A4.4.2 Integration and Differentiation of Improper Integrals with respect to a Parameter: In general, it is not true that improper integrals may be differentiated or integrated under the integral sign with respect to a parameter. In other words, these operations are not interchangeable in order with the original integration (example ).
In order to determine whether the order of integration in improper repeated integrals is reversible, we can often use the following test or else make a special investigation along the lines of the following proof.
If the improper integral
converges uniformly in the interval a £ x £ b, then
For the proof, we set
Then, by assumption, |RA(x)| < e(A), where e(A) is a number which depends only on A and not on x and tends to zero as A®¥. By virtue of the elementary theorem for ordinary integrals, we have
whence by the mean value theorem of the integral calculus
If we now let A tend to infinity, we obtain the above formula.
If the integration with respect to a parameter also takes place over an infinite integration interval, the change of order is not always possible, even though, the convergence is uniform. However, it can be performed, if the corresponding improper double integral exists . Thus, for example,
if there exists the double integral 
over the entire first quadrant.
The proof of this follows from the fact that the improper double integral is independent of the mode of approximation to the region of integration. In one case, we perform this approximation by means of infinite strips parallel to the x-axis, in the other case, by strips parallel to the y-axis.
A similar result also holds, if the interval of integration is finite, but the integrand is discontinuous along a finite number of straight lines y = const or on a finite number of more general curves in the region of integration. The corresponding theorem is: If when x lies in the interval a £ x £ b the function f(x, y) is discontinuous only along a finite number of straight lines y=a1, y= a2, ··· , y = ar, and if the integral
converges uniformly in x, then it represents in this interval a continuous function of x, and
Thus, under these assumptions, the order of integration can be changed. The proof of the theorem is analogous to that given above.
It is equally easy to extend the rules to differentiation with respect to a parameter. One has the theorem: If the function f(x,y) has a sectionally continuous derivative with respect to x in the interval a £ x £ b and the two integrals
converge uniformly, then
Thus, under these conditions, the order of the integration and differentiation with respect to a parameter can be interchanged. In fact, if we set
then, using the theorem of interchangeability just proved, we have
The integrand on the right hand side has the value
whence
hence, if we differentiate and then replace x by x, we obtain
as was to be proved.
Similarly, we can extend the rule for differentiation when one of the limits depends on the parameter x. In fact, we can write
where a is any fixed value in the interval of integration; we can then apply previously proved rules to each of the two terms on the right hand side.
As above, our rules of differentiation also hold for improper integrals with finite intervals of integration.
A4.4.3 Examples: 1. Consider the integral
If x ³ 1, this integral converges uniformly, since we have for positive values of A
where the right hand side no longer depends on x and can be made as small as we please, if we choose A sufficiently large. The same is true of the integrals of the partial derivatives of the function with respect to x. By repeated differentiation, we thus obtain
In particular, if we set x = 1, we have
This formula has already been established in a different way in Volume I, 4.8.4.
2. Next, consider the integral
Again, it is easy for us to convince ourselves that if x £ a, where a is any positive number, all the assumptions required for differentiation under the integral sign are satisfied. Hence, we find by repeated differentiation the sequence of formulae
From these formulae, we can derive another proof of Wallis's product for p. In fact, if we set 
, we
have
As n increases, the left-hand side tends to the integral
In fact, the difference
satisfies the inequality
However, if we choose T so large that
and then n so large that
as can be done by virtue of the uniform convergence of the process,
it follows at once that
This establishes the relation
which is equivalent to Wallis's product.
