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Mar 26, 1992, 8:30:26 AM3/26/92

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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...

--

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)

Mar 27, 1992, 8:03:10 AM3/27/92

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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

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