Integral Tables
Dennis Haag
lot of differential equations and need the answers more than the
practice. Any information would be appreciated.
=-=-=-=-=-=
Dennis R. Haag
University Computing Services
Brigham Young University
Denni...@byu.edu
David Hill
>Date: Fri, 18 Sep 92 18:44:35 MDT
>Subject: Integral Tables
>From: Dennis Haag <Denni...@byu.edu>
You could use one of the symbolic software packages like Derive which can
solve many differential equations.
Marc Roussel
(David Hill) writes:
>In article <d##@byu.edu> Dennis Haag <Denni...@byu.edu> writes:
>>I am looking for a very extensive table of integrals. I'm working with a
>>lot of differential equations and need the answers more than the
>>practice. Any information would be appreciated.
>
>solve many differential equations.
There are two problems with this advice:
1) None of the symbolic packages can really be trusted for
integration. They produce correct answers MOST of the
time, but that's just not good enough for research use.
Sometimes, the errors in the answers produced are so
subtle that they might even pass several tests of
verisimilitude, especially if these tests are performed
with the same symbolic algebra program.
2) Even when they produce answers, the form in which this
answer is given is often far more complex than is
necessary. Computers don't have the sense of aesthetics
that we do, and aesthetics plays a far greater role in
mathematics than many of us realize.
Symbolic algebra systems are useful tools in research. I
personally don't even want to think about going back to doing everything
by hand. In particular, symbolic algebra programs are very good at
automating the process of result verification. I just don't think that
any of them can integrate well enough to trust them. At least in my
opinion, tables are still the preferred way of obtaining definite and
indefinite integrals.
Marc R. Roussel
mrou...@alchemy.chem.utoronto.ca
Gerald Edgar
> 1) None of the symbolic packages can really be trusted for
> integration. They produce correct answers MOST of the
> time, but that's just not good enough for research use.
The makers of one of the symbolic packages went through some of the
standard integral tables to compare and find errors in the symbolic
program. They reported that about ten percent of the integrals published in
the standard tables are wrong...
(I forget where I read this.)
--
Gerald A. Edgar Internet: ed...@mps.ohio-state.edu
Department of Mathematics Bitnet: EDGAR@OHSTPY
The Ohio State University telephone: 614-292-0395 (Office)
Columbus, OH 43210 -292-4975 (Math. Dept.) -292-1479 (Dept. Fax)
John Harper
>opinion, tables are still the preferred way of obtaining definite and
>indefinite integrals.
>
some other renowned table-compiler who said that a formula for some integral
in a table should be used merely as evidence that some formula vaguely
resembling the given one is true, which it is now your job to find; there
were a great many errors in all such tables. (I quote this when teaching
calculus of residues and would be grateful for a correct reference.)
John Harper Mathematics Dept. Victoria University Wellington New Zealand
Steve Linton
Steve Linton
Jens Tingleff
> In article <19q00a...@function.mps.ohio-state.edu> ed...@function.mps.ohio-state.edu (Gerald Edgar) writes:
> >The makers of one of the symbolic packages went through some of the
> >standard integral tables to compare and find errors in the symbolic
> >program. They reported that about ten percent of the integrals published in
> >the standard tables are wrong...
> >
> >(I forget where I read this.)
I don't know about a ref. in writing for this, but that example was part of the
sales/demo talk that Stephen Wolfram gave when vers. 2 came out. Apparently, they
had scanned several standard integral table books (i.e. put each page of the book
into a scanner and "read" the integrals into the computer). Wolfram said that
there were surprisingly few errors in the books... But then again, he would ;-)
--
Mr Jens Tingleff, M.Sc.EE. PhD student at
Imperial College, Dept of EE, Exhibition Road, London SW7 2BT, England
jens...@ic.ac.uk or jens...@titan.ee.ic... (used to be jens...@diku.dk)
"I was just a child then, now I'm only a man" Pink Floyd, 'The Final Cut'
Marc Roussel
I think that Mathematica (and Maple and Macsyma and ...) all use
Risch integration at one point or another. The problem is that in order
to use the Risch algorithm it is normally necessary to do some symbolic
manipulations first; that is where these programs (including Reduce and,
I suspect, Axiom) normally fail. For instance, how you should simplify
sqrt((1-x)^2) depends on the range of the integral; abs(1-x) is always
correct but causes the symbolic integrator fits. This is just one
example, albeit typical of a common class of potential difficulties.
I have been thinking about this matter a little in the last year or
two and have come to the conclusion that symbolic algebra systems should
be more conservative than they now are. The attitude among the
designers of such programs right now seems to be that they should
endeavour to return an answer as often as possible. I think rather that
they should only return an unqualified answer when every step leading to
the solution is correct beyond any doubt; it should be permissible to
return an answer otherwise only if the user is clearly warned that
the answer may not be correct for all possible values of the parameters
(or, in many cases, for all values of the parameters). I realize that
there are marketing reasons which prevent this from ever happening.