3. With a view to calculating the integral
we shall discuss the function
This integral converges uniformly if x ³ 0, while the integral
converges uniformly if x ³ d > 0, where d is an arbitrarily small positive number. Both these statements will be proved below. Hence, F(x) is continuous if x ³ 0 and, if x ³ d , we have
We can easily evaluate this last integral by integrating twice by parts; we obtain
Hence, we can find the value of F(x) by integration; this value is
where C is a constant. By virtue of the relation
which holds if x ³ d,
we see that 
Since lim 
C must also be 0 and we find
On account of the continuity of F(x) for x ³ 0,
which, since
yields the required formula
We now return to the proof that
converges uniformly if x ³ 0. If A is an arbitrary number and kp is the least multiple of p which exceeds A., we can write the remainder of the integral in the form
The terms of the series on the right hand side have alternating signs and their absolute value tends monotonically to zero, whence, by Leibnitz's convergence test, the series converges and the absolute value of its sum is less than that of its first term and we have the inequality
in which the right-hand side is independent of x and can be made as small as we please. This establishes the uniformity of the convergence. The uniform convergence of
follows at once from the relation
In A4.4.2, we have learned that uniform convergence of integrals is a sufficient condition for interchangeability of the order of integration. Mere convergence is not sufficient, as the following example shows:
If we set f(x, y) = (2— xy)xye-xy, then, since
the integral 
exists for every x
in the interval 0 £ x £ 1 and, in fact, for every such value of x
it has the value 0, whence
On the other hand, since
for every y ³ 0, we have
and therefore
Hence,
A4.4.4 Evaluation of Fresnel Integrals: The integrals
which are of importance in optics, are known as Fresnel integrals. In order to evaluate them, we apply the substitution t²=t and obtain
Now, set
(this follows from the substitution 
and change the order of
integration, as you may do by our rules. Then,
The inner integrals are easily evaluated by integration by parts, and F1 and F2 reduce to the elementary rational integrals
These integrals may be evaluated by the methods given in Volume
I, 4.6; the second integral can be reduced to the first by
means of the substitution x' = 1/x and
both have the value 
i.e.
1. Evaluate
2. How must one choose a, b, c in order that
3. Evaluate
where J0 denotes the Bessel function .
5*. Prove that 
is of the order of log n when n
is large and that
6. Replace the statement "the integral 
is
not uniformly convergent " by an equivalent statement not
involving any form of the words "uniformly
convergent". (Volume I)
A4.5.1 Introduction: The theory of A4.4.2 is illustrated by the important example known as the Fourier integral theorem. You will remember that Fourier Series give a representation of a sectionally smooth, but otherwise arbitrary periodic function in terms of trigonometric functions. The Fourier integral gives a corresponding trigonometrical representation of a function f(x) which is defined in the entire interval -¥ < x < +¥ and is not subject to any condition of periodicity.
We shall make about the function f(x) the assumptions:
(1) f{x) is sectionally smooth,
i.e., the function f(x) and its first derivative
are continuous in any finite interval, except possibly for a
finite number of jump discontinuities.
(2) The integral
converges.
(3) At a discontinuity x of the function, it is assumed
that f(x) is the arithmetic mean of the limits on the
right and left hand side. Thus
One has then the Fourier integral theorem:
or, in complex notation, we have the equivalent formula
We may also restate the theorem in the form: If
then
The last two formulae are reciprocal equations for f(x) and g(x), each equation being the solution of the other. If the variable r = t/2p is introduced and finally replaced by t again, we can express the Fourier integral theorem by means of the two reciprocal formulae
where
We shall give some examples to illustrate this theorem and then proceed to its proof. First of all, note that, if f(x) is an even function, i.e., if f(x) = f(-x), then a short calculation shows that the theorem may be stated in the simplified form
If, on the other hand, f(x) is an odd function, i.e., if
we obtain in the same way
1. Let f(x) = 1 when x² < 1, f(x) = 0 when x² > 1. Then,
The integral on the right hand side has appeared in the mathematical literature as Dirichlet discontinuous factor.
2. Let f(x) = e-kx (k > 0) when x > 0 and f(x) =f(-x). It is easy to show that
However, if we set f(-x) = -f(x), we obtain
Hence, we obtain the two integral formulae
3. The function f(x) = e-x²/2 yields an interesting illustration of the reciprocal formulae. Since
(Exercise 4.7, 4a), the two reciprocal formulae for g(t) and f(x) coincide.
A4.5.2 Proof of the Fourier Integral Theorem: The essential steps in the proof of the Fourier integral formula are a transformation and a simple limit operation applied to Dirichlet's limit formula
which holds for arbitrary positive values of a. We shall first prove this formula, although the substance of the proof has been given in Volume I, 9.5.1 . We will relyl on the elementary limit formula given there
which holds when s(t) ia continuous or sectionally continuous in the interval a £ t £ b, but is otherwise arbitrary.
Consider first the interval from 0 to a. In this interval,
is a sectionally continuous function which, by the assumptions regarding f(x), must have the limit f '(x + 0) as t tends to zero. Thus,
and the elementary formula given above shows that the last integral on the right hand side tends to zero as l tends to infinity.
