4.3 Double integrals and repeated integrals.

Let \(f(x, y)\) be a function which is continuous with regard to both of the variables \(x\) and \(y\), when \(a \leq x \leq b\), \(\alpha \leq y \leq \beta\).

By §4.2 example 2, it is clear[1] that \[\int_a^b \!\left\{\int_\alpha^\beta \! f(x,y)\, dy \right\}dx=\int_\alpha^\beta \!\left\{\int_a^b \! f(x,y)\, dx \right\}dy\] both exist. These are called repeated integrals.

[1]Editor’s Note: It is only clear that both integrals exist. To show they are equal requires an argument. We can use the existence of both integrals, the first mean value theorem, and the uniform continuity \(f(x,y)\) on the region \([a,b] \times [\alpha, \beta]\) to demonstrate the equality. Or we can, as the authors do later, show that they are both equal the double integral defined below. ↩

Also, as in §3.62, \(f(x, y)\), being a continuous function of both variables, attains its upper and lower bounds.

Consider the range of values of \(x\) and \(y\) to be the points inside and on a rectangle in a Cartesian diagram; divide it into \(n\nu\) rectangles by lines parallel to the axes.

Let \(U_{m,\:\! \mu}, \, L_{\,m,\:\! \mu}\) be the upper and lower bounds of \(f(x, y)\) in one of the smaller rectangles whose area is, say, \(A_{m,\:\! \mu}\); and let \[\sum_{m=1}^n \sum_{\mu=1}^\nu U_{m,\:\! \mu}A_{m,\:\! \mu}=S_{n,\:\! \nu}, \quad \sum_{m=1}^n \sum_{\mu=1}^\nu L_{\,m,\:\! \mu}A_{m,\:\! \mu}=s_{n,\:\! \nu} .\]

Then \(S_{n,\:\! \nu} \geq s_{n,\:\! \nu}\), and, as in §4.11, we can find numbers \(\underline{S}_{n,\:\! \nu}\), \(\overline{s}_{n,\:\! \nu}\), which are the lower and upper bounds of \(S_{n,\:\! \nu}\), \(s_{n,\:\! \nu}\), respectively, the values of \(\underline{S}_{ n,\:\! \nu}\), \(\overline{s}_{n,\:\! \nu}\) depending only on the number of the rectangles and not on their shapes; and \(\underline{S}_{n,\:\! \nu} \geq \overline{s}_{n,\:\! \nu}\). We then find the lower and upper bounds (\(S\) and \(s\)) respectively of \(\underline{S}_{n,\:\! \nu}\), \(\overline{s}_{n,\:\! \nu}\) qua[2] functions of \(n\) and \(\nu\); and \(S_{n,\:\! \nu} \geq S \geq s \geq s_{n,\:\! \nu}\), as in §4.11.

[2]Editor’s Note: In this usage, the word qua means treated as …  ↩

Also, from the uniformity of the continuity of \(f(x, y)\), given \(\epsilon\), we can find \(\delta\) such that \[U_{m,\:\! \mu} - L_{\,m,\:\! \mu} < \epsilon,\] (for all values of \(m\) and \(\mu\)) whenever the sides of all the small rectangles are less than the number \(\delta\) which depends only on the form of the function \(f(x, y)\) and on \(\epsilon\).

And then \[(S_{n,\:\! \nu}-s_{n,\:\! \nu} < \epsilon (b - a) (\beta - \alpha),\] and so \[(S-s < \epsilon (b - a) (\beta - \alpha).\]

But \(S\) and \(s\) are independent of \(\epsilon\), and so \(S = s\).

The common value of \(S\) and \(s\) is called the double integral of \(f(x, y)\) and is written \[\int_a^b \!\int_\alpha^\beta \! f(x,y)(dxdy).\]

It is easy to shew that the repeated integrals and the double integral are all equal when \(f(x, y)\) is a continuous function of both variables.

For let \(\Upsilon_m,\, \Lambda_m\) be the upper and lower bounds of \[\int_\alpha^\beta \! f(x,y)\, dy\] as \(x\) varies between \(x_{m-1}\) and \(x_m\).

