## Symbolic integration bug with Mathematica 7?

January 19th, 2009 | Categories: math software, mathematica | Tags:

Someone at my university reported the following Mathematica 7 bug to me and I thought I would share it with all of you in case someone was interested or could offer insight that we don’t have.  Needless to say this has been reported to Wolfram via the official channels and I am sure that it will be sorted out soon.

If you do

Integrate[k^2 (k^2 – 1)/((k^2 – 1)^2 + x^2)^(3/2), {k, 0, Infinity},Assumptions -> x > 0]

you get

EllipticK[(2*x)/(-I + x)]/(2*Sqrt[-1 – I*x])

Which is complex for all x whereas the integrand is purely real.  As a  particular example let us put x=2.  Evaluating numerically:

NIntegrate[k^2 (k^2 – 1)/((k^2 – 1)^2 + 2^2)^(3/2), {k, 0, Infinity}]

gives

0.706094

But if I plug x=2 into the symbolic result given earlier

N[EllipticK[(2*x)/(-I + x)]/(2*Sqrt[-1 – I*x]) /.x->2]

Then I get

-5.35054*10^-17 + 0.568541 I

Which is obvioulsy incorrect.  Would someone with a copy of Maple 12 mind seeing what that makes of this integral please?  Extra kudos to anyone who can do it by hand!

1. Hello,

As our multi-years research based on a failure prediction oracle named
the VM machine reveals, unfortunately, as of Jan 20, 2009, the current
commercial computer algebra systems like Maple 12, Mathematica 7 as well
as the engineering packages like MATLAB 2008b are really crammed full
with various types of defects, from crashes to hangs to invalid, often

For your general case, Maple 12 returns sophisticated output

limit(1/2*(-k*((1+I*x)/(1+x^2))^(1/2)+(-(-1-x^2+k^2+I*k^2*x)/(1+x^2))^(1/2)*((1+x^2-k^2+I*k^2*x)/(1+x^2))^(1/2)*EllipticF(k*((1+I*x)/(1+x^2))^(1/2),(-(-1+x^2+2*I*x)/(1+x^2))^(1/2))+2*PIECEWISE([(-signum(-1+I*x))^(1/2)*infinity*(-(-I*signum((1-I*x)^(1/2)*x))^(1/2)+(I*signum((1-I*x)^(1/2)*x))^(1/2))/(I*signum((1-I*x)^(1/2)*x))^(1/2)/(-I*signum((1-I*x)^(1/2)*x))^(1/2), 0 < (1-I*x)^(1/2)],[0, otherwise])*((1+I*x)/(1+x^2))^(1/2)*(k^4-2*k^2+1+x^2)^(1/2)+2*PIECEWISE([signum(1+I*x)^(1/2)*infinity*(-(I*signum((1+I*x)^(1/2)*x))^(1/2)+(-I*signum((1+I*x)^(1/2)*x))^(1/2))/(-I*signum((1+I*x)^(1/2)*x))^(1/2)/(I*signum((1+I*x)^(1/2)*x))^(1/2), 0 < (1+I*x)^(1/2)],[0, otherwise])*((1+I*x)/(1+x^2))^(1/2)*(k^4-2*k^2+1+x^2)^(1/2)+2*PIECEWISE([(-signum(-1+I*x))^(1/2)*infinity*(-(-I*signum((1-I*x)^(1/2)*x))^(1/2)+(I*signum((1-I*x)^(1/2)*x))^(1/2))/(I*signum((1-I*x)^(1/2)*x))^(1/2)/(-I*signum((1-I*x)^(1/2)*x))^(1/2), 0 < -(1-I*x)^(1/2)],[0, otherwise])*((1+I*x)/(1+x^2))^(1/2)*(k^4-2*k^2+1+x^2)^(1/2)+2*PIECEWISE([signum(1+I*x)^(1/2)*infinity*(-(I*signum((1+I*x)^(1/2)*x))^(1/2)+(-I*signum((1+I*x)^(1/2)*x))^(1/2))/(-I*signum((1+I*x)^(1/2)*x))^(1/2)/(I*signum((1+I*x)^(1/2)*x))^(1/2), 0 int(z^2*(z^2 – 1)/((z^2 – 1)^2 + 4)^(3/2), z= 0..infinity);

1/10*5^(3/4)*EllipticK(1/10*(50+10*5^(1/2))^(1/2))

> evalf(%);

.7060937855

> evalf(Int(z^2*(z^2 – 1)/((z^2 – 1)^2 + 4)^(3/2), z=0..infinity));

.7060937855

Cheers,

CEO, Mathematical Director

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2. Maple 12 computes both symbolic and numeric integrals well.

> t:=int(k^2* (k^2 – 1)/((k^2 – 1)^2 + x^2)^(3/2),k=0..infinity) assuming x>0; # compute the integral symbolically

t := 1/2*1/(1+x^2)^(1/4)*EllipticK(1/2*2^(1/2)*(((1+x^2)^(1/2)+1)/(1+x^2)^(1/2))^(1/2))

> evalf(subs(x=2,t)); # substitute x=2 in symbolic result

.7060937851

> int(k^2* (k^2 – 1)/((k^2 – 1)^2 + 2^2)^(3/2),k=0.0..infinity); # compute integral numerically at x=2

.7060937855

3. @VolMike Thanks for that – I really must get a copy of Maple it seems.

@Vladimir – Thanks also but I wonder why the result you get is so large? Did you include the assumption that x>0?

4. “@Vladimir – Thanks also but I wonder why the result you get is so large? Did you include the assumption that x>0?”

Oooops, woe unto me, there was a typo. My input was “assuming k>0” while
it should be “assuming x>0”.

I confirm the result by VolMike.

In the given case Maple 12 beats Mathematica 7.

Why my typo occurred – and I did not spot it?

Maybe, it’s because of the following. To tell the truth, by and large, as
long as it is about integration, I got accustomed to see that Mathematica
7 beats Maple 12.

But in this case, Maplesoft got even Wolfram Research :)

5. No problem Vladimir – we all make mistakes ;)
Thanks again for your contribution.

6. Not surprisingly, Mable-based Matlab 7.02 gives the same (correct) result as Maple 12, while mupad based Matlab R2008b returns “Explicit integral could not be found.”

7. Hi Robert

Thanks for the feedback. In this case I think Mupad/MATLAB wins over Mathematica. I’d rather be told that it can’t find a result than be given an incorrect one. Not all gaffs are as obviously wrong as this one.

Cheers,
Mike

8. The simple way to avoid (decrease the probability of) such errors is comparing results in CAS’s based on different symbolic (numeric) engines.