Curious definite integral related to sophomores dream

[lamarche]

The integral I am seeking to evaluate is
int_{-\inf}^{+\inf) exp(-|w|/2) exp( i w [ln(|w|)/pi-x]) dw
a definite integral which is a function of x... at x=0, it is nearly the same
as the "sophomore dream" integral...

I don't know why, but this function seems to have a singularity at -pi^2/6
This does have a physical application: it would be the minimum phase impulse
response of a cable having losses linear with frequency,,,

 

[JJacquelin]

Probably, there is a misunderstanding.
The integral eq.(1) is the sum of two regular functions Eq.(2) and (3).
I cannot see any singularity.

 

[lamarche]

Hi! JJ,

Cos(a+b) = cos(a) cos(b) - sin(a) sin(b)

so actually what you want to plot is (2) minus (3), not plus...

which is such that the function abruply goes to zero (and stays exactly there) at -pi^2/6

It is a singularity in the same way that the function f=(0,x<=0;exp(-1/x),x>0) has

a singularity at 0: all derivatives vanish... but only exactly on the real axis. On the

complex plane the series diverges/oscillated like crazy (no Taylor expansion possible...).

 

[JJacquelin]

Cos(a+b) = cos(a) cos(b) - sin(a) sin(b)
so actually what you want to plot is (2) minus (3), not plus...

a = w*ln(w)/pi
b = -w*x ... (not +w*x)
Cos(a+b) = cos(w*ln(w)/pi) cos(w*x) + sin(w*ln(w)/pi) sin(w*x)
The sum is (2) plus (3), not minus.
And, as a matter of fact, the red curve ploted in not { (2) plus (3) }, but it is (1), before any transformation.
Also I ploted first { (2) plus (3) } but the curve is not visible because it is exactly the same as the curve (1).

the function abruply goes to zero (and stays exactly there) at -pi^2/6

I have doubts about that. It seems that, when x is positive and large, both fonctions Fc(x) and Fs(x) tend to zero, Fc(x) with positive values and Fs(x) with negative values, both tending to a common absolute value : i.e. Fc(x) equivalent to -Fs(x) . Equivalent, is not equal : the sum isn't exacty 0, but very close to 0.
So, Fc(x)+Fs(x) tend to zero very quickly, as it is often the case, due to exponentials.
The computation of the sum of two very low numerical values of oposite sign is hasardous.
I would not swear to it, but I think that the appearance of a threshold is related to the precision of the numerical integrator. If the threshold is crossed, the numerical result is 0, instead of a correct very low but not nil value, which moreover tends even more quickly to 0 when x increases.
I think that pi^2/6 is an empirical result, only due to hasard and without signifiance.

 

[lamarche]

Hi! JJ,

caution, in English you would say "chance" not "hasard"... "hasard" is like "danger"

You are right, I flipped the sign of x in what I sent you (here I actually FFT/IFFT
with MATLAB), so on your graph what I call a singularity is at +1.6..., not -1.6...
and the impulse response is time-flipped...

Merci pour les corrections... je suppose que vous vivez en France? Je pourrais
vous expliquer comment j'en suis arrive la et ce que je compte faire, mais ce serait
mieux par telephone que dans ce forum...

 

[JJacquelin]

Je pourrais vous expliquer comment j'en suis arrive la et ce que je compte faire, mais ce serait mieux par telephone que dans ce forum...

Bonjour,
En effet, il vaut mieux ne pas encombrer le forum par des sujets devenant trop spécifiques.
Dans ce cas, la messagerie privée de ce site peut être un moyen de communication plus discret.

 

[JJacquelin]

Hi! Lamarche,
The document showing that there is no singularity around x=-pi²/6 is in attachment.
The sign of x has been flipped.

 

[HallsofIvy]

Hi! JJ,

caution, in English you would say "chance" not "hasard"... "hasard" is like "danger"

And, in English, the spelling is "hazard"!

You are right, I flipped the sign of x in what I sent you (here I actually FFT/IFFT
with MATLAB), so on your graph what I call a singularity is at +1.6..., not -1.6...
and the impulse response is time-flipped...

Merci pour les corrections... je suppose que vous vivez en France? Je pourrais
vous expliquer comment j'en suis arrive la et ce que je compte faire, mais ce serait
mieux par telephone que dans ce forum...

 

[lamarche]

Hi! Lamarche,
The document showing that there is no singularity around x=-pi²/6 is in attachment.
The sign of x has been flipped.

Your data looks convincing. I had the wrong intuition.
Do you think the function actually reaches zero at some negative value (e.g. -2 ?)?
or do you think it is entire? I tried a Taylor expansion on it, and I actually got some hints
of an infinite radius of convergence!