Then \[\sum_{m=1}^n \Upsilon_m (x_{m}-x_{m-1}) \geq \int_a^b \!\left\{\int_\alpha^\beta \! f(x,y)\, dy \right\}dx \geq \sum_{m=1}^n \Lambda_m (x_{m}-x_{m-1}). \] But[3] \[\sum_{\mu=1}^\nu U_{m,\:\! \mu} (y_{\mu}-y_{\mu-1}) \geq \Upsilon_m \geq \Lambda_m \geq \sum_{\mu=1}^\nu L_{\,m,\:\! \mu} (y_{\mu}-y_{\mu-1}).\]

[3]The upper bound of \(f(x, y)\) in the rectangle \(A_{m,\:\! \mu}\) is not less than the upper bound of \(f(x, y)\) on that portion of the line \(x = \xi\) which lies in the rectangle. ↩

Multiplying these last inequalities by \((x_{m}-x_{m-1})\), using the preceding inequalities and summing, we get \[\sum_{m=1}^n \sum_{\mu=1}^\nu U_{m,\:\! \mu} A_{m, \:\! \mu} \geq \int_a^b \!\left\{\int_\alpha^\beta \! f(x,y)\, dy \right\}dx \geq \sum_{m=1}^n \sum_{\mu=1}^\nu L_{\,m,\:\! \mu} A_{m, \:\! \mu};\] and so, proceeding to the limit, \[S \geq \int_a^b \!\left\{\int_\alpha^\beta \! f(x,y)\, dy \right\}dx \geq s.\] But \[S=s=\int_a^b \!\int_\alpha^\beta \! f(x,y)(dxdy),\] and so one of the repeated integrals is equal to the double integral. Similarly the other repeated integral is equal to the double integral.

Corollary. If \(f(x, y)\) be a continuous function of both variables, \[\int_0^1 \! dx \left\{\int_0^{1-x}\! f(x,y)\,dy \right\}=\int_0^1 \! dy \left\{\int_0^{1-y}\! f(x,y)\,dx \right\}\]

4.4 Infinite integrals.

If \(\lim\limits_{b\rightarrow \infty} \left( \int_a^b f(x)\,dx \right)\) exists, we denote it by \(\int_a^\infty f(x)\,dx\); and the limit in question is called an infinite integral.[4]

[4]This phrase, due to Hardy, Proc. London Math. Soc. xxxiv. (1902), p. 16, suggests the analogy between an infinite integral and an infinite series. ↩

Examples. \[ \begin{align*} &(1) \quad\int_a^\infty \!\frac{dx}{x^2}=\lim_{b \rightarrow \infty} \left(\frac{1}{b}-\frac{1}{a}\right)=\frac{1}{a}.\\ &(2) \quad\int_0^\infty \!\frac{xdx}{(x^2+a^2)^2}=\lim_{b \rightarrow \infty}\left(\frac{1}{2(b^2+a^2)}+\frac{1}{2a^2}\right)=\frac{1}{2a^2}\\ &(3) \quad \text{By integrating by parts, shew that } \int_0^\infty \! t^n e^{-t}dt=n!. &(\mathrm{Euler.}) \end{align*} \]

Similarly we define \( \int_{-\infty}^b f(x)\,dx\) to mean \(\lim\limits_{a \rightarrow -\infty} \int_a^b f(x)\,dx\), if this limit exists; and \( \int_{-\infty}^\infty f(x)\,dx\) is defined as \( \int_{-\infty}^a f(x)\,dx + \int_{a}^\infty f(x)\,dx\). In this last definition, the choice of \(a\) is a matter of indifference.

4.41 Infinite integrals of continuous functions. Conditions for convergence.

A necessary and sufficient condition for the convergence of \(\int_a^\infty f(x)\,dx\) is that, corresponding to any positive number \(\epsilon\), a positive number \(X\) should exist such that \(\left|\,\int_{x'}^{x' '} f(x)\,dx\,\right| < \epsilon\) whenever \(x'' \geq x' \geq X\).

