\int_a^b g(x)\,\mathrm{d}x} \le M\]

Since \(f\) is continuous on \([a,b]\), the intermediate value theorem tells us that \(f\) attains every value between its minimum \(m\) and maximum \(M\), i.e. there exists some \(\xi \in [a,b]\) such that

\[f(\xi) = \frac{\int_a^b f(x)g(x)\,\mathrm{d}x}{\int_a^b g(x)\,\mathrm{d}x}\]

Multiplying by \(\int_a^b g(x)\,\mathrm{d}x\), we obtain:

\[\int_a^b f(x) g(x) \,\mathrm{d}x = f(\xi)\int_a^b g(x) \,\mathrm{d}x\]

The Fundamental Theorem of Calculus (Part I)

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function .

If \(f\) is Riemann-integrable on a compact interval \([a,b] \subseteq \mathcal{D}\), then the function \(F: [a,b] \to \mathbb{R}\) defined as

\[F(x) \overset{\text{def}}{=} \int_a^x f(t) \mathop{\mathrm{d}t}\]

is continuous. Moreover, if \(f\) is continuous at some \(c \in (a,b)\), then \(F\) is an antiderivative of \(f\) at \(c\), i.e. \(F'(c) = f(c)\).

Proof

TODO

The Fundamental Theorem of Calculus (Part II)

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(F: \mathcal{D}_F \subseteq \mathbb{R} \to \mathbb{R}\) be real functions.

If \(f\) is Riemann-integrable on a compact interval \([a,b] \subseteq \mathcal{D}_f \cap \mathcal{D}_F\), \(F\) is continuous on \([a,b]\) and is an antiderivative of \(f\) on \((a,b)\), then

\[\int_a^b f(x) \mathop{\mathrm{d}x} = F(b) - F(a).\]

Notation

The expression \(F(b) - F(a)\) is usually shortened to just \(F(x)\big|_a^b\).

Proof

TODO

Theorem: Riemann Integration by Parts

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions and let \([a,b] \subseteq \mathcal{D}_f \cap \mathcal{D}_g\) be a compact interval.

If \(f\) and \(g\) are continuous on \([a,b]\), differentiable on \((a,b)\) and if \(f'\) and \(g'\) can be extended to be Riemann-integrable on \([a,b]\), then

\[\int_a^b f(x) g'(x) \mathop{\mathrm{d}x} = f(b)g(b) - f(a)g(a) - \int_a^b f'(x)g(x) \mathop{\mathrm{d}x}.\]

Proof

TODO

Example: \(\int_0^{\pi} x \sin x \mathop{\mathrm{d}x}\)

We want to find the following Riemann-integral:

\[\int_0^{\pi} x \sin x \mathop{\mathrm{d}x}\]

We have

\[\int_0^{\pi} x \sin x\mathop{\mathrm{d}x} = \int_0^{\pi} f(x) g'(x)\]

with \(f(x) = x\), \(f'(x) = 1\), \(g'(x) = \sin x\) and \(g(x) = - \cos x\).

Therefore:

\[\begin{aligned}\int_{0}^{\pi}x\sin x\mathop{\mathrm{d}x} & =\int_0^\pi f(x)g'(x) \mathop{\mathrm{d}x} \\& = f(x)g(x)|_0^\pi - \int_0^\pi f'(x)g(x) \mathop{\mathrm{d}x} \\& =\left. -x\cos x \right|_0^\pi+\int_0^\pi\cos x \mathop{\mathrm{d}x} \\ & = \left.-x\cos x\right|_0^\pi+\sin x|_0^\pi \\ & =-\pi\cos\pi+\sin\pi-\sin0=\pi.\end{aligned}\]

Theorem: Riemann Integration by Substitution

Let \(f: \mathcal{D}_f \subseteq \mathbb{R} \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R} \to \mathbb{R}\) be real functions and let \([a,b] \subseteq \mathcal{D}_g\) be a compact interval.

