# Integral representation of the delta function

1. Apr 5, 2012

### thomas49th

1. The problem statement, all variables and given/known data
http://gyazo.com/7b2a903b6b3165595b8766d3540f43d9

What is this really saying? I can see that a functino is the inverse fourier transform of the fourier transform... and it doesn't matter which way round you integrate. Is that all it's saying. What's the difference between f(t) and f(t')

Thanks
Thomas

2. Apr 5, 2012

### sunjin09

Fourier transform is a linear integral transform that maps a function f to F, inverse FT maps back to f. So iFt{Ft{f}}=f for all f (that satisfies blah blah blah ...), therefore iFt{Ft{.}} is an identity transformation, i.e., delta function.

3. Apr 5, 2012

### thomas49th

I get what you mean all upto

therefore iFt{Ft{.}} is an identity transformation, i.e., delta function.

?? what is the dot meant to represent? How does it become a delta func?

4. Apr 5, 2012

### sunjin09

Sorry my bad notation. FT{.} is a integral operator that acts on functions, where the dot is replaced by a function f, i.e., FT{f} means Fourier transform of function f. iFT{FT{.}} is the composition of the operator iFT and FT, i.e., successively applying the two operators. Since successively applying the two ops to f gives back f, the composition of the two ops is the identity operator, i.e., delta function. Remember delta function is defined as an integral operator which, when applied to a function f, gives back f.

5. Apr 5, 2012

### thomas49th

ahhh
"delta function is defined as an integral operator which, when applied to a function f, gives back f."

i didn't really think of it like that. But the delta function only works as an identity function a specific instant in time. So it only works for a function at a particular point agreed?

edit: We can aproximate a function using a train of diracs (shannon sampling or some baloni like that). But to say it's a true identity ... only for a specific instant

6. Apr 5, 2012

### sunjin09

If you have to think pointwise, consider ∫ δ(t-t')f(t') dt'=f(t), agreed?
And iFT{FT{f(t')}}(t)=∫ dω exp(iωt) $\frac{1}{2\pi}$∫ exp(-iωt')f(t') dt'=f(t), agreed?
Now note ∫ dω exp(iωt) $\frac{1}{2\pi}$∫ exp(-iωt')f(t') dt'=∫ [$\frac{1}{2\pi}$∫ dω exp(iω(t-t'))]*f(t') dt', comparing the two, you find δ(t-t')=$\frac{1}{2\pi}$∫ dω exp(iω(t-t'))