The condition is obviously necessary; to prove that it is sufficient, suppose it is satisfied; then, if \(n \geq X - a \) and \(n\) be a positive integer and \(S_n = \int_a^{a+n} f(x)\,dx\), we have \[\left|\, S_{n+p} - S_n \right| < \epsilon.\] Hence, by §2.22, \(S_n\) tends to a limit, \(S\); and then, if \(\xi > a + n\), \[ \begin{align*} \left|\,S - \int_a^\xi \! f(x)\,dx \,\right| &\leq \left|\,S - \int_a^{a+n} \! f(x)\,dx \, \right|+\left|\int_{a+n}^\xi \! f(x)\,dx \,\right|\\ \\ \\ & < 2\epsilon; \end{align*} \] and so \(\lim\limits_{\xi \rightarrow \infty}\int_a^\xi f(x)\, dx = S\); so that the condition is sufficient.

4.42 Uniformity of convergence of an infinite integral.

The integral \(\int_a^\infty f(x, \boldsymbol{\alpha})\, dx\) is said to converge uniformly with regard to \(\boldsymbol{\alpha}\) in a given domain of values of \(\boldsymbol{\alpha}\) if, corresponding to an arbitrary positive number \(\epsilon\), there exists a number \(X\) independent of \(\boldsymbol{\alpha}\) such that \[\left|\,\int_{x'}^\infty \! f(x,\boldsymbol{\alpha})\,dx \,\right| < \epsilon.\] for all values of \(\boldsymbol{\alpha}\) in the domain and all values of \(x' \geq X\).

The reader will see without difficulty on comparing §2.22 and §3.31 with §4.41 that a necessary and sufficient condition that \(\int_a^\infty f(x, \boldsymbol{\alpha})\, dx\) should converge uniformly in a given domain is that, corresponding to any positive number \(\epsilon\), there exists a number \(X\) independent of \(\boldsymbol{\alpha}\) such that \[\left|\,\int_{x'}^{x''} \! f(x,\boldsymbol{\alpha})\,dx \,\right| < \epsilon.\] for all values of \(\boldsymbol{\alpha}\) in the domain whenever \(x'' \geq x' \geq X\).

4.43 Tests for the convergence of an infinite integral.

There are conditions for the convergence of an infinite integral analogous to those given in Chapter II for the convergence of an infinite series.

The following tests are of special importance.

(I) Absolutely convergent integrals. It may be shewn that \(\int_a^\infty f(x)\,dx\) certainly converges if \(\int_a^\infty \left|\,f(x)\right|\,dx\) does so; and the former integral is then said to be absolutely convergent. The proof is similar to that of §2.32.

Example. The comparison test. If \(\left|\,f(x)\right| \leq g(x)\) and \(\int_a^\infty g(x)\,dx\) converges, then \(\int_a^\infty f(x)\,dx\) converges absolutely.

[5]Dirichlet’s example was \(f(x)=\sin x^2\); Journal für Math. xvii. (1837), p. 60.  ↩

[Note. It was observed by Dirichlet[5] that it is not necessary for the convergence of \(\int_a^\infty f(x)\,dx\) that \(f(x) \rightarrow 0\) as \(x \rightarrow \infty\): the reader may see this by considering the function \[ \begin{align*} &f(x)=0 \qquad\qquad \text{ when}\quad n \leq x \leq n+1-(n+1)^{-2},\\ &f(x)=(n+1)^4(n+1-x)\{x-(n+1)+(n+1)^{-2}\}\\ &\hphantom{f(x) = 0} \qquad\qquad \,\text{ when}\quad n+1-(n+1)^{-2} \leq x \leq n+1, \end{align*} \] where \(n\) takes all integral values.

For \(\int_0^\xi f(x)\,dx\) increases with \(\xi\) and \(\int_n^{n+1} f(x)\,dx=\frac{1}{6}(n + 1)^{-2}\); whence it follows without difficulty that \(\int_a^\infty f(x)\,dx\) converges. But when \(x=n+1-\frac{1}{2}(n + 1)^{-2}\), \(f(x) = \frac{1}{4}\); and so \(f(x)\) does not tend to zero.]