The first integral on the right hand side has the limit
If we now apply the corresponding argument to the integral from -a to 0, we obtain Dirichlet's formula.
The next step in the proof of Fourier's theorem is the substitution of
in Dirichlet's formula. We also introduce the notation
Dirichlet's formula then states that
Since this limit is independent of a, we may write
If it were permissible to interchange the order of the limit operations in this formula, i.e., if we might take the limit as a tends to infinity under the integral, we should at once have
This yields immediately the Fourier integral formula, if we write x + t = t' and then replace t' by t. Thus, the proof will be complete, if we justify the change of the order of the limit operations
Our previous work in A4.4.2 and A2.2.1 shows that it is sufficient to prove that the limit
exists uniformly with respect to l.
In order to prove this, we must show that, if e is given in advance, we can find A independent of l such that
whenever both a and b exceed A. However,
It follows at once that
so that it is only necessary to take A = 2C/e. This yields the proof of uniform convergence, and completes the proof of the Fourier integral theorem.
A4.6 THE EULERIAN INTEGRALS (GAMMA FUNCTION)
A discussion closely related to the present one is given by E. Artin, Einführung in die Theorie der G-Funktion - Introduction to the theory of the G-Function (Leipzig, 1931).
One of the most important examples of a function defined by an improper integral involving a parameter is the Gamma-function G(x). We shall give a fairly detailed discussion of this function.
A4.6.1 Definition and Functional Equation: The function G(x) is defined for every x > 0 by the improper integral
We have studied this integral for integral arguments x = n in Volume I, 4.8.4,. The method used there shows at once that the integral converges for any x > 0, the convergence being uniform in every closed interval of the positive x-axis which does not include the point x = 0. The function G(x) is therefore continuous for x > 0.
By simple substitutions, we can transform the integral for G(x) into other forms which are often used. Here we only mention the substitution t = u² which transforms it into
Thus, the frequently occurring integral
can be expressed in terms of the G-function as
Integration by parts shows, as in Volume I, 4.8.4, that the relation
holds for any x > 0. This equation is called the functional equation of the Gamma function.
Of course, G(x) is not uniquely defined by the property of being a solution of this functional equation. In fact, we obtain another solution merely by multiplying G(x) by an arbitrary periodic function p(x) with period unity. On the other hand, the functions
represent the aggregate of all solutions of the equation; in fact, if u(x) is any solution, the quotient
which can always be formed since G(x) ¹ 0, satisfies the equation
Instead of the function G(x), it is frequently more convenient to consider the function u(x) = log G(x); since G(x) > 0 for x > 0, this is always defined. The function satisfies the functional equation (difference equation)
We obtain other solutions of this equation by adding to log G(x) an arbitrary periodic function with period unity. In order to specify log G(x) uniquely, we must therefore supplement the functional equation by other conditions. One very simple condition of this type is given by H. Bohr's theorem:
Every convex solution of the difference equation
is identical with the function log G(x) in the interval 0 < x < ¥, except perhaps for an additive constant.
A4.6.2 Convex Functions: Proof of Bohr's Theorem: We say that a function f(x) is convex in a region a £ x £ b if for every two points x1 and x2 of the region and every two positive numbers a, b, where a + b = 1, the expression
never changes sign or, intuitively speaking, if the chord joining two points of the curve y = f(x) either never lies beneath or never above the arc of the curve itself between x1and x2 (Fig. 19)(1.1.4, A2.1.3)
Before proving this theorem, we shall establish certain properties of convex functions. We restrict ourselves to functions which are convex downwards, for which
holds; functions which are convex upwards can always be changed into functions which are convex downwards by multiplication by —1. If a convex function f{x) is twice continuously differentiable, the expression
can be represented by the double integral
as is easily verified. Thus, the inequality of the definition is certainly satisfied, provided that
On the other hand, the passage to the limit x1 ® x2 shows that this condition is also necessary, and it is therefore a characteristic property of convex functions which are twice continuously differentiable.
