asymptotics of an integral
Gerald Edgar
1
/
| 1
| ----------------------------------- dy
| 4 3 2 2
/ y - 2 y + 2 I r y - 2 I r y - r
-1
Here I = sqrt(-1), and r > 0 is small.
For r=0, the integral diverges. For r positive, but small, the poles
are off the real axis. How does this integral behave as r -> 0 ?
Numerical evidence leads me to conjecture that it is asymptotically
- Pi
---- + O(1)
2 r
But I did not see how to get Maple to verify this for me...
--
Gerald A. Edgar Internet: ed...@mps.ohio-state.edu
Department of Mathematics Bitnet: EDGAR@OHSTPY
The Ohio State University telephone: 614-292-0395 (Office)
Columbus, OH 43210 -292-4975 (Math. Dept.) -292-1479 (Dept. Fax)
Gerald Edgar
Three responses to my problem "Asymptotics of an integral"
The statement of the problem:
>
> Consider the integral
>
> 1
> /
> | 1
> | ----------------------------------- dy
> | 4 3 2 2
> / y - 2 y + 2 I r y - 2 I r y - r
> -1
>
> Here I = sqrt(-1), and r > 0 is small.
>
> For r=0, the integral diverges. For r positive, but small, the poles
> are off the real axis. How does this integral behave as r -> 0 ?
> Numerical evidence leads me to conjecture that it is asymptotically
>
> - Pi
> ---- + O(1)
> 2 r
>
> But I did not see how to get Maple to verify this for me...
Reply From: <l...@paradigm.com>
The denominator factors into two quadratics, so you can use a partial
fraction decomposition and integrate each term. Since the quadratics
have no real zeroes for r#0, no singularity exists to hinder the
integration and substitution of limits. The y^2+Ir and y^2-2y+Ir
terms need a little care to make sure the right log/atan branch is taken,
since the values at -1 and 1 are the same in the first term, but the
latter term goes through 0 at y=0, and the atan may need to pick up a
%pi at this crossing. It probably helps a lot to convert all expressions,
such as the partial fraction decomposition, to rectangular form. This
immediately gave me a form with two terms out of 4 being odd, so they
don't count. Then I have one real and one imaginary term. The real
term is (y-2)/((y^2-2*y)^2+r^2)/2, which leads to log(y^2-2*y +/- %i*r)
and with log cut along the negative axis, there is a jump from
log(-1+%i*r) to log(-1-%i*r) at y=1 which is not apparent when you
taylor the log, but if you convert to rectangular form, the
difference is 2*%i*atan2(r,-1) which is 2*%pi*%i-2*%i*r+...
That is the probably what is missing from a plain taylor of the log
form, which would yield only the -2*%i*r piece and forget the 2*%pi*%i.
Since the manipulation of the logs may tend to mask or eliminate this,
you probably can't do too many operations without invalidating the
result.
Reply From: Carmelo Pisani <MA...@lure.latrobe.edu.au>
Dear Gerald,
I have just taken a quick look at the problem and think that
I have a way to prove your conjecture. let y=ru in the integral. With some
simple manipulation one can show that, after taking the r tends to zero
limit in the integrand, you end up with (-1/2Ir)*the integral from -1/r
to 1/r of 1/(u + (1/2I)). This integrates to a difference of logs
which you can readily expand as a power series in r. One term involves
log(-1+O(r))= log (-1) + log(1-O(r))= plus or minus I*pi + O(r).
One has to think about what branch of the log function to take, this
decides whether one has the plus or the minus sign, though you have
numerical evidence for which sign it is. (I just took another look
and decided how to determine the sign. I thus end up with your conjectured
behaviour). There appears to be no problem whatsoever is generating
higher order terms via this method. Hope this suggestion helps.