(II) The Maclaurin-Cauchy[6] test. If \(f(x) > 0\) and \(f(x)\rightarrow 0\) steadily, \(\int_1^\infty f(x)\,dx\) and \(\sum_{n=1}^\infty f(x)\) converge or diverge together.
[6]Maclaurin (Fluxions, i. pp. 289, 290) makes a verbal statement practically equivalent to this result. Cauchy’s result is given in his Oeuvres (2), vii, p. 269.  ↩

For \[f(m) \geq \int_m^{m+1} \! f(x)\,dx \geq f(m+1),\] and so \[\sum_{m=1}^n f(m) \geq \int_1^{n+1} \! f(x)\,dx \geq \sum_{m=2}^{n+1} f(m). \]

The first inequality shews that, if the series converges, the increasing sequence \( \int_1^{n+1} f(x)\,dx\) converges (§2.2) when \(n \rightarrow \infty\) through integral values, and hence it follows without difficulty that \( \int_1^{x'} f(x)\,dx\) converges when \(x' \rightarrow \infty\); also if the integral diverges, so does the series.

The second shews that if the series diverges so does the integral, and if the integral converges so does the series (§2.2).

[7]Journal de Math. vii. (1842), pp. 38, 39. ↩
(III) Bertrand’s[7] test. If \(f(x) = O(x^{\:\!\lambda-1})\), \(\int_a^\infty f(x)\,dx\) converges when \(\lambda < 0\); and if \(f(x) = O(x^{-1} \{\log x\}^{\lambda-1})\), \(\int_a^\infty f(x)\,dx\) converges when \(\lambda < 0\).

These results are particular cases of the comparison test given in (I).

(IV) Chartier’s test[8] for integrals involving periodic functions.

[8]Journal de Math. xviii. (1853), pp. 201–212. It is remarkable that this test for conditionally convergent integrals should have been given some years before formal definitions of absolutely convergent integrals.  ↩

If \(f(x) \rightarrow 0 \) steadily as \(x \rightarrow \infty\) and if \(\left| \int_a^x \phi(x)\,dx \right|\) is bounded as \(x \rightarrow \infty\), then \( \int_a^x f(x)\phi(x)\,dx \) is convergent.

For if the upper bound of \(\left| \int_a^x \phi(x)\,dx \right|\) be \(A\), we can choose \(X\) such that \(f(x) < \left.\epsilon \middle/2A\right.\) when \(x\geq X\); and then by the second mean value theorem, when \(x'' \geq x' \geq X\), we have \[ \begin{align*} \left|\int_{x'}^{x''} \! f(x)\phi(x)\,dx\,\right| &= \left|\,f(x')\int_{x'}^\xi \!\phi(x)\,dx \,\right| \\ &= f(x')\ \left|\int_{a}^\xi \!\phi(x)\,dx - \int_{a}^{x'} \!\phi(x)\,dx \,\right|\\ \\ \\ &\leq 2A\,f(x')< \epsilon. \end{align*} \] which is the condition for convergence.

Example 1. \(\displaystyle \int_0^\infty \!\frac{\sin x}{x} dx\) converges.

Example 2. \(\displaystyle \int_0^\infty \! x^{-1} \sin (x^3- ax) \,dx\) converges.

4.431 Tests for uniformity of convergence of an infinite integral.[9]
[9]The results of this section and of §4.44 are due to de la Vallée Poussin, Ann. de la Soc. Scientifique de Bruxelles, xvi. (1892), pp. 150–180.  ↩
[10]This name is due to Osgood.  ↩
(I) De la Vallée Poussin’s test.[10] The reader will easily see by using the reasoning of §3.34 that \(\int_a^\infty f(x,\boldsymbol{\alpha})\,dx\) converges uniformly with regard to \(\boldsymbol{\alpha}\) in a domain of values of \(\boldsymbol{\alpha}\) if \(\left| \, f(x,\boldsymbol{\alpha})\, \right| < \mu(x)\), where \(\mu(x)\) is independent of \(\boldsymbol{\alpha}\) and \(\int_a^\infty \mu(x)\,dx\) converges. ( For, choosing \(X\) so that \(\int_{x'}^{x''} \mu(x)\,dx < \epsilon\) when \(x'' \leq x' \leq X\), we have \(\left|\,\int_{x'}^{x''} f(x,\boldsymbol{\alpha})\,dx\right| < \epsilon\), and the choice of \(X\) is independent of \(\boldsymbol{\alpha}\). )