A fact worth noting and useful in applications is that we need not assume the continuity of a convex function f(x); on the contrary, this property follows from the definition of convexity. In fact, we can replace the above inequality by an apparently weaker one, which, however, is equivalent to it and expressed in the theorem:
If for every x and h, for which the arguments x ± h still lie in the region of definition, the bounded function f(x) satisfies the inequality
i.e., if the mid-point of every chord of the curve y = f(x) lies above or on the curve, then f(x) is convex.
We first show that every bounded function f(x) which satisfies the inequality
is continuous.
In order to prove this, we rewrite this condition in the form
from which we derive the inequalities
valid for every integer n ³ 0. If we add these for values of n from n = 0 to n = n - 1, we obtain the estimate
whence, if we assume that |f(x)| £ C,
Here, n can be any positive integer such that the argument x ± h lies in the interval of definition. If we let h tend to zero, the largest possible number n increases without limit, i.e., the expression f(x + h) - f(x) tends to zero. This proves the continuity of f(x).
From the continuity of f(x), we can now easily prove its convexity, i.e., we can establish the inequality
From the inequality
by means of the substitution
we obtain
and hence, in general,
If we put x + nh = x1, a few transformations yield
which is exactly the inequality we require for rational values of a and b. We then deduce its validity for any values of a and b from the continuity of f(x).
From the inequalities
the validity of which for any numbers c £ a £ b differing from zero is a direct consequence of the definition of convexity, we see that the difference quotient [f(x + h) - f(x)]/a is bounded and monotonic if a tends to zero through positive values or through negative values, whence it has a limit. Thus, a convex function has a derivative on the right and on the left hand side at every point.
Finally, we apply the following inequality, which is obvious from the geometrical interpretation of a convex function:
where d and h are two positive numbers, d £ h.
This is proved by adding the two relations
We now return to Bohr' theorem.We see at once that log G(x) is convex. In fact, if we write G(x) in the form
where h has any positive value and x any value larger than h, and apply Schwarz's inequality, we have
whence
This fact is a special case of a general theorem. If the functions fn(x), n = 1, 2, ··· , n satisfy the conditions
so that the functions log fn(x)
are convex, then the sum 
also satisfies these conditions.
In fact, if we write in
the form
and use the relation
we obtain the inequality
if we now apply Schwarz's inequality to the right hand side, we have
An analogous theorem holds for integrals of the form
if for all values of the parameter t the functions fn(x, t) satisfy the conditions
The gamma function is of this type.
Again, if f(x) and g(x) are two continuous convex solutions of the functional equation
the difference 
is
a continuous periodic function of period unity. Moreover, since 
f(x)
satisfies the relation 
Since f(x) is convex,
the inequality 
holds
for every h in the range 0 < h £ 1 (cf. above). We likewise
obtain 
whence
If we now let x
increase beyond all bounds, the expression log x/(x
- 1) tends to zero, and so does the function 
Since this function is periodic, we obtain the equation
valid for every x > 0.
A continuous periodic function f(x), which satisfies such a condition for every positive value of h and every value of x > h, must be a constant. However, this proves that any continuous convex solution of the equation u(x + 1) - u(x) = log.x can only differ from log G(x) by an additive constant.
If, say, f(1) = f(2) = a, then, by the equation f(3/2) = [f(l) + f(2)]/2, we have f(3/2) = a, and likewise f(xn) = a at all points xn of the interval 1 £ x £ 2, which are obtained by repeated bi-section of the sub-intervals. Since these points xn are everywhere dense, it follows from the continuity of f(x) that f(x) = a throughout the interval 1 £ x £ 2, and by the periodicity of f(x) this holds for every x > 0.
A4.6.3 The Infinite Product for the Gamma Function: We shall give here the infinite products for the G-function found by Gauss and Weierstrass.
We first show that the relation
holds, where
This statement is plausible, since for integers x = n, we have
and, as n increases, this obviously tends to the value (n - 1)!
We must show in general that the sequence Gn(x) converges for every x ¹ 0, -1, -2, ··· , and then that the limit function G(x) coincides with the G-function for positive values of x. In order to prove this last statement, we note that if x > 0 the
By Bohr's theorem, we need only show that log G(x) is convex.