Marc R. Roussel
mrou...@alchemy.chem.utoronto.ca
John Prentice
> I have been thinking about this matter a little in the last year or
>two and have come to the conclusion that symbolic algebra systems should
>be more conservative than they now are. The attitude among the
>designers of such programs right now seems to be that they should
>endeavour to return an answer as often as possible. I think rather that
>they should only return an unqualified answer when every step leading to
>the solution is correct beyond any doubt; it should be permissible to
>return an answer otherwise only if the user is clearly warned that
>the answer may not be correct for all possible values of the parameters
>(or, in many cases, for all values of the parameters).
I strongly endorse what Marc is saying here! What is amazing is that it
has to be said, it is simple good software design. We, the consumers,
should not be tolerating bad mathematical software, we should be raising
hell with these developers.
John
--
Dr. John K. Prentice
Partner, Quetzal Computational Associates
3200 Carlisle N.E., Albuquerque, NM 87110-1664
Phone: 505-889-4543 E-mail: jo...@aquarius.unm.edu -or- jkp...@cs.sandia.gov
D. J. Bernstein
> to use the Risch algorithm it is normally necessary to do some symbolic
> manipulations first; that is where these programs (including Reduce and,
> I suspect, Axiom) normally fail. For instance, how you should simplify
> sqrt((1-x)^2) depends on the range of the integral; abs(1-x) is always
> correct but causes the symbolic integrator fits.
Gee, was it just my imagination that Macsyma would pester me with ``is
1-x positive or negative?'' whenever I gave it such an integral? Sure,
Mathematica thinks that sqrt((-1)^2) = -1, but just because the new
technology produces wrong answers doesn't mean that the old technology
did.
---Dan
Richard Fateman
do, versus what it actually does in the context of a symbolic math system.
Let's take the simplest problem imaginable, integrate(0,x). The answer
we would expect from the Risch algorithm is 0 (plus a constant, as they say).
Let us say that someone offers you another algorithm A that is alleged
to compute indefinite integrals just like the Risch algorithm.
Being a skeptic, you
test the A algorithm, starting with the simplest example, integrate(0,x).
Say A gives for an answer, f(x):=arctan(x)+arctan(1/x). You say that
is absurd, it should be 0 (plus a constant). Try evaluating f(x) though.
f(1.0) = f(2.0) = f(3.0) --- IT IS ZERO PLUS A CONSTANT (pi/2).
Try differentiating f(x). You get a rational function that is equivalent to
zero... so it has the right derivative...f(x) is an anti-derivative of zero,
so maybe it is right???
Look again (maybe graph f(x))... and you notice that
f(-1.0) = f(-2.0) = ... for negative arguments f(x) is ZERO PLUS A CONSTANT...
BUT IT IS A DIFFERENT CONSTANT (-pi/2). oops...
And what about f(0) ? It too is a constant, though we don't know which...
Is this answer f(x) correct? According to the Risch rules, f(x) IS AN
ANTIDERIVATIVE of 0.
Now this may seem to you rather ridiculous -- why return f(x) when a perfectly
good alternative representation (namely 0) is a much better answer? The
answer f(x) is discontinuous, it has a singularity at zero (and another at
infinity, but that you may not have noticed).
The fact of the matter is, the Risch algorithm may return answers that are
just as ridiculous as A's f(x). They may not be defined everywhere you want
them to be defined and they may be discontinuous. They may even be
defined nowhere.
You might not notice these problems because the answer would be more
complicated that atan(x)+atan(1/x). Systems like Mathematica (especially)
would not notice these problems either. That's been shown over and over
again. Maple and Macsyma may be more careful, but they are not careful
enough. (Some research here at Berkeley is pursuing how to be careful.)
In a nutshell, any algorithm that only guarantees algebraic
"closed-form antiderivatives", like the Risch algorithm, is perfectly
entitled to give you f(x) instead of 0 for an answer to the "antiderivative
of 0". If you don't like this answer, it is not the fault of the Risch
algorithm. If Mathematica (etc.) gives you the wrong integral (indefinite
or definite) because it uses the Risch algorithm, then don't complain to
Robert Risch. He solved Liouville's conjecture on "integrability in
terms of elementary functions" in part by restating it in terms of
differential fields. He didn't tell anyone to write a program.
--
Richard J. Fateman
fat...@cs.berkeley.edu 510 642-1879
Richard Bumby
>In article <1992Sep24.1...@alchemy.chem.utoronto.ca> mrou...@alchemy.chem.utoronto.ca (Marc Roussel) writes:
>> I have been thinking about this matter a little in the last year or
>>two and have come to the conclusion that symbolic algebra systems should
>>be more conservative than they now are. . . .
>I strongly endorse what Marc is saying here! What is amazing is that it
>has to be said, it is simple good software design.
My own experience here is mostly limited to use of MapleV on a small
machine. However, I have also quotes of integral tables or other
symbolic systems in connection with problems proposed in the American
Mathematical Monthly. These sources are often not very careful about
selecting the base point (the point where the given integral will be
zero). The most annoying effect is having a solution produced in a
portion of the complex plane which has been slit along the very
interval that you are considering. In Maple, the base point seems to
be chosen to be infinity wherever possible, even when integrating
something involving sqrt(1-x^2), so this effect is fairly common.
