Sifting property of a Dirac delta inverse Mellin transformation

In summary, the conversation discusses the sifting property of a function that is the inverse Mellin transformation of the Dirac delta function. The sifting property states that as the variable n approaches infinity, the integral of the function times the Dirac delta function approaches the value of the function at a specific point. The conversation also includes a discussion on how to prove this property using an integral.
  • #1
happyparticle
406
20
Homework Statement
Sifting property of a Dirac delta inverse Mellin transformation
Relevant Equations
##f(t) = \delta(t-a) ##
##\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw = f(t) = \delta(t-a)##
##\int_{-\infty}^{\infty} f(x) \delta(x) dx = f(0)##
Hi,
I have to verify the sifting property of ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds## which is the inverse Mellin transformation of the Dirac delta function ##f(t) = \delta(t-a) ##.

let ##s = iw## and ##ds = idw##

##\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw = f(t) = \delta(t-a)## (1)

The sifting property says that if n -- > ##\infty## then ##\int_{-\infty}^{\infty} f(x) \delta(x) dx --> f(0)##

From (1),
##2 \pi \delta(t-a) = \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw ##

Thus, my guess is that

##\int_{\infty}^{-\infty} f(t) \delta(t-a) dt = f(0)##

However, I don't see how to prove it.
 
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  • #2
EpselonZero said:
I have to verify the sifting property of ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds##
Shouldn't you just put this integral inside another integral and show that it does the sifting just like δ(t-a)?
So if you have f(t) in the integrand then the result of the whole expression should be f(a).
EpselonZero said:
The sifting property says that if n -- > ##\infty## then ##\int_{-\infty}^{\infty} f(x) \delta(x) dx --> f(0)##
I can't see n anywhere.
 
  • #3
Philip Koeck said:
I can't see n anywhere.
This is a reason why I'm not sure.
Philip Koeck said:
Shouldn't you just put this integral inside another integral and show that it does the sifting just like δ(t-a)?
So if you have f(t) in the integrand then the result of the whole expression should be f(a).

I'm not so sure to follow.
 
  • #4
EpselonZero said:
This is a reason why I'm not sure.I'm not so sure to follow.
The sifting property of the Dirac function is
∫f(t) δ(t-a) dt = f(a), where the integration can be from -∞ to +∞ or it can just be in a small range that includes the point t = a.

Now simply replace δ(t-a) with the Mellin transform you give in the first post and see if you can carry out the integration and get f(a).
 
  • #5
I'm still confuse because in my textbook I have this. where I have f(0).
ACj1PF0.png

thus, ##\int_{-\infty}^{\infty} f(t) \frac{1}{2\pi} e^{iw(t-a)} dw##
I still have a ##2\pi## and f is a function of t but I have ##\int dw##
 
  • #6
EpselonZero said:
I'm still confuse because in my textbook I have this. where I have f(0).
View attachment 298781
thus, ##\int_{-\infty}^{\infty} f(t) \frac{1}{2\pi} e^{iw(t-a)} dw##
I still have a ##2\pi## and f is a function of t but I have ##\int dw##
In the textbook it's the special case for a = 0.

You ought to have an integral in the integrand!

Do this:
Write the left hand side of my version of the sifting property.
Replace the delta-function by the mellin transform (the whole integral).
See what you can do.
 
  • #7
There's probably something I just don't know, because my question still remains. I don't see how to get rid of the ##\frac{1}{2 \pi}##

I get ##\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(t) dt \cdot \frac{e^{iw (t-a)}}{ i(t-a)} |_{-\infty}^{\infty}##
If this is correct, I don't see how this could be equal to f(a), because if t = a, I have a 0 in the denominator.
 
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  • #8
EpselonZero said:
There's probably something I just don't know, because my question still remains. I don't see how to get rid of the ##\frac{1}{2 \pi}##

I get ##\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(t) dt \cdot \frac{e^{iw (t-a)}}{ i(t-a)} |_{-\infty}^{\infty}##
If this is correct, I don't see how this could be equal to f(a), because if t = a, I have a 0 in the denominator.
I don't see how (t-a) can end up in a denominator.

If you write the whole derivation you did, somebody might be able to help.
 
  • #9
##\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw##
u = iw(t-a) , du = i(t-a)dw
##\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(t) dt \cdot \int_{-\infty}^{\infty} e^u \frac{du}{i(t-a)}##Do you have an idea how can I have a ##2 \pi## on the numerator?
 
  • #10
EpselonZero said:
##\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw##
u = iw(t-a) , du = i(t-a)dw
##\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(t) dt \cdot \int_{-\infty}^{\infty} e^u \frac{du}{i(t-a)}##Do you have an idea how can I have a ##2 \pi## on the numerator?
I believe what you are supposed to do is this:

∫ f(t) MT(t-a) dt = ...

