I'm going to embarrass myself by asking how to evaluate the integral
given above. Currently, if intanalysis is false, maxima converts
integral(x^k*log(x)/(x+3),x,0,inf) to the above integral.[1] If k+1>0, k
< 0, and k is not an integer, maxima says the above integral is
-(log(3)^2+%pi^2)*%e^(log(3)*kk)*(%i*sin(%pi*kk)+cos(%pi*kk))/2
But imagpart(%) is
-(log(3)^2+%pi^2)*%e^(log(3)*kk)*sin(%pi*kk)/2
which is not zero. That's silly since the integrand is always real
for real x.
Maxima evaluates this integral using a rectangular region with corners
at (-R,0) to (R,2*%i*%pi) and lets R -> inf. The derivation is given
in Wang, p 98-100. I've done the integration by hand, and get the
same result as maxima. But something is wrong, and I don't know what
it is.
A quick summary of the approach is in the comments for rectzto%pi2 in
defint.lisp:
;; This is a very brief description of the algorithm. Basically, we
;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
;; function and p(x) is a polynomial.
;;
;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
;; Then consider a contour integral of R(exp(z))*q(z) over a
;; rectangular contour. Opposite corners of the rectangle are (-R,
;; 2*%i*%pi) and (R, 0).
;;
;; Wang shows that this contour integral, in the limit, is the
;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
;; exactly the integral we're looking for.
;;
;; Thus, to find the value of the contour integral, we just need the
;; residues of R(exp(z))*q(z). The only tricky part is that we want
;; the log function to have an imaginary part between 0 and 2*%pi
;; instead of -%pi to %pi.
Can someone evaluate this integral and give a correct answer? Bonus
points if you can do it by "hand". :-)
There must be some subtle point that maxima and I are missing.
Ray
Footnotes:
[1] I'm going to fix it so that integral(x^k*log(x)/(x+3),x,0,inf) is
evaluated correctly according to Wang's thesis. The answer turns out
to be
3^k*(psi[0](k+1)-psi[0](-k)+log(3))*beta(k+1,-k)
Numerical evaluations for k=-1/2 and k=-1/3 indicates that this result
is probably correct.