If \(g\) is continuous on \([a,b]\) and is differentiable on \((a,b)\) such that \(g'\) can be extended to be Riemann-integrable on \([a,b]\) and if \(f\) is continuous on \(g([a,b])\), then the function \((f\circ g)g'\) is Riemann-integrable on \([a,b]\) with

\[\int_a^b f(g(x))g'(x) \mathop{\mathrm{d}x} = \int_{g(a)}^{g(b)} f(u) \mathop{\mathrm{d}u}\]

Example: \(\int_а^b f(cx+r)\, \mathrm{d}x\)

We can use this theorem to find a formula for Riemann integrals of the following form:

\[\int_a^b f(cx + r) \,\mathrm{d}x\]

If we define \(g: \mathbb{R} \to \mathbb{R}\) as \(g(x) = cx + r\), then the above Riemann integral becomes:

\[\int_a^b f(g(x)) \,\mathrm{d}x\]

We see that \(g\) is polynomial and thus fulfills all the requirements in the theorem. In particular, \(g\) is continuous on \([a,b]\) and, since \(f\) is continuous on \(g([a,b])\), the composition \(f \circ g\) is continuous on \([a,b]\). This means that \(f\circ g\) is Riemann-integrable on \([a,b]\) and we can write the following:

\[\int_a^b f(g(x)) \,\mathrm{d}x = \frac{1}{c} \int_a^b f(g(x))\cdot c \,\mathrm{d}x\]

We notice that \(g'(x) = c\) and so we have:

\[\int_a^b f(g(x)) \,\mathrm{d}x = \frac{1}{c} \int_a^b f(g(x))g'(x) \,\mathrm{d}x\]

We can now apply the theorem:

\[\int_a^b f(g(x)) \,\mathrm{d}x = \frac{1}{c} \int_{g(a)}^{g(b)} f(u) \,\mathrm{d}u\]

\[\int_a^b f(cx + r) \,\mathrm{d}x = \frac{1}{c} \int_{ca+r}^{cb+r} f(u) \,\mathrm{d}u\]

Example: \(\int_0^{\frac{\pi}{2}} \mathrm{e}^{\sin x} \cos {x} \,\mathrm{d}x\)

We want to determine the following Riemann integral:

\[\int_{0}^{\frac{\pi}{2}} \mathrm{e}^{\sin x} \cos {x} \,\mathrm{d}x\]

We notice that

\[\int_{0}^{\frac{\pi}{2}} \mathrm{e}^{\sin x} \cos {x} \,\mathrm{d}x = \int_{0}^{\frac{\pi}{2}} f(g(x))g'(x) \,\mathrm{d}x\]

with \(g(x) = \sin x\), \(g'(x) = \cos x\) and \(f(u) = \mathrm{e}^u\). We can verify that \(f\) and \(g\) satisfy the requirements of the theorem. Therefore, we have:

\[\begin{aligned}\int_{0}^{\frac{\pi}{2}} \mathrm{e}^{\sin x} \cos {x} \,\mathrm{d}x & = \int_{g(0)}^{g\left(\frac{\pi}{2}\right)} f(u)\, \mathrm{d}u \\ & = \int_{0}^{1} \mathrm{e}^u \, \mathrm{d}u \\ & = \mathrm{e}^u \vert_0^1 = \mathrm{e}^1 - \mathrm{e}^0 = \mathrm{e} -1 \end{aligned}\]

Proof

TODO

Theorem: King's Rule

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

If \(f(x)\) is Riemann-integrable on \([a,b] \subseteq \mathcal{D}\), then so is \(f(a + b - x)\):

\[\int_a^b f(x) \,\mathrm{d}x = \int_a^b f(a + b - x) \,\mathrm{d}x.\]

If \(\frac{f(x)}{f(x) + f(a + b - x)}\) is Riemann-integrable on \([a,b] \subseteq \mathcal{D}\), then

\[\int_a^b \frac{f(x)}{f(x) + f(a + b - x)}\,\mathrm{d}x = \frac{b - a}{2}.\]

Proof

TODO

Theorem: Integrability of the Absolute Value

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

If \(f\) is Riemann-integrable on a compact interval \([a,b]\), then so is \(|f|\) and

\[\left|\int_a^b f(x) \mathop{\mathrm{d}x}\right| \le \int_a^b |f(x)| \mathop{\mathrm{d}x}.\]

Proof

TODO

Improper Riemann Integrals#

The notion of Riemann integrals can be extended to open and semi-open intervals using [limits](../Limits%20(Real%20Functions.md).

Definition: Improper Riemann Integrals

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

We say that \(f\) is improperly Riemann-integrable on:

\([a, b) \subseteq \mathcal{D}\), where \(-\infty \lt a \lt b \lt +\infty\), if \(f\) is locally Riemann-integrable on \([a, b)\) and the following [limit](../Limits%20(Real%20Functions.md) exists and is finite:

\[\int_a^b f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \lim_{\beta \to b^-} \int_a^\beta f(x) \mathop{\mathrm{d}x}\]

\([a, +\infty) \subseteq \mathcal{D}\), where \(-\infty \lt a \lt +\infty\), if \(f\) is locally Riemann-integrable on \([a, +\infty)\) and the following [limit](../Limits%20(Real%20Functions.md) exists and is finite:

\[\int_a^{+\infty} f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \lim_{\beta \to +\infty} \int_a^\beta f(x) \mathop{\mathrm{d}x}\]

