Tricky DeltaFunction integration

[sukharef]

Hey!
There is a question.
Here is the integral:

What I'm trying to do is starting with the third integral over zi and with the help of integral definition of DeltaFunction i want to calculate it. As you can see f(yi) has no influence on the integral. Am i right here?

 

[ShayanJ]

No, you're wrong

First, in the step $\frac{1}{2\pi}\int_0^{f(y_i)} dz_i \int_{-\infty}^{\infty} dk \exp{(i(z^{'}-z_i)k)}=\frac{1}{2\pi}\int_0^{f(y_i)} dz_i \exp{(i(z^{'}-z_i)k)} \int_{-\infty}^{\infty} dk$, you can't take the exponential out of the integral w.r.t. k because it contains a k.

Second, you don't need such calculations. The integral $\int_0^{f(y_i)} dz_i \delta(z^{'}-z_i)$ is zero if $-z^{'} \epsilon\!\!/\ (0,f(y_i))$ and is one otherwise.

 

[sukharef]

No, you're wrong

First, in the step $\frac{1}{2\pi}\int_0^{f(y_i)} dz_i \int_{-\infty}^{\infty} dk \exp{(i(z^{'}-z_i)k)}=\frac{1}{2\pi}\int_0^{f(y_i)} dz_i \exp{(i(z^{'}-z_i)k)} \int_{-\infty}^{\infty} dk$, you can't take the exponential out of the integral w.r.t. k because it contains a k.

I don't think you are right. If you've got something like that:

you can calculate the integral over x thinking that y is a constant. Then you come back to integral over y. Maybe i misunderstood something?

No, you're wrong
Second, you don't need such calculations. The integral $\int_0^{f(y_i)} dz_i \delta(z^{'}-z_i)$ is zero if $-z^{'} \epsilon\!\!/\ (0,f(y_i))$ and is one otherwise.

I know, but i'd like to write it down analytically. That's why i use integral definition of DeltaFunction.

 

[ShayanJ]

I don't think you are right. If you've got something like that:

you can calculate the integral over x thinking that y is a constant. Then you come back to integral over y. Maybe i misunderstood something?

Oh yeah...you're right. I thought you were going to integrate w.r.t. k.

I know, but i'd like to write it down analytically. That's why i use integral definition of DeltaFunction.

Well, then you certainly should know that there is something wrong with your calculations. Because $f(y_i)$ does affect the value of the integral. But I don't see any problem!
I should say such calculations can be very tricky but since we know that your result is wrong, you should either find what's wrong with your calculations or abandon them.

EDIT:
One more piece of information.
Dirac delta is an even distribution meaning $\delta(-x)=\delta(x)$, so you can write $\int_0^{f(y_i)} dz_i \delta(z^{'}-z_i)=\int_0^{f(y_i)} dz_i \delta(z_i-z^{'})$. The first is non-zero only when $-z^{'}\epsilon [0,f(y_i)] \Rightarrow z^{'}\epsilon [-f(y_i),0]$ and the second one is non-zero only when $z^{'}\epsilon[0,f(y_i)]$, So the integral is non-zero only when $z^{'}=0$. That's what you should assume from the first place, before doing any other calculation.

 

[sukharef]

EDIT:
One more piece of information.
Dirac delta is an even distribution meaning δ(−x)=δ(x) \delta(-x)=\delta(x) , so you can write ∫f(yi)0dziδ(z′−zi)=∫f(yi)0dziδ(zi−z′) \int_0^{f(y_i)} dz_i \delta(z^{'}-z_i)=\int_0^{f(y_i)} dz_i \delta(z_i-z^{'}) . The first is non-zero only when −z′ϵ[0,f(yi)]⇒z′ϵ[−f(yi),0] -z^{'}\epsilon [0,f(y_i)] \Rightarrow z^{'}\epsilon [-f(y_i),0] and the second one is non-zero only when z′ϵ[0,f(yi)] z^{'}\epsilon[0,f(y_i)] , So the integral is non-zero only when z′=0 z^{'}=0 . That's what you should assume from the first place, before doing any other calculation.

if the integral is non-zero only when z′=0 z^{'}=0 , it is a little bit weird . and f(yi) has no influence on the integral as it was supposed, but i doubt it.