Definite Integration tests

1
Definite Integration tests

• tests suggested by Dave Rusin
• 5/7/2002
• sci.math.symbolic

2
(No Transcript)
3
The story

• A colleague teaching a graduate complex analysis
  course mentioned
• that one of his better students reported that he
  liked computing definite
• (real) integrals with the standard
  line-integral-and-residue techniques,
• but that, in the end, these were nothing that
  Mathematica could not do.
• So the instructor asked me for some challenge
  problems for the student
• to feed to Mma. I found some in old notes (taken
  from this newsgroup
• and elsewhere). Since software evolves, it is
  reasonable to
• expect that old challenge problems are challenges
  no longer
• indeed some progress seems to be evident. (I
  tested with Mathematica
• version 4.1 and Maple version 7 on a Sun
  Solaris.)

4
The story

• The case of definite integrals is particularly
  interesting to me
• since the great hype which accompanied early
  releases of Mathematica
• induced people to gleefully report its errors,
  many of which were
• indeed definite integrals computed in "closed
  form".
• So this became an afternoon project of tracking
  down some definite
• integrals which show software struggling. I am
  interested in
• (i) other examples of definite integrals which
  can be done by
• a grad student in complex analysis, but not
  by these packages
• (Examples from the residue calculus
  particularly welcome.)
• (ii) more examples of definite integrals which
  the packages (still) do wrong
• (iii) lessons about the science -- as opposed to
  the art -- of computing
• closed-form expressions for definite
  integrals what's the algorithm?

5
The story

• I am aware of the Risch "algorithm" for computing
  _indefinite_
• integrals, so my interest is in definite
  integrals which can
• be computed only because the limits of
  integration are "special".
• I don't know, for example, how to determine
  whether the
• integrals of rational functions of coordinates
  over _closed loops_
• on a Riemann surface can be expressed in some
  closed form.
• (For example, it was recently discussed in this
  group that the
• length of the curve yx2(4-x) from x0 to x4
  can be given
• in a "closed form" which is quite lengthy.)
• For all but the third, fourth, and fifth of the
  examples below, the
• antiderivatives actually _may_ be written using
  the primitives known
• to the software, but -- particularly for some
  important choices of
• the parameters a, b -- the definite integrals
  are clearly simpler
• than the antiderivatives in form. (I'm not
  wholly sure how to
• quantify "simpler".) Also, sometimes the software
  knows the general
• antiderivative but cannot seem to compute the
  definite integral correctly!

6
The story

• Attached below are the integrals I considered,
  showing input and output
• for first Maple and then Mathematica. Here they
  are in TeX form in
• case you're writing a final exam right about
  now...
• \int_01 (1-xa)1/a \,
  dx
• \int_-\infty\infty ei a x\overx2b2
  \, dx
• \int_0\infty x-b (\log(x))a
  \log(1xx2) \, dx
• \int_-\pi\pi (1-a \cos(x)-b \cos(2
  x))-1/2 \, dx
• \int_0\pi/2 1\over1(\tan(x))a
  \, dx
• \int_02\pi \log(a\sin(x))
  \, dx
• \int_0\pi/2 1\over(a\sqrt\cos(x))
  2 \, dx
• dave

7
The story

• int((1-xa)(1/a), x0 .. 1)
• Definite integration Can't determine if the
  integral is convergent.
• Need to know the sign of --gt -1/a
• Will now try indefinite integration and then take
  limits.
• (1 - 2/a) 1/2
• 2 Pi
  GAMMA(1/a 1)
• --------------------------
  -----
• (2/a - 1) GAMMA(1/a -
  1/2)
• Hmm if it needs to know the sign of -1/a,
  shouldn't that be
• in the answer somewhere?
• ................
• Integrate(1-xa)(1/a), x, 0, 1
• 
  1
• SqrtPi
  Gamma1 -

8
The story

• int(exp(Iax)/(x2b2), x -infinity ..
  infinity)
• Gives the answer 0 but if I add the
  assumptions agt0 and bgt0 I get
• exp(-a b) Pi
• --------------
• b
• ................
• IntegrateExpI a x/(x2b2), x, -Infinity,
  Infinity
• 
  -2
• 2
  Sqrtb Pi
• ( otherwise no answer )

