# [Maxima] new defint.lisp bug - was: new defint.lisp andradexpand:false?

Richard Hennessy rich.hennessy at verizon.net
Thu Jan 12 17:28:45 CST 2012

```Actually a more general answer that works for positive and negative x is

integrate(exp(x^7),x);
(-%e^-(%i*%pi/7)*gamma_incomplete(1/7,-x^7)/7+%e^-(%i*%pi/7)*gamma(1/7)/7+gamma(1/7)/7)*(signum(x)+1)/2
+gamma_incomplete(1/7,-x^7)*(1-signum(x))/14

which is very complicated but more correct.  Maybe someone can suggest a
better form or we have to fix the -1 problem.  Maybe if assume(x>0) should
affect the answer.  Then integrate would have to check for that.

Rich

-----Original Message-----
From: Richard Hennessy
Sent: Thursday, January 12, 2012 6:04 PM
To: Edwin Woollett ; Richard Fateman
Cc: maxima mailing list ; Aleksas Domarkas
Subject: Re: [Maxima] new defint.lisp bug - was: new defint.lisp

Since maxima does simplify this factor to 1.0 then we could fix that by
using the value of polarform(-1) in the answer as suggested by Aleksas.

integrate(exp(x^7),x); should give

-%e^-(%i*%pi/7)*gamma_incomplete(1/7,-x^7)/7

and

integrate(exp(x^5),x); should give

-%e^-(%i*%pi/5)*gamma_incomplete(1/5,-x^5)/5

etc . . .

That would fix the problem, since Maxima's simplifier does not simplify away
the exponential term.

Aleksas has already shown the proof but his method will not work in
integrate(exp(x^n),x) when n is a known integer.  The reason is that Maxima
has already converted the (-1) expression to 1.

Rich

-----Original Message-----
From: Edwin Woollett
Sent: Thursday, January 12, 2012 3:28 PM
To: Richard Fateman
Cc: maxima mailing list
Subject: Re: [Maxima] new defint.lisp bug - was: new defint.lisp

On Jan. 11, 2012, Richard Fateman wrote:
----------------------------------
>> (%i3) integrate(exp(x^5),x,1,2);
>> (%o3) (gamma_incomplete(1/5,-32)-gamma_incomplete(1/5,-1))/5
>
>for what it is worth, Mathematica 7 returns
>
>1/5 (-1)^(4/5) (Gamma[1/5, -32] - Gamma[1/5, -1])}
>
>
>Notice the factor of (-1)^(4/5).
-------------------------------------------
This is a very important observation, and explains
a difference between Maxima and Mathematica results
for these type of integrals.  Maxima evaluates the
factor (-1)^(4/5) numerically to 1.0, which differs
from Mathematica's evaluation (see below).

Here I consider a different (but related via a simple
change of variables)  integral, and compare in detail
the results from Maxima and Mathematica (using the
Wolfram alpha site).

-----------------------------------------------------
Maxima 5.25.1 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.8 (a.k.a. GCL)

(%i1) integrate(exp(y)/y^(4/5),y,1,2);
(%o1) gamma_incomplete(1/5,-2)-gamma_incomplete(1/5,-1)

/*  MMa NIntegrate[Exp[y]/y^(4/5),{y,1,2}] yields

(-1)^(4/5)*(Gamma[1/5,-2] - Gamma[1/5,-1])

does not agree with Maxima, since in Mathematica
(-1)^(4/5) float value is not 1.0, whereas
is does have value 1.0 in Maxima.  */

(%i2) expand(float(%o1));
(%o2) -1.951024982232477*%i-2.685355511932373

/* this Maxima float value agrees with the first 14 digits of

Mma's   N[Gamma[1/5,-2] - Gamma[1/5,-1]] which gives

-2.685355511932382... -1.951024982232483... i

which translates into

-2.685355511932382 - 1.951024982232483*%i     */

/* the approximate (real) value of the integral

(%o3) [3.319281956502171,3.6851432533315483E-14,21,0]

/* Maxima gives value 1.0 to (-1)^(4/5)  */

(%i4) expand(float(-1)^(4/5));
(%o4) 1.0

/* whereas Mathematica gives

N[(-1)^(4/5)]  --->

-0.8090169943749474... + 0.5877852522924731... i

which translates into

-0.8090169943749474 + 0.5877852522924731*%i

If we multiply this factor by %o2 and expand we get the
correct numerical answer to within floating point roundoff
errors:    */

(%i5) expand((-0.8090169943749474 + 0.5877852522924731*%i)*%o2);

(%o5) 4.4408920985006262E-16*%i+3.319281956502161

/* the above was with 5.25.1 defint.lisp.
The updated version makes no difference here  */

(%o6) "c:/work2/defint-new.lisp"

(%i7) integrate(exp(y)/y^(4/5),y,1,2);
(%o7) gamma_incomplete(1/5,-2)-gamma_incomplete(1/5,-1)

(%i8) expand(float(%));
(%o8) -1.951024982232477*%i-2.685355511932373

--------------------------------------
Ted

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