kind regards
Carmelo Pisani
Dept. of Maths
Latrobe University
Melbourne Australia
Reply From: wmsical!jjca...@watmath.waterloo.EDU (Jacques Carette)
Neat problem. Here is some Maple output:
#This is the function to be studied:
> f := Int(1/(y^4-2*y^3+2*I*r*y^2-2*I*r*y-r^2),y=-1..1);
1
/
| 1
f := | ----------------------------------- dy
| 4 3 2 2
/ y - 2 y + 2 I r y - 2 I r y - r
-1
>
# but Maple doesn't like it - it probably chokes on the I.
# so, make its life easier:
> ff := eval(subs(r=s/I,f));
1
/
| 1
ff := | ------------------------------- dy
| 4 3 2 2
/ y - 2 y + 2 s y - 2 s y + s
-1
>
> with(student):
> ans := value(ff);
ln(s + 1) - ln(s - 1)
ans := 1/4 ---------------------
s
1/2 1/2 2
ln(s + 1) (s - 1) - ln(s + 3) (s - 1) - 2 arctan(----------)
1/2
(s - 1)
- 1/4 ------------------------------------------------------------------
1/2
s (s - 1)
#and we get an answer!!!!
>
# And we want to see its asymptotic behavior at 0:
> series(ans, s);
-1
(- 1/4 ln(-1) + 1/4 %1) s + (1/8 %1 - 1/8 ln(3) + 1/2)
+ (3/32 %1 - 1/16 ln(3) + 1/4 %2) s
2
+ (- 3/64 ln(3) + 5/64 %1 + 1/4 (7/18 I + 1/16 ln(3) I) I + 1/8 %2 + 1/6) s
35
+ (1/8 (7/18 I + 1/16 ln(3) I) I - 5/128 ln(3) + --- %1
512
/ 1109 \ 3 63
+ 1/4 |- ---- I + 5/128 ln(3) I| I + 3/32 %2) s + (1/10 + ---- %1
\ 2592 / 1024
35
+ 3/32 (7/18 I + 1/16 ln(3) I) I - ---- ln(3)
1024
/ 1109 \ / 37 \
+ 1/8 |- ---- I + 5/128 ln(3) I| I + 5/64 %2 + 1/4 |--- I + 7/256 ln(3) I| I
\ 2592 / \144 /
4 5
) s + O(s )
%1 := (2 arctanh(2) I - ln(3) I) I
%2 := (1/8 ln(3) I - 5/6 I) I
>
# We really want to know about r, not s, so:
> subs(s = r*I, ");
(- 1/4 ln(-1) + 1/4 %2) I
- ------------------------- + 1/8 %2 - 1/8 ln(3) + 1/2
r
+ (3/32 %2 - 1/16 ln(3) + 1/4 %1) r I
2
- (- 3/64 ln(3) + 5/64 %2 + 1/4 (7/18 I + 1/16 ln(3) I) I + 1/8 %1 + 1/6) r
35
- (1/8 (7/18 I + 1/16 ln(3) I) I - 5/128 ln(3) + --- %2
512
/ 1109 \ 3 63
+ 1/4 |- ---- I + 5/128 ln(3) I| I + 3/32 %1) r I + (1/10 + ---- %2
\ 2592 / 1024
35
+ 3/32 (7/18 I + 1/16 ln(3) I) I - ---- ln(3)
1024
/ 1109 \ / 37 \
+ 1/8 |- ---- I + 5/128 ln(3) I| I + 5/64 %1 + 1/4 |--- I + 7/256 ln(3) I| I
\ 2592 / \144 /
4 5
) r + O(r I)
%1 := (1/8 ln(3) I - 5/6 I) I
%2 := (2 arctanh(2) I - ln(3) I) I
> evalc(");
Pi / 25\ 2 35 3
- 1/2 ---- - 1/8 ln(3) + 1/2 + 3/32 Pi r - |- 5/64 ln(3) + ---| r - --- Pi r
r \ 144/ 512
/ 12223 63 \ 4 5
+ |------ - ---- ln(3)| r + O(r I) + (- 1/8 Pi + (- 3/32 ln(3) + 5/24) r
\103680 1024 /
2 / 1415 35 \ 3 63 4
+ 5/64 Pi r - |----- - --- ln(3)| r - ---- Pi r ) I
\10368 512 / 1024
>
# and the conjecture was right...
> quit
Jacques Carette
jjca...@daisy.waterloo.edu