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MathGroup Archive 2007

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a suprising result from Integrate (Null appeared in the result!)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74466] a suprising result from Integrate (Null appeared in the result!)
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Thu, 22 Mar 2007 01:13:45 -0500 (EST)

Hello to all.

I discover (I think) a serious bug in Integrate.

The results are (at least!) surprisingly so I checked many times
before
I post anything!

Quit

$Version
5.2 for Microsoft Windows (June 20, 2005)

Consider the integral

f = HoldForm[Integrate[(1 - Sin[x])^(y/4), {x, 0, 4}]]

Then

ReleaseHold[f /. y -> 1]
4 + Null*(-4 - (4*Sqrt[1 - Sin[4]])/(Cos[2] - Sin[2])) - 4*(1 -
Sin[4])^(1/4)

I believe you noticed immediately the appearance of Null!

Information[Null]

"Null is a symbol used to indicate the absence of an expression or a
result. When it appears as an output expression, no output is
printed."
Attributes[Null] = {Protected}

Note also

DeleteCases[4 + Null*(-4 - (4*Sqrt[1 - Sin[4]])/(Cos[2] - Sin[2])) -
4*(1 - Sin[4])^(1/4), Null, Infinity]
N[%]
ReleaseHold[f /. y -> 1 /. Integrate[x___] :> NIntegrate[x,
MaxRecursion -> 12]]

-4*(1 - Sin[4])^(1/4) - (4*Sqrt[1 - Sin[4]])/(Cos[2] - Sin[2])
-0.6051176113064889
3.0908946898699625

Note also that Null appears also for other values of y

(ReleaseHold[f /. y -> #1] & ) /@ Range[2, 5]
{-2 + 4*Sqrt[2] + (2*(Cos[2] + Sin[2])*Sqrt[1 - Sin[4]])/(Cos[2] -
Sin[2]),
  (2/3)*(-2 + (2*2^(1/4)*Pi^(3/2))/Gamma[3/4]^2 -
2^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -1] +
    (Cos[4]*(2^(1/4)*Hypergeometric2F1[1/2, 3/4, 3/2, Cos[2 - Pi/4]^2]
+ 2*(1 - Sin[4])^(1/4)))/Sqrt[1 - Sin[4]]), 3 + Cos[4],
  (4*((Cos[2] - Sin[2])*(5 + (-6 + Cos[4])*(1 - Sin[4])^(1/4)) -
6*Null*(Cos[2] - Sin[2] + Sqrt[1 - Sin[4]])))/
   (5*(Cos[2] - Sin[2]))}

It appears for y=5 and when it appears the result is wrong

DeleteCases[%, Null, Infinity]
N[%]

{-2 + 4*Sqrt[2] + (2*(Cos[2] + Sin[2])*Sqrt[1 - Sin[4]])/(Cos[2] -
Sin[2]),
  (2/3)*(-2 + (2*2^(1/4)*Pi^(3/2))/Gamma[3/4]^2 -
2^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -1] +
    (Cos[4]*(2^(1/4)*Hypergeometric2F1[1/2, 3/4, 3/2, Cos[2 - Pi/4]^2]
+ 2*(1 - Sin[4])^(1/4)))/Sqrt[1 - Sin[4]]), 3 + Cos[4],
  (4*((Cos[2] - Sin[2])*(5 + (-6 + Cos[4])*(1 - Sin[4])^(1/4)) -
6*(Cos[2] - Sin[2] + Sqrt[1 - Sin[4]])))/(5*(Cos[2] - Sin[2]))}
{2.670553068935302, 2.454058460511143, 2.346356379136388,
-2.128162283559219}

(ReleaseHold[f /. y -> #1 /. Integrate[x___] :> NIntegrate[x,
MaxRecursion -> 12]] & ) /@ Range[2, 5]
{2.6705535326036554, 2.4540619799501173, 2.3463563791363886,
2.3070526406806264}

So Null took the place of something?

BTW, Here are plots of the integrands

(Plot[(1 - Sin[x])^(#1/4), {x, 0, 4}] & ) /@ Range[5]

Even the following setting does not fix the situation

Integrate[(1 - Sin[x])^(1/4), {x, 0, Pi/2, 4}]
4 + Null*(-4 - (4*Sqrt[1 - Sin[4]])/(Cos[2] - Sin[2])) - 4*(1 -
Sin[4])^(1/4)

The next is funnier!

Integrate[(1 - Sin[x])^(1/4), {x, 0, 1, Pi/2, 2, 4}]
Simplify[%]

Null*(-4 + (4*Sqrt[1 - Sin[1]])/(Cos[1/2] - Sin[1/2])) +
  4*(1 + Null - (1 - Sin[1])^(1/4) - (Null*Sqrt[1 - Sin[1]])/(Cos[1/2]
- Sin[1/2])) + 4*(1 - Sin[1])^(1/4) +
  Null*(-4 - (4*Sqrt[1 - Sin[2]])/(Cos[1] - Sin[1])) - 4*(1 -
Sin[2])^(1/4) +
  4*((1 - Sin[2])^(1/4) + (Null*Sqrt[1 - Sin[2]])/(Cos[1] - Sin[1]) -
(1 - Sin[4])^(1/4) -
    (Null*Sqrt[1 - Sin[4]])/(Cos[2] - Sin[2]))
-((4*((Cos[2] - Sin[2])*(-1 + (1 - Sin[4])^(1/4)) + Null*(Cos[2] -
Sin[2] + Sqrt[1 - Sin[4]])))/(Cos[2] - Sin[2]))

What about indefinite integrals?

Integrate[(1 - Sin[x])^(1/4), x]
-4*(1 - Sin[x])^(1/4) - (4*Null*Sqrt[1 - Sin[x]])/(Cos[x/2] - Sin[x/
2])

Integrate[(1 - Sin[x])^(1/5), x]
-5*(1 - Sin[x])^(1/5) - (4*Null*Sqrt[1 - Sin[x]])/(Cos[x/2] - Sin[x/
2])

Trying also the setting

(Integrate[(1 - Sin[x])^(1/4), {x, #1[[1]], #1[[2]]}] & ) /@
Partition[{0, Pi/2, 4}, 2, 1]
{4, Null*(-4 - (4*Sqrt[1 - Sin[4]])/(Cos[2] - Sin[2])) - 4*(1 -
Sin[4])^(1/4)}

Of course you can get for Mathematica the correct answer by working as
follows

Integrate[(1 - Sin[x])^a, {x, 0, 4}, Assumptions -> a > 0]
% /. a -> 1/4
N[%]

2^a*(-2*Hypergeometric2F1[1/2, 1 + a, 3/2, -1] +
(2*Pi^(3/2)*Sec[a*Pi])/(Gamma[1/2 - a]*Gamma[1 + a]) +
   Cos[4]*Hypergeometric2F1[1/2, 1/2 - a, 3/2, Cos[2 - Pi/4]^2]*Sqrt[2/
(1 - Sin[4])])
2^(1/4)*((2*Sqrt[2]*Pi^(3/2))/(Gamma[1/4]*Gamma[5/4]) -
2*Hypergeometric2F1[1/2, 5/4, 3/2, -1] +
   Cos[4]*Hypergeometric2F1[1/2, 1/4, 3/2, Cos[2 - Pi/4]^2]*Sqrt[2/(1
- Sin[4])])
3.0908948436614616

But I believe the appearance of Null is a matter of headache for the
developers!

Dimitris



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