Example. \(\int_0^\infty x^{\alpha-1}e^{-x} dx\) converges uniformly in any interval \([A, B]\) such that \(1 \leq A \leq B\).

(II) The method of change of variable.

This may be illustrated by an example.

Consider \(\displaystyle \int_0^\infty \!\frac{\sin \alpha x}{x} dx \) where \(\alpha\) is real. We have \[\int_{x'}^{x''} \!\frac{\sin \alpha x}{x} dx=\int_{\alpha x'}^{\alpha x''} \!\frac{\sin y}{y} dy. \]

Since \(\displaystyle \int_0^\infty \!\frac{\sin y}{y} dy \) converges we can find \(Y\) such that \(\displaystyle \int_{y'}^{y''} \!\frac{\sin y}{y} dy\) when \(y'' \geq y' \geq Y\). So \(\displaystyle \left| \int_{x'}^{x''} \!\frac{\sin \alpha x }{x}dx\, \right| < \epsilon\) whenever \(\left|\, ax' \right | \geq Y\); if \(\left|\, \alpha \,\right| \geq \delta > 0\), we therefore get \[\displaystyle \left| \int_{x'}^{x''} \!\frac{\sin \alpha x }{x}dx\, \right| < \epsilon\]when \(x'' \geq x' \leq X= \left.Y\right/\delta \:\!\); and this choice of \(X\) is independent of \(\alpha\). So the convergence is uniform when \(\alpha \geq \delta > 0\), and when \(\alpha \leq -\delta < 0\),.

Example. \(\displaystyle \int_1^\infty \!\left\{\int_0^\alpha \!\sin (\beta^2 x^3)\, d\beta \right\} dx\) is uniformly convergent in any range of real values of \(\alpha\). (de la Vallée Poussin.)

[ Write \(\beta^2x^3 = z\) and observe that \( \left|\int_0^{\alpha^2 x^3} z^{-\frac{1}{2}}\sin z\, dz \, \right|\) does not exceed a constant independent of \(\alpha\) and \(x\) since \(\int_0^\infty z^{-\frac{1}{2}}\sin z\, dz\) converges.]

(III) The method of integration by parts.

If \[ \int \! f(x, \boldsymbol{\alpha})\,dx = \phi (x, \boldsymbol{\alpha}) + \int \!\chi( x, \boldsymbol{\alpha} )\, dx\] and if \(\phi(x,\boldsymbol{\alpha}) \rightarrow 0\) uniformly as \(x \rightarrow \infty\) and \( \int_a^\infty \chi(x, \boldsymbol{\alpha})\,dx \) converges uniformly with regard to \(\boldsymbol{\alpha}\), then obviously \( \int_a^\infty f(x, \boldsymbol{\alpha})\,dx \) converges uniformly with regard to \(\boldsymbol{\alpha}\).

(IV) The method of decomposition.

Example. \[\int_0^\infty \!\frac{\cos x \sin \alpha x}{x} dx=\frac{1}{2}\int_0^\infty \!\frac{\sin(\alpha+1)x}{x}dx+\frac{1}{2}\int_0^\infty \!\frac{\sin(\alpha-1)x}{x}dx; \] both of the latter integrals converge uniformly in any closed domain of real values of a from which the points \(\alpha = \pm 1\) are excluded.

