${}^{\ast }$Department of Mathematics, University of Texas, Edinburg, TX, 78540, USA, e-mail: paul.bracken@utrgv.edu

1 Introduction.

Invariance of certain types of integrals under special transformations is a subject of great interest for both theoretical and practical reasons [ 1 , 2 ] . These are often referred to as Landen transformations [ 3 , 4 ] . Suppose $I(\mathbf{p})$ represents a definite integral which depends on a set of parameters $\mathbf{p}=(p_1,\cdots ,p_m)$. A Landen transformation for integral $I(\mathbf{p})$ is a map $\phi$ defined on the parameters of $I$ such that

The classical example of a Landen transformation is given by [ 5 ]

It can be shown [ 6 ] that this operation preserves the elliptic integral,

This transformation of the parameters $u$ and $v$ can also be represented in the form,

It can be shown that this process defines a sequence or dynamical system $(u_n,v_n)$ inductively by

with $(u_0,v_0)=(u,v)$ is known to converge to a limit [ 6 ] ,

This limit is called the arithmetic-geometric mean of $u$ and $v$, and is denoted $\sigma =\text{rm AGM}(u,v)$. It also of interest in that it can be applied to calculate the elliptic integral $G$ by iteration quickly and efficiently.

It is the objective here to study the invariance of the specific integral

under two different but related types of transformation of parameters. Each transformation maps the integral back to its original form 1 but under a redefinition of the parameters $a,b,c$. Each of these procedures defines a different dynamical system, each with its own rate of convergence. After defining the dynamical system, some of the associated properties, such as its limit and convergence behavior, will be studied and several new proofs of these aspects will be presented. It will turn out to be more advantageous to work with 1 in trigonometric form. To obtain this form, make the substitution $x=\tan \vartheta$ in 1 to obtain,

For convergence of integral 1, it must be that $4ac-b^2>0$. Since $b^2\ge 0$, $a$ and $c$ must have the same sign. When they are both negative, a factor of negative one can be extracted from the denominator. It suffices then to restrict attention to the case in which $a$ and $c$ are strictly positive.

2 The Transformation of the Integral.

The first step is to develop some algebraic techniques which are required for the transformation. These will be common to both transformations investigated here. The integrand will be scaled by multiplying both numerator and denominator by an appropriate polynomial. To this end, a basis set of polynomials $U(x)$ and $V(x)$ is introduced as follows,

These two polynomials have the following property

It is required to determine coefficients $z_i$ and $g_i$ which depend on $a$, $b$ and $c$ such that the following identity holds

The coefficients of $x$ obtained from 5 generate a system of linear equations in the set of unknowns $\{z_i,g_i\}$. The solution set for the $\{z_i\}$ is given as follows

Two equations at order zero and one in $x$ remain, hence substituting the $z_i$ into them, $g_1$ and $g_2$ can be obtained in terms of $g_0$

Putting 6 back into the set of $z_i$ gives

This procedure transforms the integrand of 1 into the form

Integral 2 can now be expressed as

where $h_0(\vartheta )={\sin }^2\vartheta$, $h_1(\vartheta )=\sin \vartheta \cos \vartheta$ and $h_2(\vartheta )={\cos }^2\vartheta$. Now $h_0(\vartheta )$ and $h_2(\vartheta )$ can be replaced using the trigonometric identities $(1\mp \cos (2\vartheta ))/2$ and $2h_1(\vartheta )=\sin (2\vartheta )$. It is shown the following two integrals appearing in 9 vanish,

Consider first the integral $I_S$ under the transformation of the variable $\vartheta \to \vartheta +\pi /2$ and use the identities $\sin (2(\vartheta +\pi /2))=-\sin (2\vartheta )$ and $\cos (2(\vartheta +\pi /2))=-\cos \vartheta$. Next repeat this by shifting the variable $\vartheta \to \vartheta +3\pi /2$ and use the identities $\sin 2(\vartheta +3\pi /2)=-\sin (2\vartheta )$ and $\cos 2(\vartheta +3\pi /2)\to -\cos (2\vartheta )$. Substituting these results into $I_S$, the following two equivalent expressions for $I_S$ are obtained

Adding these two results for $I_S$, we find that

Now transform the original form of $I_S$ by means of $\vartheta \to \vartheta -\pi$ so that

Adding this result to the original form for $I_S$ and using periodicity, it is also the case that

Comparing 11 and 12, it follows that $I_S=-I_S$ which implies that $I_S=0$. A similar analysis shows that $I_C=0$. Therefore, it follows that only the constant terms contribute and 9 takes the form,