If the MT(t-a) has the sifting property then you should get f(a).

MT(t-a) stands for the integral you state in the first post (the inv. Mellin transform of the delta-function).
 
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  • #11
Philip Koeck said:
If the MT(t-a) has the sifting property then you should get f(a).
Exactly, but clearly I don't know how to verify that since what I did so far doesn't get me anywhere.
I thought that to have the sifting property MT should be equal f(a) if t = a and 0 otherwise.
 
  • #12
EpselonZero said:
Exactly, but clearly I don't know how to verify that since what I did so far doesn't get me anywhere.
I thought that to have the sifting property MT should be equal f(a) if t = a and 0 otherwise.
No, MT is not f(a).
You have to show that MT(t-a) behaves like δ(t-a).

Follow my instructions from post 10.
 
  • #13
What exactly is MT ? Is it delta(t-a) ? In this case I just get back to beginning where I have to do like in my textbook in post #5. And this is exactly what I did in post #7, but again if t = a I have 0 in the denominator or
##
\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw
##
If t = a
##\int_{-\infty}^{\infty} f(a) dt \cdot \frac{1}{2 \pi} (\infty+\infty)##
 
  • #14
EpselonZero said:
What exactly is MT ? Is it delta(t-a) ? In this case I just get back to beginning where I have to do like in my textbook in post #5. And this is exactly what I did in post #7, but again if t = a I have 0 in the denominator or
##
\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw
##
If t = a
##\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} (\infty+\infty)##
MT(t-a) is just a shorthand I introduced for the inverse Mellin transform you state in post 1.
I define it in post 10.
 
  • #15
EpselonZero said:
I have to verify the sifting property of ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds## which is the inverse Mellin transformation ...
There you write what MT(t-a) is.
 
  • #16
However, I think what I wrote isn't correct.
##f(t) \neq \delta(t-a)##
 
  • #17
EpselonZero said:
However, I think what I wrote isn't correct.
##f(t) \neq \delta(t-a)##
Just follow my instructions in post 10.
 
  • #18
That is what I did, but I got what I wrote in post #13. I'm more confused than ever. I doubt about everything I wrote.
 
  • #19
I knew that ##\int_{-\infty}^{\infty} \delta(t-a)f(t) dt = f(a)##
And that's what I tried.
##
\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw
##

The only way I see to have f(a) from the left hand side is if t = a
However, I get what I wrote in post #13.

I was probably unclear, but that's what I don't know.

I just running around since the beginning.
 
  • #20
EpselonZero said:
I knew that ##\int_{-\infty}^{\infty} \delta(t-a)f(t) dt = f(a)##
And that's what I tried.
##
\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw
##

The only way I see to have f(a) from the left hand side is if t = a
However, I get what I wrote in post #13.

I was probably unclear, but that's what I don't know.

I just running around since the beginning.
In post 10 I put MT(t-a) inside the integral over t. I think you place MT(t-a) as a factor outside this integral.
That might be what's wrong.
 
  • #21
EpselonZero said:
I knew that ##\int_{-\infty}^{\infty} \delta(t-a)f(t) dt = f(a)##
And that's what I tried.
##
\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw
##

The only way I see to have f(a) from the left hand side is if t = a
However, I get what I wrote in post #13.

I was probably unclear, but that's what I don't know.

I just running around since the beginning.
If you just follow the instructions in post 10 you'll have it in 3 lines.
 
  • #22
I really tried, but I didn't find. As I said, it is probably something I don't know, so I can't solve it.

Right now it feels like a riddle.

no matter how I put the expression, I don't see how to get f(a).
##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} e^{iw (t-a)}dwdt
##

As I said, if t = a I don't get f(a).

##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} e^{iw (0)}dwdt
##

##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} 1dwdt
##

##
\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} (\infty + \infty)dt
##

##\int_{-\infty}^{\infty} f(a) \cdot \infty dt##
 
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  • #23
EpselonZero said:
I really tried, but I didn't find. As I said, it is probably something I don't know, so I can't solve it.

Right now it feels like a riddle.

no matter how I put the expression, I don't see how to get f(a).
##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} e^{iw (t-a)}dwdt
##
You have too many dt-s.
 
  • #24
It's a typo. I didn't kept the error below.

I don't think you understand my problem.

How do you get f(a) in 3 lines?
 
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  • #25
EpselonZero said:
It's a typo. I didn't kept the error below.

I don't think you understand my problem.

How do you get f(a) in 3 lines?
Write the whole calculation without the typo, then we'll see.
 