\((a, b] \subseteq \mathcal{D}\), where \(-\infty \lt a \lt b \lt +\infty\), if \(f\) is locally Riemann-integrable on \((a, b]\) and the following [limit](../Limits%20(Real%20Functions.md) exists and is finite:

\[\int_a^b f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \lim_{\alpha \to a^+} \int_\alpha^b f(x) \mathop{\mathrm{d}x}\]

\((-\infty, b] \subseteq \mathcal{D}\), where \(-\infty \lt b \lt +\infty\), if \(f\) is locally Riemann-integrable on \((-\infty, b]\) and the following [limit](../Limits%20(Real%20Functions.md) exists and is finite:

\[\int_{-\infty}^b f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \lim_{\alpha \to -\infty} \int_\alpha^b f(x) \mathop{\mathrm{d}x}\]

\((a, b) \subseteq \mathcal{D}\), where \(-\infty \lt a \lt b \lt +\infty\), if \(f\) is improperly Riemann-integrable on \((a, c]\) and \([c, b)\) for any \(c \in (a, b)\):

\[\int_a^b f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \int_a^c f(x) \mathop{\mathrm{d}x} + \int_c^b f(x) \mathop{\mathrm{d}x}\]

\((-\infty, b) \subseteq \mathcal{D}\), where \(-\infty \lt b \lt +\infty\), if \(f\) is improperly Riemann-integrable on \((-\infty, c]\) and \([c, b)\) for any \(c \in (-\infty, b)\):

\[\int_{-\infty}^b f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \int_{-\infty}^c f(x) \mathop{\mathrm{d}x} + \int_c^b f(x) \mathop{\mathrm{d}x}\]

\((a, +\infty) \subseteq \mathcal{D}\), where \(-\infty \lt a \lt +\infty\), if \(f\) is improperly Riemann-integrable on \((a, c]\) and \([c, +\infty)\) for any \(c \in (a, +\infty)\):

\[\int_a^{+\infty} f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \int_a^c f(x) \mathop{\mathrm{d}x} + \int_c^{+\infty} f(x) \mathop{\mathrm{d}x}\]

\((-\infty, +\infty) \subseteq \mathcal{D}\), if \(f\) is improperly Riemann-integrable on \((-\infty, c]\) and \([c, +\infty)\) for any \(c \in (-\infty, +\infty)\):

\[\int_{-\infty}^{+\infty} f(x) \mathop{\mathrm{d}x} \overset{\text{def}}{=} \int_{-\infty}^c f(x) \mathop{\mathrm{d}x} + \int_c^{+\infty} f(x) \mathop{\mathrm{d}x}\]

Notation

Given an arbitrary interval \(I\), the expression

\[\int_I f(x)\,\mathrm{d}x\]

denotes either the Riemann integral of \(f\) on \(I\) (whenever \(I\) is a compact interval) or \(f\)'s corresponding improper Riemann integral.

Example: \(\int_0^1 \frac{1}{x^r} \,\mathrm{d}x\)

We want to determine

\[\int_0^1 \frac{1}{x^r} \,\mathrm{d}x.\]

For \(r = 1\), we have:

\[\begin{aligned}\int_0^1 \frac{1}{x} \,\mathrm{d}x & = \lim_{\alpha \to 0^{+}} \int_{\alpha}^1 \frac{1}{x} \,\mathrm{d}x \\ & = \lim_{\alpha \to 0^{+}} \left. \ln x \right\vert_\alpha^1 \\ & = \lim_{\alpha \to 0^{+}} (\ln 1 - \ln \alpha) = +\infty \end{aligned}\]

However, \(\frac{1}{x}\) is not improperly Riemann-integrable on \((0, 1]\) because this value is not finite.

For \(r \ne 1\), we have:

\[\begin{aligned}\int_0^1 \frac{1}{x^r} \,\mathrm{d}x & = \lim_{\alpha \to 0^{+}} \int_{\alpha}^1 \frac{1}{x^r} \,\mathrm{d}x \\ & = \lim_{\alpha \to 0^{+}} \left. \frac{1}{1-r}x^{1-r}\right\vert_\alpha^1 \\ & = \lim_{\alpha \to 0^{+}} \frac{1-\alpha^{1-r}}{1-r} = \begin{cases}\frac{1}{1-r}, & \text{if} & r \lt 1 \\ +\infty, & \text{if} & r \gt 1\end{cases} \end{aligned}\]

