## #1645 possible integration bug

closed
nobody
None
5
2009-08-19
2009-04-14
Anonymous
No

I tried below.

(%i2) declare(n,integer) ;
(%o2) done
(%i3) integrate(sin(n*%pi*x/a)*(sin(%pi*x/a))^3,x,-a,a) ;
(%o3) 0

Now, (sin(sin(%pi*x/a))^3 can be converted to (3*sin(%pi*x/a)-sin(3*%pi*x/a))/4 and if I do above
integration manually, it yields that the result is 0 for any n except 1 or 3. If n=1 result should be 3/4*a
and if n=3 result should be -a/4 but not zero for both cases. Is this a bug?

Maxima version: 5.17.1
Maxima build date: 20:10 3/10/2009
host type: powerpc-apple-darwin8.11.0
lisp-implementation-type: CMU Common Lisp
lisp-implementation-version: Stage 3 20071108T014921 (19D)

## Discussion

• Nobody/Anonymous - 2009-04-14

Ok, it seems it an orthogonal problem.
(%i7) integrate(sin(n*%pi*x)*sin(%pi*x),x,-1,1) ;
(%o7) 0

• Dieter Kaiser - 2009-08-19
• status: open --> closed

• Dieter Kaiser - 2009-08-19

When n is not declared to be an integer the answer is:

(%i6) integrate(sin(n*%pi*x/a)*(sin(%pi*x/a))^3,x,-a,a);
(%o6) 12*a*sin(%pi*n)/(%pi*n^4-10*%pi*n^2+9*%pi)

This answer simplifies correctly to zero when n is an integer.

Closing this bug report as invalid.

Dieter Kaiser

• Raymond Toy - 2009-08-20

Reopening bug.

Shouldn't maxima at least ask if n = 1 or something? When n = 1, the integrand is sin(%pi*x/a)^4, so the integral can't be zero.