4.44 Theorems concerning uniformly convergent infinite integrals.

(I) Let \(\int_a^\infty f(x, \boldsymbol{\alpha}) \,dx\) converge uniformly when \(\boldsymbol{\alpha}\) lies in a domain \(S\). Then, if \(f(x, \boldsymbol{\alpha})\) is a continuous function of both variables when \(x \geq a\) and \(\boldsymbol{\alpha}\) lies in \(S\), \(\int_a^\infty f(x, \boldsymbol{\alpha})\,dx\) is a continuous function[11] of \(\boldsymbol{\alpha}\).

[11]This result is due to Stokes. His statement is that the integral is a continuous function of \(\boldsymbol{\alpha}\) if it does not ‘converge infinitely slowly’. ↩

For, given \(\epsilon\), we can find \(X\) independent of \(\boldsymbol{\alpha}\), such that \(\left|\int_\xi^\infty f(x, \boldsymbol{\alpha}) \,dx \right| < \epsilon \) whenever \(\xi \geq X\)

Also we can find \(\delta\) independent of \(x\) and \(\boldsymbol{\alpha}\), such that \[\left|\,f(x, \boldsymbol{\alpha}) - f(x,\boldsymbol{\alpha}') \, \right| < \left. \epsilon \middle/ (X- \boldsymbol{\alpha})\right. \] whenever \(\left|\,\boldsymbol{\alpha} -\boldsymbol{\alpha}' \, \right|< \delta\).

That is to say, given \(\epsilon\), we can find \(\delta\) independent of \(\boldsymbol{\alpha}\), such that \[ \begin{align*} \left|\,\int_a^\infty \!f(x,\boldsymbol{\alpha})\, dx - \int_a^\infty \!f(x,\boldsymbol{\alpha}') \,dx\,\right| &\leq \left|\, \int_a^X \!f(x, \boldsymbol{\alpha}) - f(x,\boldsymbol{\alpha}') \,dx\, \right|\\ & \quad + \left|\, \int_X^\infty \!f(x, \boldsymbol{\alpha}') \, dx \, \right|+ \left|\, \int_X^\infty \!f(x, \boldsymbol{\alpha}) \, dx \, \right|\\ \\ \\ & < 3\epsilon \end{align*} \] whenever \(\left|\,\boldsymbol{\alpha} -\boldsymbol{\alpha}' \, \right|< \delta\); and this is the condition for continuity.

(II) If \(f (x, \boldsymbol{\alpha})\) satisfies the same conditions as in (I), and if \(\boldsymbol{\alpha}\) lies in \(S\) when \(A \leq \boldsymbol{\alpha} \leq B\), then \[\int_A^B \!\left\{\int_a^\infty \!f(x,\boldsymbol{\alpha})\, dx \right\} d\boldsymbol{\alpha} = \int_a^\infty \!\left\{\int_A^B \!f(x,\boldsymbol{\alpha})\, d\boldsymbol{\alpha} \right\} dx.\]

For, by §4.3, \[\int_A^B \!\left\{\int_a^\xi \!f(x,\boldsymbol{\alpha})\, dx \right\} d\boldsymbol{\alpha} = \int_a^\xi \left\{\int_A^B \!f(x,\boldsymbol{\alpha})\, d\boldsymbol{\alpha} \right\} dx.\] Therefore \[\left|\,\int_A^B \!\left\{\int_a^\infty \!f(x,\boldsymbol{\alpha})\, dx \right\} d\boldsymbol{\alpha} - \int_a^\xi \left\{\int_A^B \!f(x,\boldsymbol{\alpha})\, d\boldsymbol{\alpha} \right\} dx\,\right|\] \[ \begin{align*} \qquad \qquad \qquad \quad \qquad &=\left|\,\int_A^B \!\left\{\int_\xi^\infty \!f(x,\boldsymbol{\alpha})\, dx \right\} d\boldsymbol{\alpha}\,\right| \\ &<\int_A^B \!\epsilon d\boldsymbol{\alpha} < \epsilon (B-A) \end{align*} \] for all sufficiently large values of \(\xi\).