There is some indication that the user will be given greater control
over this in versions currently being developed. It may be that the
promise of having a general and accurate answer available on demand is
responsible for some thought being given to what kind of answer is
possible. There is something that considers it unsatisfactory for an
"answer" to be "there are several possible answers, which do you
want?".
A good test question for any system would be Problem 6653 from the
Monthly (March 1991), whose solution is scheduled for publication in
December 1992. The integral in question is a definite integral from 0
to Pi of a complicated expression involving an inverse cosine of a
complicated trigonometric expression. The integral depends on a
parameter which is restricted to being a non-negative real number in
the statement of the problem. However, some standard techniques break
down when the parameter is equal to 1. When the parameter is greater
than 1, the result can be found in the table of Gradshteyn & Ryzhik.
Some of our more persistent solvers were able to evaluate the integral
for all nonzero real values of the parameter.
--
R. T. Bumby ** Rutgers Math || Amer. Math. Monthly Problems Editor
bu...@math.rutgers.edu || P.O. Box 10971 New Brunswick, NJ08906-0971
bu...@dimacs.rutgers.edu || Phone: [USA] 908 932 0277 * FAX 908 932 5530
Ray Zimmerman
brn...@nyu.edu writes:
>1-x positive or negative?'' whenever I gave it such an integral? Sure,
>Mathematica thinks that sqrt((-1)^2) = -1, but just because the new
>technology produces wrong answers doesn't mean that the old technology
>did.
Mathematica 2.0 on the Mac thinks that Sqrt[(-1)^2] = 1. Please check
before posting this type of information.
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Ray Zimmerman \ 397 Theory Center, Electrical Engineering
\ Cornell University, Ithaca NY 14853
Ray-Zi...@cornell.edu \ Phone: 607-254-8819 Fax: 607-255-9072
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
David Kastrup
>In article <10843.Sep2...@virtualnews.nyu.edu> D. J. Bernstein,
>brn...@nyu.edu writes:
>>1-x positive or negative?'' whenever I gave it such an integral? Sure,
>>Mathematica thinks that sqrt((-1)^2) = -1, but just because the new
>>technology produces wrong answers doesn't mean that the old technology
>>did.
>Mathematica 2.0 on the Mac thinks that Sqrt[(-1)^2] = 1. Please check
>before posting this type of information.
So both are wrong. hooray. sqrt((-1)^2) = +- 1, that is either value,
depending on the choice of the Riemann leaf. And if you choose the leaf
you would expect from the squaring, you get -1. If you want to choose the
`main' leaf, you get 1. Fixing it one way or the other can always lead
to problems.
D. J. Bernstein
> Macsyma is probably the most reliable of the symbolic packages. In
> this case however, its behaviour is unsatisfactory.
It's been long enough since I used Macsyma that I don't recall whether
it would automatically handle range checks on the definite integrals you
mention; certainly it was never as ignorant as Mathematica. I do agree
that the available symbolic math packages just don't know enough math.
I'll know computer math is improving when I see a correct graph-drawing
program. Such a program would produce a _picture_ of any graph given by
an equation in two variables; by definition, a picture of a graph is a
set (of dots, of ink, whatever) which contains the graph. Of course, a
big black square is a picture of any graph, but a high-quality program
will produce pictures which are close to minimal.
---Dan
Richard Fateman
implemented interval arithmetic. See, for example,
R. Fateman, Honest Plotting, Global Extrema, and Interval Arithmetic.
in Proc. ISSAC '92, (Assoc. for Comp. Mach., ISBN 0-89791-489-9)
pp. 216-223.
Programs for doing this in Mathematica are illustrated, modulo
Mathematica's (incorrect) interval arithmetic.
See files available on this machine: ~ftp/pub/plotinterval.m and
realinterval.m
Better programs for computing correctly with intervals in Lisp,
as well as "native" plotting are under development; perhaps some other
systems that support intervals, like Pascal SC could do a better job.
Eric Postpischil
(D. J. Bernstein) writes:
>It's been long enough since I used Macsyma that I don't recall whether
>it would automatically handle range checks on the definite integrals you
>mention; certainly it was never as ignorant as Mathematica.
Here's a plug for Derive; I've been checking the examples given in this
discussion in Derive on the HP-95 palmtop, and Derive seems to do very
well in applying only valid reductions. It does not check integrals for
singularities inside the bounds of a definite integral, and it warns of
this in the manual, but it otherwise appears quite robust.
Interestingly, I believe it accomplishes this _without_ range checks on
the definite integrals. For example, the integral of sqrt(x^2) dx
simplifies to x|x|/2. This expression is valid regardless of whether x
is positive or negative. If it is given as a definite integral from a
to b, the result is b|b|/2-a|a|/2 unless Derive can figure out that b or
a is positive or negative, in which case the result may be (when both
are non-negative) b^2/2-a^2/2. But this simplification comes after the
integral has been resolved; it is simply a general transformation that
is known to be valid, not a range check on the integral.
-- edp (Eric Postpischil)
"Always mount a scratch monkey."
e...@alien.enet.dec.com