In sum:

\[\int_0^1 \frac{1}{x^{\alpha}} \,\mathrm{d}x = \begin{cases}\frac{1}{1-r}, & \text{if} & r \lt 1 \\ +\infty, & \text{if} & r \ge 1\end{cases}\]

Example: \(\int_1^{\infty} \frac{1}{x^r} \,\mathrm{d}x\)

We want to determine

\[\int_1^{\infty} \frac{1}{x^r} \,\mathrm{d}x.\]

For \(r = 1\), we have:

\[\begin{aligned}\int_1^{\infty} \frac{1}{x} \,\mathrm{d}x & = \lim_{\beta \to +\infty} \int_1^{\beta} \frac{1}{x} \,\mathrm{d}x \\ & = \lim_{\beta \to +\infty} \left. \ln x \right\vert_1^{\beta} \\ & = \lim_{\beta \to +\infty} (\ln \beta - \ln 1) = +\infty \end{aligned}\]

For \(r \ne 1\), we have:

\[\begin{aligned}\int_1^{\infty} \frac{1}{x^r} \,\mathrm{d}x & = \lim_{\beta \to +\infty} \int_1^{\infty} \frac{1}{x^r} \,\mathrm{d}x \\ & = \lim_{\beta \to +\infty} \left. \frac{1}{1-r}x^{1-r}\right\vert_1^{\beta} \\ & = \lim_{\beta \to +\infty} \frac{\beta^{1-r}-1}{1-r} = \begin{cases}\frac{1}{r-1}, & \text{if} & r \gt 1 \\ +\infty, & \text{if} & r \lt 1\end{cases} \end{aligned}\]

In sum:

\[\int_1^{\infty} \frac{1}{x^r} \,\mathrm{d}x = \begin{cases}+\infty, & \text{if} & r \le 1 \\ \frac{1}{r-1}, & \text{if} & r \gt 1\end{cases}\]

Example: \(\int_0^{\infty} \mathrm{e}^{-3x}\,\mathrm{d}x\)

\[\begin{aligned}\int_0^{\infty} \mathrm{e}^{-3x}\,\mathrm{d}x & = \lim_{\beta \to \infty} \int_0^{\beta} \mathrm{e}^{-3x}\,\mathrm{d}x \\ & = \lim_{\beta \to \infty} \left(\left.-\frac{1}{3}\mathrm{e}^{-3x}\right\vert_0^{\beta}\right) \\ & = \lim_{\beta \to \infty} \left(-\frac{1}{3}(\mathrm{e}^{-3\beta}-1)\right) \\ & = \frac{1}{3}\end{aligned}\]

Therefore, \(\mathrm{e}^{-3x}\) is improperly Riemann-integrable on \([0, +\infty)\).

Theorem: Constant Independence of Improper Integrals

TODO

Proof

TODO

Theorem: Improper Integral \(\rightarrow\) Proper Integral

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function.

If \(f\) is improperly Riemann-integrable on \([a,b)\) and \(\tilde{f}\) is an extension of \(f\) which is Riemann-integrable on \([a,b]\), then

\[\int_a^b f(x) \,\mathrm{d}x = \int_a^b \tilde{f}(x) \,\mathrm{d}x.\]

If \(f\) is improperly Riemann-integrable on \((a,b]\) and \(\tilde{f}\) is an extension of \(f\) which is Riemann-integrable on \([a,b]\), then

\[\int_a^b f(x) \,\mathrm{d}x = \int_a^b \tilde{f}(x) \,\mathrm{d}x.\]

Proof

TODO

Theorem: Comparison Test for Improper Integrals

Let \(f: \mathcal{D} \subseteq \mathbb{R} \to \mathbb{R}\) be a real function and let \(I\) be any interval.

If \(f\) is locally Riemann-integrable on \(I\) and there exists some real function \(g\) which is improperly Riemann-integrable on \(I\) with

\[|f(x)| \le g(x)\]

for all \(x \in I\), then \(f\) is also improperly Riemann-integrable on \(I\) with

\[\left\vert\int_{I} f(x) \,\mathrm{d}x\right\vert \le \int_I g(x) \,\mathrm{d}x.\]

Example: \(\int_1^{\infty} \mathrm{e}^{-x}x^3 \,\mathrm{d}x\)

From the [limit](../Limits%20(Real%20Functions.md)

\[\lim_{x \to \infty} \frac{x^5}{\mathrm{e}^{-x}} = \lim_{x \to \infty} \frac{x^5}{\mathrm{e}^x} = 0\]

we know that \(\mathrm{e}^{-x}x^5\) is bounded on \([1; +\infty)\), i.e. there exists some \(B \gt 0\) such that

\[\mathrm{e}^{-x}x^5 \le B\]

for all \(x \ge 1\). From this inequality, we get

\[\mathrm{e}^{-x}x^3 \le Bx^{-2}\]

for all \(x \ge 1\). Since the integral

\[\int_1^{\infty} x^{-2} \,\mathrm{d}x\]

converges, we know that

\[\int_1^{\infty} Cx^{-2} \,\mathrm{d}x\]

also converges and so

\[\int_1^{\infty} \mathrm{e}^{-x}x^3 \,\mathrm{d}x\]

must converge as well.