But, from §2.1 and §4.41, this is the condition that \[\lim_{\xi \rightarrow \infty} \int_a^\xi \left\{\int_A^B \!f(x,\boldsymbol{\alpha})\, d\boldsymbol{\alpha} \right\} dx\] should exist, and be equal to \[\int_A^B \!\left\{\int_a^\infty \!f(x,\boldsymbol{\alpha})\, dx \right\} d \boldsymbol{\alpha}.\]

[12]Editor’s Note: The proof below only assumes \( \int_a^\infty \!\phi(x,\boldsymbol{\alpha})\,dx\) is convergent in a small neighborhood of \(\boldsymbol{\alpha}\) so that its derivative is meaningful. ↩
Corollary. The equation \(\displaystyle \frac{d}{d\boldsymbol{\alpha}} \int_a^\infty \!\phi(x,\boldsymbol{\alpha})\,dx = \int_a^\infty \!\frac{\partial \phi}{\partial \boldsymbol{\alpha}}dx\) is true if the integral on the right converges uniformly and the integrand is a continuous function of both variables, when \(x \geq a\) and \(\boldsymbol{\alpha}\) lies in a domain \(S\), and if the integral on the left is convergent.[12]

Let \(A\) be a point of \(S\), and let \(\frac{\partial \phi}{\partial \boldsymbol{\alpha}}=f(x, \boldsymbol{\alpha})\), so that, by §4.13 example 3, \[\int_A^\boldsymbol{\alpha} \!f(x, \boldsymbol{\alpha}')\,d\boldsymbol{\alpha}' = \phi(x,\boldsymbol{\alpha})-\phi(x,A).\] Then \( \int_a^\infty \left\{\int_A^\boldsymbol{\alpha} f(x,\boldsymbol{\alpha}')\, d\boldsymbol{\alpha}' \right\} d x\) converges,[13] that is \(\int_a^\infty \left\{ \phi(x,\boldsymbol{\alpha})-\phi(x,A) \right\} dx\) converges, and therefore, since \(\int_a^\infty \phi(x,\boldsymbol{\alpha})\,dx\) converges, so does \(\int_a^\infty \phi(x,A)\,dx\).

[13]Editor’s Note: By assumption, \( \int_a^\infty f(x,\boldsymbol{\alpha}')\, d x\) is convergent and, by (I), it is continuous. Since the conditions of (II) are satisfied, \( \int_A^\boldsymbol{\alpha} \left\{\int_a^\infty f(x,\boldsymbol{\alpha}')\, dx \right\} d\boldsymbol{\alpha}'\) is equal to \(\int_a^\infty \left\{\int_A^\boldsymbol{\alpha} f(x,\boldsymbol{\alpha}')\, d\boldsymbol{\alpha}' \right\} d x \), which must be convergent. ↩

Then \[ \begin{align*} \frac{d}{d\boldsymbol{\alpha}}\left[\int_a^\infty \!\phi(x,\boldsymbol{\alpha})\, dx \right]&=\frac{d}{d\boldsymbol{\alpha}}\left[\int_a^\infty \!\left\{\phi(x,\boldsymbol{\alpha})-\phi(x,A)\right\}\,dx\right] \\ &=\frac{d}{d\boldsymbol{\alpha}}\left[\int_a^\infty \!\left\{\int_A^\boldsymbol{\alpha}\! f(x,\boldsymbol{\alpha}')\,d\boldsymbol{\alpha}'\right\}\,dx\right] \\ &=\frac{d}{d\boldsymbol{\alpha}} \int_A^\boldsymbol{\alpha} \left\{\int_a^\infty \! f(x,\boldsymbol{\alpha}')\,dx\right\}\,d\boldsymbol{\alpha}' \\ &=\int_a^\infty \! f(x,\boldsymbol{\alpha})\, dx = \int_a^\infty \!\frac{\partial \phi}{\partial \boldsymbol{\alpha}} dx, \end{align*} \] which is the required result; the change of the order of the integrations has been justified above, and the differentiation of \(\int_A^\boldsymbol{\alpha} \) with regard to \(\boldsymbol{\alpha}\) is justified by §4.44 (I) and §4.13 example 3.