Making a change of variable $\varphi =2\vartheta$ in the integral on the right-hand side, 13 returns to its original form, but with a redefinition of parameters

Therefore, this procedure has produced an integral which has the same structure as the original $I$, however, the parameters $a$, $b$ and $c$ of the quadratic have been redefined in the process as follows

This procedure can be iterated again with 15 the first $n=1$ terms of an infinite sequence or dynamical system. Substituting 15 into $4a_1{c}_1-b_1^2$ and simplifying, it is found that it reduces to the original discriminant, $4ac-b^2$. Consequently, by induction, the resulting dynamical system will preserve the discriminant of the quadratic at each phase of the iteration.

3 Convergence Properties of First Dynamical System.

The convergence properties of the sequence given by 17 can be established. Let us define

Then 18 appears in the definition of the error ${ϵ}_n$,

It will be shown that ${ϵ}_n\to 0$ as $n\to \mathrm{\infty }$. To do this it will be convenient to introduce the transformation

Substituting 20 into 17 and solving for $x_{n+1}$ and $z_{n+1}$ gives

The sequence $(x_n,y_n,z_n)$ satisifes $4a_n{c}_n-b_n^2=x_n^2-y_n^2-z_n^2$ and is initialized by the values $(x_0,y_0,z_0)=(x,y,z)=(a+c,b,c-a)$.

To prove 22, it can be shown that it suffices to study a sequence in which the number of variables has been reduced from three to one. Note first that 22 can be put in the equivalent form

By invariance of the discriminant ${\gamma }^2=4a_n{c}_n-b_n^2=x_n^2-z_n^2-y_n^2=x^2-y^2-z^2$. Consequently, using $2x_n^2-2\gamma x_n+{\gamma }^2-x_n^2+y_n^2+z_n^2=2(x_n^2-\gamma x_n)$, the limit 23 can be expressed in the equivalent form,

Using the fact that ${\gamma }^2=x_n^2-y_n^2-z_n^2$, by invariance of the discriminant, the first equation in 21 for $\{x_n\}$ can be written in the equivalent form

By introducing the new sequence ${\alpha }_n$ obtained by scaling the $x_n$ as $x_n=\gamma {\alpha }_n$, the new dynamical relation is obtained

where $f(x)$ is defined to be

To establish convergence, note that the sequence results by iteration of $f(x)$, ${\alpha }_{n+1}=f({\alpha }_n)$. Since the derivative of $f(x)$ is negative when $x<1$ and positive when $x>1$, this function decreases monotonically on $(0,1)$, increases monotonically on $(1,\mathrm{\infty })$ and has a global minimum at $x=1$. In fact, $x=1$ is a fixed point of $f(x)$ and $x_0>0$ means ${\alpha }_0>0$. If ${\alpha }_0\in (0,1)$ then ${\alpha }_1\in (1,\mathrm{\infty })$, so it suffices to consider the case ${\alpha }_0\in (1,\mathrm{\infty })$. Since $f(x)>1$ for all $x\in (0,1)\cup (1,\mathrm{\infty })$, it follows that ${\alpha }_{n+1}=f({\alpha }_n)>1$, so ${\alpha }_n>1$ for all $n\in \mathbb{N}$ by induction. Moreover,

when ${\alpha }_n\in (1,\mathrm{\infty })$, so the sequence is bounded above and below and decreases monotonically. Therefore, the Monotone Convergence Theorem implies the sequence converges to a real number $\alpha \ge 1$. Since the limit exists, the limit can be calculated by letting $n\to \mathrm{\infty }$ on both sides of the recursion. This gives $\underset{n\to \mathrm{\infty }}{lim}(2{\alpha }_n{\alpha }_{n+1}-{\alpha }_n^2)=1$ which implies $\alpha =1$. Since this is the case, we calculate that

This means the convergence is quadratic. This also means that sequence $\{x_n\}$ converges and $\underset{n\to \mathrm{\infty }}{lim}{x}_n=\gamma$, and is consistent with 24.

4 The Second Integral Transformation.

Another transformation can be formulated for the same integral by simply changing the basis set of polynomials, $U(x),V(x)$. This change will result in a different dynamical system. This transformation has been introduced [ 7 ] , and will be used here. To do this, introduce the two polynomials

These polynomials have the following property,

It is required to determine a new set of coefficients $z_0,z_1,z_2,z_3$ and $g_0,g_1,g_2$ such that

When this is expanded out and like powers are collected, a set of six equations are obtained. Starting from order zero and solving, it is found that

This leaves two equations that arise from the two highest powers of $x$. These can be used to find the coefficients $g_i$. The fifth order equation for $g_2$ is solved in terms of $a,b,c,g_1$ and $g_0$, so that substituting this result into the equation for the leading term produces

Let us choose $g_0=a((a+3c{)}^2-3b^2)$, then it follows that $g_1=-b(b^2-3(a-c{)}^2)$, and this yields $g_2=-c(3b^2-(3+c{)}^2)$.