In order to prove the convergence of the sequence Gn(x) for x ¹ 0, -1, -2, ···, we introduce the expression
for the number n, and accordingly write
By a test, proved in Volume I, A8.3, the product
converges absolutely and uniformly, provided that the series
converges uniformly. If we use the Taylor expansion in powers of 1/n up to the terms of second order, the general term of this series can be written in the form
where q is a number between 0 and 1, whence
where C is a constant independent of n. In every closed region which contains none of the points x = -1, -2, ··· we can replace the estimate C by a number which is also independent of x. In. every such region, the series converges uniformly, and therefore so does the product.
The limit function
is continuous for every x ¹ 0, -1, -2, ··· and, as we see at once, satisfies the functional equation
In order to show that, if x > 0, the function G(x) is identical with the function G(x), we consider the function log G(x) for x>1. It is the limit function of the sequence
For any positive value of h and any value of x larger than h, the functions log Gn(x) satisfy the convexity condition for
which consequently also applies to the function log G(x). Moreover, since
by the general theorem, G(x) must be identical withG(x). Hence we have obtained Gauss's infinite product forG(x):
The theoretical importance of this expression arises from the fact that we can regard it as not only defining the G-function for all positive values of x, but also for all negative non-integral values of x.
This product can easily be given a somewhat different form. If we substitute in the expression
for log n the value
where g is Euler's constant and let en tend to zero as n ® ¥, we obtain for 1/G(x)
Since the factor 
tends to 1 as n increases, the
product 
also converges and yields Weierstrass'
infinite product for1/G(x)
from which we see at once that 1/G(x) has zeros of the first order at the points x = 0, -1, -2, ··· .
A4.6.4 The Function log G(x) and its Derivatives:. If we form the logarithm of Weierstrass' infinite product
we obtain for the function log G(x):
By the relation
whence
the right.hand side of the equation for log
G(x) is dominated by the series 
, and therefore converges absolutely and uniformly
in every closed interval of the positive x-axis.
The derivatives of the function log G(x) are of particular interest, since they provide an explicit representation of the values of the series
s 
If we differentiate the expression for log G(x) term by term with respect to x, we again obtain a series which, since
converges absolutely and uniformly in every closed interval of the positive x-axis. Hence, by known theorems on the differentiation of infinite series,
If we differentiate again term by term, we obtain similarly
and, finally, forming the higher derivatives,
A4.6.5. The Extension Theorem:. The values of the G-function for negative values of x are easily obtained from its values for positive x by means of the so-called extension theorem. If we form the product G(x)G(-x)
and combine the two limiting processes into one, we obtain
But by the infinite product for the sine in Volume I, 9.4.7,
we have
Hence,
We can put this relation in a somewhat different form by calculating the product G(x)G(l - x). Since
we obtain the extension theorem
Thus, if we set x = ½ , we find 
Since
this yields a new proof of the fact that the integral 
has
the value 
In addition, we can calculate the G-function for the arguments x= n
+ ½, where n is any positive integer:
A4.6.6 The Beta Function:. Another important function defined by an improper integral involving a parameter is Euler's beta function. The beta function is defined by
If either x or y is less than 1, the integral is improper. However, by the criterion of A4.4, it converges uniformly in x and y, provided we restrict ourselves to intervals x £ e, y £ h, where e and h are arbitrary positive numbers. It therefore represents a continuous function for all positive values of x and y.
We obtain a somewhat different expression by using the substitution t = t + ½:
or, in general, if we now put t = t/2s, where s > 0,
Finally, if we set t = sin²f in the original formula, we obtain
We shall now show how the Beta function can be expressed in terms of the Gamma function by using a few transformations which may seem strange at first sight. If we multiply both sides of the equation
by e-2s and integrate with respect to s from 0 to A, we have
The double integral on the right hand side may be regarded as an integral of the function
the region of integration being the isosceles triangle bounded by the lines
If we now apply the transformation
this integral becomes
We now have as the region of integration the triangle in the st-plane bounded by the lines s = 0, t = 0 and s+t = 2A. If we now let A increase beyond all bounds, the left hand side tends to the function
The right-hand side must therefore also converge and its limit is the double integral over the entire first quadrant of the st-plane, the quadrant being approximated to by means of isosceles triangles. Since the integrand is positive in this region and the integral converges for a monotonic sequence of regions, by 4.5.5, this limit is independent of the mode of approximation to the quadrant.
