Computing an Integral with a limiting parameter

In summary, the author of the paper presents an integral relating the function J(2t - frac{2x+s_1+s_2}{c}) to the functions f(s_1) and f(s_2). For sufficiently large c, the integration can be carried out to yield f(s_1) vert.^{ct-x}_{-\infty} f(s_2) vert.^{ct-x}_{-\infty}. However, the term $$\frac{2x}{c}$$ "survives" the limiting process as c goes to infinity.
  • #1
166
1
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
 
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  • #2
muzialis said:
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
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
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
muzialis said:
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.

muzialis said:
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 interested 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
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
muzialis said:
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 :
 

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  • #8
Many thanks, it is really appreciated

Best Regards
 

1. What is a limiting parameter in computing an integral?

A limiting parameter in computing an integral refers to a variable or constant that is used in the upper or lower limit of integration. It is used to indicate the range of values over which the integral is being calculated. The limiting parameter is usually denoted by a letter, such as "x" or "t".

2. How does the limiting parameter affect the integral?

The limiting parameter plays a crucial role in computing an integral as it determines the range of values over which the integral is being evaluated. The value of the limiting parameter affects the final result of the integral, as it determines the boundaries of the integration and the function being integrated over that range.

3. Can the limiting parameter be changed during the integration process?

Yes, the limiting parameter can be changed during the integration process. This is known as a variable limit of integration. In this case, the value of the limiting parameter is not fixed and can vary within a certain range, resulting in a different integral value.

4. What is the purpose of using a limiting parameter in an integral?

The use of a limiting parameter in an integral allows for a more flexible and general solution. It allows us to evaluate the integral over a specific range of values, rather than a fixed value. This is especially useful when the function being integrated is dependent on a variable or when we need to calculate the integral for different values of the limiting parameter.

5. Are there any limitations to using a limiting parameter in an integral?

One limitation of using a limiting parameter in an integral is that it can only be used for integrals with a finite range of integration. If the range of integration is infinite, then the limiting parameter cannot be used. Additionally, the choice of the limiting parameter can also affect the complexity of the integral and the ease of computation.

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