Using 29, the integrand of 2 is transformed into

where $h_k(\vartheta )={\sin }^{4-k}\vartheta {\cos }^k\vartheta$. Substituting this into integral 2, it follows that

The following identities

transform 32 into a linear combination of the following integrals,

Both $\sin (3\vartheta )$ and $\cos (3\vartheta )$ are invariant under shifts of $2\pi /3$ and $4\pi /3$, hence

The numerator reduces to either $3$ or $0$ depending on whether $3$ divides $k$ or not. This means that only the terms that contribute to $I$ are the constants in the identities at $k=0,2,4$. Therefore, by periodicity, $I$ becomes

Changing the variable according to $\vartheta =3u$ leads to the same structure for $I$ as in 2,

This is the original form of the integral, but with different set of parameters. Substituting the constants $z_i$ and $g_i$ in terms of $a$, $b$ and $c$, the new coefficents in 35 are given as

Moreover, substituting 36 into $4a_1{c}_1-b_1^2$ and simplifying, the original discriminant is recovered, $4ac-b^2$. This can be summarized in the following theorem.

5 Convergence Properties of the Second Dynamical System.

The convergence properties of the resulting dynamical system 37 will be studied. Define the error as was done in 19 and show that it approaches zero. Introduce the change of variables

so that the dynamical system 37 goes into the form

with initial conditions $x_0=x$, $y_0=y$ and $z_0=z$. Again, the discriminant of the quadratic is preserved by transformation 39.

System 40 can be expressed equivalently as,

The invariance of the discriminant can be exploited to put the first equation of 39 in the equivalent form

where $x_0=a+c$. Therefore, this invariance or symmetry of the discriminant has allowed us again to reduce the number of variables from three to one giving a simpler recursion. Thus, 41 is equivalent to

As before, let us scale the sequence $x_n$ and define a new sequence $\{{\alpha }_n\}$ by setting $x_n=\gamma {\alpha }_n$, so sequence 42 becomes

The sequence is then generated by means of the iteration ${\alpha }_{n+1}=h({\alpha }_n)$ where

In this instance, both $x=0,1$ are fixed points of function 45 which has derivative

when $x\ne 1$. It follows that $h(x)$ is strictly increasing on $(0,1)\cup (1,\mathrm{\infty })$.

Suppose first that ${\alpha }_0>1$. Then since $h$ is strictly increasing $h(x)>1$ when $x>1$ and so ${\alpha }_n>1$ then implies that ${\alpha }_{n+1}=h({\alpha }_n)>1$ for all $n\in \mathbb{N}$ and

This implies that the sequence $\{{\alpha }_n\}$ is monotonically decreasing and is bounded below by one. Therefore, it must converge to a limit in $[1,\mathrm{\infty })$ by the Monotone Convergence Theorem.

Next, suppose now that $0<{\alpha }_n<1$, then since $h:(0,1)\to (0,1)$, we have ${\alpha }_{n+1}=h({\alpha }_n)<1$ and by 47, it follows that ${\alpha }_{n+1}-{\alpha }_n>0$. Therefore, the sequence is monotonically increasing and bounded above by one, so the sequence again converges by the Monotone Convergence Theorem. Since the limit $\alpha =\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n$ exists and satisfies the equation $\alpha =h(\alpha )$, the only positive solution of this gives $\alpha =1$. Although the sequence in the first case was shown to have quadratic convergence, under this second transformation, it is found that

Therefore, this procedure generates a sequence in which the convergence is cubic. Moreover, from the definition of the $x_n$ in terms of the ${\alpha }_n$, it follows that $\underset{n\to \mathrm{\infty }}{lim}{x}_n=\gamma$ in agreement with 43.

It is worth stating in summary that these kinds of transformations are going to be useful in fast numerical evaluation of integrals. For either transformation considered here, moving the limit $n\to \mathrm{\infty }$ inside the integral, for example 16, allows the explicit calculation of the integral itself

It may be conjectured that this iterative process can be continued over other basis sets of polynomials, for example Chebyshev polynomials, which could result in a hierarchy of dynamical systems.