Example: \(\int_0^{\infty} \frac{\sin x}{x}\,\mathrm{d}x\)

We want to determine if the integral

\[\int_0^{\infty} \frac{\sin x}{x}\,\mathrm{d}x\]

converges. We split it into two:

\[\int_0^{\infty} \frac{\sin x}{x}\,\mathrm{d}x = \int_0^1 \frac{\sin x}{x}\,\mathrm{d}x + \int_1^{\infty}\frac{\sin x}{x}\,\mathrm{d}x\]

The function \(f: \mathbb{R} \to \mathbb{R}\) defined as

\[f(x) = \begin{cases} 1 & \text{if} x = 0 \\ \frac{\sin x}{x} & \text{if} x \ne 0\end{cases}\]

is an extension of \(\frac{\sin x}{x}\). It is continuous, since \(\lim_{x \to 0}\frac{\sin x}{x} = 1\), and so it is also Riemann-integrable on \([0,1]\). We thus have:

\[\int_0^1 \frac{\sin x}{x} \,\mathrm{d}x = \int_0^1 f(x) \,\mathrm{d}x \in \mathbb{R}\]

For \(\int_1^{\infty}\frac{\sin x}{x}\,\mathrm{d}x\), we have:

\[\int_1^{\infty}\frac{\sin x}{x}\,\mathrm{d}x = \lim_{\beta \to +\infty} \int_1^{\beta} \frac{\sin x}{x} \,\mathrm{d}x\]

We use integration by parts:

\[\begin{aligned}\int_1^{\infty}\frac{\sin x}{x}\,\mathrm{d}x & = \lim_{\beta \to +\infty} \int_1^{\beta} \frac{\sin x}{x} \,\mathrm{d}x \\ & = \lim_{\beta \to +\infty} \left(\left.-\frac{\cos x}{x} \right\vert_1^{\beta} - \int_{1}^{\beta} \frac{\cos x}{x^2} \,\mathrm{d}x\right)\end{aligned}\]

For the first term:

\[\begin{aligned}\lim_{\beta \to +\infty} \left.-\frac{\cos x}{x} \right\vert_1^{\beta} = \lim_{\beta \to +\infty} \left( -\frac{\cos \beta}{\beta} + \frac{\cos 1}{1}\right) = \cos 1\end{aligned}\]

For the second, term \(\lim_{\beta \to +\infty} -\int_{1}^{\beta} \frac{\cos x}{x^2} \,\mathrm{d}x\) is just the improper integral \(\int_1^{\infty} -\frac{\cos x}{x}\, \mathrm{d}x\). Since

\[\left\vert \frac{\cos x}{x} \right\vert \le x^{-2}\]

for all \(x \in [1, \infty)\) and since \(\int_1^{\infty} x^{-2}\, \mathrm{d}x\) converges, we know that

\[\int_1^{\infty} -\frac{\cos x}{x}\, \mathrm{d}x\]

must also converge.

Therefore, the [limit](../Limits%20(Real%20Functions.md)

\[\int_1^{\infty} \frac{\sin x}{x}\,\mathrm{d}x = \lim_{\beta \to +\infty} \int_1^{\beta} \frac{\sin x}{x} \,\mathrm{d}x = \lim_{\beta \to +\infty} \left.-\frac{\cos x}{x} \right\vert_1^{\beta} + \lim_{\beta \to +\infty} -\int_{1}^{\beta} \frac{\cos x}{x^2} \,\mathrm{d}x\]

must also exist.

Proof

TODO

Theorem: Integral Test

Let \(f: [a, \infty) \subseteq \mathbb{R} \to \mathbb{R}\) be a non-negative, decreasing real function.

The improper integral

\[\int_a^{\infty} f(x) \,\mathrm{d}x\]

[converges](../Limits%20(Real%20Functions.md) if and only if the series

\[\sum_{n=\lceil a \rceil}^{\infty} f(n)\]

converges.

Proof

TODO