Computing an Integral with a limiting parameter

  • Thread starter muzialis
  • Start date
  • #1
166
1

Main Question or Discussion Point

Hi All,

I am reading a paper in which the following integral is presented
$$\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x+s_1+s_2}{c}) \frac{\mathrm{d}f (s_1)}{\mathrm{d}s_1} \frac{\mathrm{d}f(s_2)}{\mathrm{d}s_2} \mathrm{d}s_1 \mathrm{d}s_2$$
where J and f are unknwon functions.
Now, it seems that for "sufficiently large c the integration can be carried out to yield"
$$\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x}{c})\big \lbrack f(s_1) \vert^{ct-x}_{-\infty} f(s_2) \vert^{ct-x}_{-\infty} \big \rbrack$$

but I can not follow, why would the term $$\frac{2x}{c}$$ "survive" the limiting process as c goes to infinity?

Many thanks as usual
 

Answers and Replies

  • #2
1,796
53
Hi All,

I am reading a paper in which the following integral is presented
$$\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x+s_1+s_2}{c}) \frac{\mathrm{d}f (s_1)}{\mathrm{d}s_1} \frac{\mathrm{d}f(s_2)}{\mathrm{d}s_2} \mathrm{d}s_1 \mathrm{d}s_2$$
where J and f are unknwon functions.
Now, it seems that for "sufficiently large c the integration can be carried out to yield"
$$\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x}{c})\big \lbrack f(s_1) \vert^{ct-x}_{-\infty} f(s_2) \vert^{ct-x}_{-\infty} \big \rbrack$$

but I can not follow, why would the term $$\frac{2x}{c}$$ "survive" the limiting process as c goes to infinity?

Many thanks as usual
Are we to consider:

[tex]
\begin{align*}\lim_{c\to\infty}\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x+s_1+s_2}{c}) \frac{\mathrm{d}f (s_1)}{\mathrm{d}s_1} \frac{\mathrm{d}f(s_2)}{\mathrm{d}s_2} \mathrm{d}s_1 \mathrm{d}s_2\overset{?}{=} \\
&\hspace{-100pt}\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x}{c})\big \lbrack f(s_1) \vert^{ct-x}_{-\infty} f(s_2) \vert^{ct-x}_{-\infty} \big \rbrack ds_1 ds_2
\end{align*}
[/tex]

If so, and if that was my problem and I couldn't initially show it analytically, I would first make up some easy functions for J and f, then numerically integrate and numerically take the limit and see if indeed the limit approaches the form you are suggesting. If numerical results tend to this, then I would put in the effort to proving it rigorously.

I should also point out the very act of studying it numerically with sample functions may, and sometimes does, provide you with additional insight allowing you to subsequently show it analytically. I know this is true from experience. :)
 
Last edited:
  • #3
166
1
Jackmell,
tahnk you for your answer. I will not say that I agree the slightest with your approach, but I performed the check you suggested, and yes, the results of numerical integration with the analytical result. Well "agree", it depends what is meant, as the concept of "limit" refuses to be captured by numerics: numerical results agree equally well if one takes the fraction with c at the denominator our altogether.

Is there anybody else who might give an advice? I suspect there is some easy (for somebody else) analytical approximation step undertaken.

Thanks a lot
 
  • #4
166
1
I still could not solve this dilemma of mine: as it is crucial for me, could I maybe be pushy and try to stimulate further answers?

Many, many thanks
 
  • #5
798
34
it seems that for "sufficiently large c the integration can be carried out to yield"
$$\int_{-\infty}^{ct-x} \int_{-\infty}^{ct-x} J (2t - \frac{2x}{c})\big \lbrack f(s_1) \vert^{ct-x}_{-\infty} f(s_2) \vert^{ct-x}_{-\infty} \big \rbrack$$
Hi !
I am surprized because I think (but I am not sure) that for sufficiently large c the integration can be carried out to yield
$$\ J(2t - \frac{2x}{c})\big \lbrack f(s_1) \vert^{ct-x}_{-\infty} f(s_2) \vert^{ct-x}_{-\infty} \big \rbrack$$
i.e. without the integrals because they are already taken into account in the right part of the formula.

but I can not follow, why would the term $$\frac{2x}{c}$$ "survive" the limiting process as c goes to infinity?
One could eliminate this term if we are not intrested to the behaviour when x tends to infinity (i.e. if we suppose that x is always a finite number sufficiently small compare to c.
But, if we are interested to the asympotic behaviour when x tends to infinity, we have to keep this term which might be important, depending on the relative magnitudes of x and c.
 
  • #6
166
1
J. Jaquelin,

thanks for your reply.
Yes indeed my equation is wrong, the integrals should not be there after the integration is carried out as you noted.

I understand your last comment but how can that be justified rigorously, formally?
If c is large enough for us to look at the asymptotics but not to make all the fractional term in the argument of J function zero, why should one not account for the other terms in the fraction such as $$\frac{s_1}{c}$$?
Many thanks again
 
  • #7
798
34
If c is large enough for us to look at the asymptotics but not to make all the fractional term in the argument of J function zero, why should one not account for the other terms in the fraction such as $$\frac{s_1}{c}$$?
If we consider how the lower limits of the integrals are defined, I am afraid that it is not possible to justify why we don't take account of s1/c and s2/c in the argument of J for any kind of function J. It should be possible to justify it if the integrals were written as below :
 

Attachments

  • #8
166
1
Many thanks, it is really appreciated

Best Regards
 

Related Threads on Computing an Integral with a limiting parameter

  • Last Post
Replies
3
Views
2K
Replies
4
Views
2K
Replies
2
Views
710
  • Last Post
Replies
8
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
20
Views
1K
Replies
7
Views
2K
Top