In particular, we can use squares of side A and accordingly write
Hence, we obtain the important relation
We see from this relation that the beta function is related to the binomial coefficients
in roughly the same manner as the G-function is related to the numbers n! In fact, for integers x= n, y = m, the function
has the value
This equation can also be obtained from Bohr's theorem. We first show that B(x, y) satisfies the functional equation
so that the function
considered as a function of x, satisfies the functional equation of the G-function
Since it follows from the earlier theorem that log u(x, y) is a convex function of x, we have
and, finally, if we set x = 1, a(y) = G(y).
Finally, we note that the definite integrals
which are identical with the functions
can be simply expressed in terms of the G-function:
1. Prove that the volume of the positive octant bounded by the planes x = 0, y = 0, x = h and the surface
2. Prove that
taken throughout the positive octant of the ellipsoid
is equal to
(Hint: Introduce new variables x, h, z by writing
and perform the integration with respect to x, h, z )
3. Find the x-co-ordinate of the centre of mass of the solid
4. Find the moment of inertia of the area enclosed by the astroid
with respect to the x-axis.
6. Prove that
A4.7 DIFFERENTIATION AND INTEGRATION TO FRACTIONAL ORDER. ABEL'S INTEGRAL EQUATION
Using our knowledge of the G-function, we shall now carry out a simple process of generalization of the concepts of differentiation and integration. We have already seen that the formula
yields the n-times repeated integral of the function f(x)
between the limits 0 and x. If D denotes
symbolically the differentiation operator and D-1
the operators 
which is the inverse of differentiation, we may
write
The mathematical statement conveyed by this formula is that the function F(x) and its first (n - 1) derivatives vanish at x=0 and the n-th derivative of F(x) is f(x). It is now very natural to construct a definition for the operator D-l even when the positive number l is not necessarily an integer. The integral of order l of the function f(x) between the limits 0 and x is defined by
This definition may now be used to generalize nth-order differentiation, symbolized by the operator Dn or dn /dxn, to m-th order differentiation, where m is an arbitrary non-negative number. Let m be the smallest integer larger than m, so that m = m - p, where 0 < p < 1. Then our definition is
A reversal of the order of the two processes would yield the definition
It may be left as an exercise for the reader to employ the formulae for the G-function to prove that
where a and b are arbitrary real numbers. He should show that these relations and the generalized process of differentiation have a meaning whenever the function f(x) is differentiable in the ordinary way to a sufficiently high order. In general, Dnf(x) exists if f(x) has continuous derivatives up to and including the m-th order.
In this context, we may mention Abel's
integral equation, which has important
applications. Since 
the integral of
a function f(x) to the order ½ is given by the
formula
If we assume that the function y(x) on the right-hand side is given and that it is required to find f(x), then the above formula is Abel's integral equation. If the function y(x) is continuously differentiable and. vanishes at x = 0, the solution of the equation is given by
A4.8. NOTE ON THE DEFINITION OF THE AREA OF A CURVED SURFACE
In 4.6.3, we have defined the area of a curved surface in a way somewhat dissimilar to that in which we have defined the length of arc in 5.2.5. In the definition of length, we started with inscribed polygons, while in the definition of area, we used tangent planes instead of inscribed polyhedra.
In order to see why we cannot use inscribed polyhedra, we may consider a cylindrical surface in xyz-space with the equation x² + y² = 1, lying between the planes z = 0 and z = 1. The area of this cylindrical surface is 2p. We now inscribe in it a polyhedral surface, all faces of which are identical triangles: We first subdivide the circumference of the unit circle into n equal parts and consider on the cylinder the m equidistant horizontal circles z = 0, z = h, z = 2h, ··· , z = (m - 1)h, where h = l/m. We perform the subdivision of each of these circles into n equal parts in such a way that the points of subdivision of each circle lie above the centres of the arcs of the preceding circle. We now consider a polyhedron inscribed in the cylinder the edges of which consist of the chords of the circles and of the lines joining the neighbouring points of subdivision of the neighbouring circles. The faces of this polyhedron are congruent isosceles triangles and, if n and m are chosen sufficiently large, this polyhedron will lie as close as we please to the cylindrical surface. If we now keep n fixed, we can choose m so large that each of the triangles is as nearly parallel as we please to the xy-plane and therefore makes an arbitrarily steep angle with the surface of the cylinder. Then we can no longer expect that the sum of the areas of the triangles will be an approximation to the area of the cylinder.