  • #26
That's what I did. that is exactly what I do since monday.

It must be the fifth time you ask me to type the same thing without further explanations.
##\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(t) \cdot \frac{1}{2 \pi} e^{iw (t-a)}dwdt##
To have f(a), t --> a
##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} e^{iw (0)}dwda
## 0 because a-a =0
##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} 1dwda
## 1 because ##e^{0} = 1##
##
\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} (\infty + \infty)da
## because ##\int_{-\infty}^{\infty} dw = \infty + \infty##
##
\int_{-\infty}^{\infty} f(a) \cdot \infty da
##
 
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  • #27
EpselonZero said:
That's what I did. that is exactly what I do since monday.

It must be the fifth time you ask me to type the same thing without further explanations.
##\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(t) \cdot \frac{1}{2 \pi} e^{iw (t-a)}dwdt##
To have f(a), t --> a
##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} e^{iw (0)}dwda
## 0 because a-a =0
##
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} 1dwda
## 1 because ##e^{0} = 1##
##
\int_{-\infty}^{\infty} f(a) \cdot \frac{1}{2 \pi} (\infty + \infty)da
## because ##\int_{-\infty}^{\infty} dw = \infty + \infty##
##
\int_{-\infty}^{\infty} f(a) \cdot \infty da
##
First line is right.
You can't let t become a because t is the integration variable.
Use one of the given formulas in stead.
 
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  • #28
Philip Koeck said:
Use one of the given formulas in stead.
That's probably what I miss.
The only formula I can think off is ##\int_{-\infty}^{\infty} \delta(x)dx = 1##, but I can't use it since there's a f(t).
 
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  • #29
EpselonZero said:
Relevant Equations::
##\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-iwa}e^{iwt} dw = \delta(t-a)##Hi,
I have to verify the sifting property of ##\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} e^{-sa}e^{st} ds##
Do I understand correctly that you are allowed to use any of the relevant equations you list and that you should show the sifting property of the last integral in the above quote?
I used the relevant equation shown above.
 
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  • #30
Since I have to verify that, I don't think I can use that right away. I see what you meant now.
However, if this is the thing I have to show it makes no sense to use it.
 
  • #31
EpselonZero said:
Since I have to verify that, I don't think I can use that right away. I see what you meant now.
However, if this is the thing I have to show it makes no sense to use it.
Exactly what are you supposed to verify?
Please state it once more.
 
  • #32
EpselonZero said:
Since I have to verify that, I don't think I can use that right away. I see what you meant now.
However, if this is the thing I have to show it makes no sense to use it.
I wouldn't say that I'm using the result I'm trying to show.
It's true that the two integrals follow from each other by a change of variable, but I guess that's the point of the exercise. I agree it's not a very good exercise.

I would also question that the integral you quote as the inverse Mellin transform of the delta function is actually that. This integral is clearly just another way of writing the integral representation of the delta function.
So to me it seems this integral is the delta function and not its Mellin transform.
In short the point of the exercise seems to be to show that the integral in question is the delta function and therefore it has the sifting property.
Unless, of course, you're supposed to verify the sifting property of the delta function, but that's a completely different problem and not what you stated in your original post.
 

Related to Sifting property of a Dirac delta inverse Mellin transformation

1. What is the sifting property of a Dirac delta inverse Mellin transformation?

The sifting property of a Dirac delta inverse Mellin transformation states that when the inverse Mellin transform is applied to a Dirac delta function, the resulting function will be equal to the original function evaluated at the singularity of the inverse Mellin transform.

2. How is the sifting property used in mathematical analysis?

The sifting property is often used in mathematical analysis to simplify integrals and solve differential equations. It allows for the transformation of complex functions into simpler forms, making them easier to manipulate and analyze.

3. Can you provide an example of the sifting property in action?

Sure, for example, if we have the inverse Mellin transform of the function f(x) given by F(s), then applying the sifting property will result in the function f(x) evaluated at the singularity of F(s). This can be written as: ∫0 f(x)δ(s - x)dx = f(s).

4. What is the significance of the Dirac delta function in this property?

The Dirac delta function is a mathematical construct that represents an infinitely narrow and tall spike at a specific point. It has the property that when integrated with any function, it will return the value of that function at the point where the spike is located. This makes it a useful tool in simplifying mathematical calculations involving inverse Mellin transforms.

5. Are there any limitations to the sifting property of a Dirac delta inverse Mellin transformation?

While the sifting property is a powerful tool in mathematical analysis, it does have some limitations. It can only be applied to functions that have a singularity at the point where the delta function is located. Additionally, it may not always work for functions that are not well-behaved or do not have a well-defined inverse Mellin transform.

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