9
The story

• Note the integrals converge, and closed-form
  solutions exist, for
• int( x(-b)log(x)alog(1xx2), x0 ..
  infinity)
• Integrate x(-b) Logxa Log1xx2,
  x,0,Infinity
• for all b in (1, 2) and all counting numbers a
  this was in a sci.math
• thread last November. (Hint differentiation with
  respect to b allows
• us to assume a0. The other logarithm term is
  ln(1-x3) - ln(1-x). )
• But both systems leave these integrals
  unevaluated. So I restricted to
• the case b3/2. First Maple
• int( x(-3/2)log(x)alog(1xx2), x0 ..
  infinity)
• infinity
• Uh oh... It even gives this response when a
  2, and yet it decides
• evalf( Int( x(-3/2)log(x)2log(1xx2), x0 ..
  infinity) )
• 208.8558862
• ................

10
The story

• int(1/(1-acos(x)-bcos(2x))(1/2),x-Pi .. Pi)
• Lengthy answer is slighly shortened by
  simplify() but not pretty!
• It is expressed using lim_x-gt0 and
  lim_x-gt0- , and
• runs about 16 lines using lprint .
• Here is the simplified result of running the
  computation with a1/2, b2/5
• / 1/2
  1/2 1/2
• 1/2 2 (11
  473 )
• 8 5 2 EllipticK(1) - EllipticF(1/2
  ---------------------, 1)
• 
  1/2 1/2
• \ (-5
  473 )
• 1/2 1/2 1/2 \
• 2 (21 473 )
  / 1/2 1/2
• - EllipticF(1/2 ---------------------, 1)
  / (44 2 473 )
• 1/2 1/2
  /

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The story

• int(1/(1(tan(x))a), x 0.. Pi/2)
• Integrate1/(1(Tanx)a), x, 0, Pi/2
• No answer from either of them. (Hint replace x
  by pi/2 - x .)
• (In a Jan. 1999 thread it was shown that MuPad
  can do this one
• without help, and the others can be coaxed into
  giving an answer.)
• Note that antiderivates _can_ be given for many
  values of a but
• that this is neither necessary nor particularly
  helpful, and as far
• as I can tell the antiderivative is not
  expressible in closed form
• for all a.
• int(log(asin(x)), x0..2Pi)
• The answer, after using simplify() , is

12
The story

• int(log(asin(x)), x0..2Pi)
• The answer, after using simplify() , is
• 2 1/2
  2 1/2
• a I (-a 1) I
  a - I (-a 1) I
• 2 Pi ln(a) - 2 Pi ln(----------------------) - 2
  Pi ln(----------------------)
• 2 1/2
  2 1/2
• a I (-a 1)
  a - I (-a 1)
• But note that branches of the logarithms must be
  chosen carefully!
• The answer must be real if a is real and
  greater than 1. In that case,
• the integeral 2 pi (cosh-1(a) - log 2).
  This is zero if a5/4, so try it
• int(log(5/4sin(x)), x0..2Pi)
• 
  2
• 2 Pi ln(5) - 2 Pi ln(1 2 I) - 2 Pi
  ln(2 I) I Pi

13
The story

• int(1/(asqrt(cos(x)))2, x0..Pi/2)
• Here is the answer in compressed form (obtained
  very quickly)
• 2((a4-1)(1/2)arctan((a21)/((a21)(a2-1))(1/2
  ))a4arctan(
• (a21)/((a21)(a2-1))(1/2)))/(a21)/(a2-1)/(a4-1
  )(1/2)a/(a21)
• 2(1/2)EllipticK(1/22(1/2))2a/(a21)/(a2-1)2
  (1/2)
• EllipticK(1/22(1/2))-2a/(a21)/(a2-1)2(1/2)E
  llipticE(1/22(1/2))
• -a(a41)/(a21)/(a2-1)22(1/2)EllipticPi(-1/(a2
  -1),1/22(1/2))
• This function has a singularity at a1 the
  numerical value when
• a.99999 is about .5406924657909949240642166 -
  24836532.758693 i, and
• that imaginary part grows proportionally to
  1/(1-a) for a lt 1.
• On the other hand, the integrals are all real
  and nicely behaved at a1
• the real part of the previous expression seems
  to get the integral.
• Maple will do the case a1 separately if asked
  numerically it's OK