Sifting property of a Dirac delta inverse Mellin transformation

[happyparticle]

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.

 

[Philip Koeck]

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).

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.

 

[happyparticle]

I can't see n anywhere.

This is a reason why I'm not sure.

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.

 

[Philip Koeck]

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).

 

[happyparticle]

I'm still confuse because in my textbook I have this. where I have f(0).

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 $\int dw$

 

[Philip Koeck]

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.

 

[happyparticle]

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.

 

[Philip Koeck]

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.

 

[happyparticle]

$\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw$
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?

 

[Philip Koeck]

$\int_{-\infty}^{\infty} f(t) dt \cdot \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{iw (t-a)}dw$
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).

 

[happyparticle]

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.

 

[Philip Koeck]

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.

 

[happyparticle]

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

 

[Philip Koeck]

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.

 

[Philip Koeck]

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.

 

[happyparticle]

However, I think what I wrote isn't correct.
$f(t) \neq \delta(t-a)$

 

[Philip Koeck]

However, I think what I wrote isn't correct.
$f(t) \neq \delta(t-a)$

Just follow my instructions in post 10.

 

[happyparticle]

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.

 

[happyparticle]

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.

 

[Philip Koeck]

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.

 

[Philip Koeck]

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.

 

[happyparticle]

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).

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

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

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

$\int_{-\infty}^{\infty} f(a) \cdot \infty dt$

 

[Philip Koeck]

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.

 

[happyparticle]

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?

 

[Philip Koeck]

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.

 

[happyparticle]

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 $\int_{-\infty}^{\infty} dw = \infty + \infty$
$\int_{-\infty}^{\infty} f(a) \cdot \infty da$

 

[Philip Koeck]

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.

 

[happyparticle]

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).

 

[Philip Koeck]

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.

 

[happyparticle]

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.

 

[Philip Koeck]

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.

 

[Philip Koeck]

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.