#4743 integrate(1/(1-x^5), x, 0, inf) commented out of rtestint

[Barton Willis]

This test is commented out (line 1048 of rtestint)

* principal value integral */
/* [ 657382 ] defint/limit infinite loop */

/* Because of changes to timesin this does no longer work 
integrate(1/(1-x^5), x, 0, inf);
0-(2*sqrt(2*sqrt(5)+10)*atan((sqrt(5)-3)*sqrt(2*sqrt(5)+10)/(4*sqrt(5)))
        +2*sqrt(10-2*sqrt(5))*atan(sqrt(10-2*sqrt(5))*(sqrt(5)+3)/(4*sqrt(5)))
        -sqrt(2)*sqrt(sqrt(5)+5)*%pi-sqrt(2)*sqrt(5-sqrt(5))*%pi)
        /20;

Current Maxima gives

(%i2) ans : integrate(1/(1-x^5), x, 0, inf);
Principal Value
(%o2) -((2^(3/2)*sqrt(5-sqrt(5))
                *atan((sqrt(5-sqrt(5))*(sqrt(2)*sqrt(5)+3*sqrt(2)))
                       /(4*sqrt(5)))
 +2^(3/2)*sqrt(sqrt(5)+5)
         *atan((sqrt(sqrt(5)+5)*(sqrt(2)*sqrt(5)-3*sqrt(2)))/(4*sqrt(5)))
 -sqrt(2)*sqrt(sqrt(5)+5)*%pi-sqrt(2)*sqrt(5-sqrt(5))*%pi)
 /20)
(%i3) float(ans);

(%o3) 0.8648062659772102

Possibly, the sign is wrong--I'm not sure. Also when domain : complex, either the calculation takes a long time, or it is in a infinite loop.

[Robert Dodier]

For the record, the aforementioned defint/limit bug is now known as:
https://sourceforge.net/p/maxima/bugs/200/

[Raymond Toy]

1/(1-x^5) can be factored as -1/(x-1)/(x^4+x^3+x^2+x+1), and we can use quad_qawc to compute the integral like so:

(%i29) quad_qawc(-1/(x^4+x^3+x^2+x+1),x,1,0,100);
(%o29)       [0.8648062684772098, 2.8888229703493868e-11, 385, 0]

And for [100,inf]`:

(%i30) quad_qagi(1/(1-x^5),x,100,inf);
(%o30)     [- 2.500000000111111e-9, 2.7250808692222103e-18, 225, 0]
(%i31) %o29[1]+%o30[1];
(%o31)                        0.8648062659772098

That's pretty close.