- #1
sirona
- 3
- 0
Hi this is my first post here so I'm sorry if my question seems trivial.
I haven't worked a lot with the dirac delta function before, so i always thought that the shifting property would only work as:
[tex]\int\delta(x-h)\;f(x)\;dx=f(h)[/tex]
Now I've been reading some articles and I came across expressions like:
[tex]\int\delta(h-x)\;f(x)\;dx=f(h)[/tex]
which didn't make sense to me so I checked on the internet and saw that the delta function is supposed to be an even function.
Now I know it is not entirely a true function, but still the only description I know is that it's zero everywhere except for x. If I can give values to x other than zero, than
[tex]\delta(x)=\delta(-x)[/tex] means for example
[tex]\delta(5)=\delta(-5)[/tex] which doesn't make sense to me.
So I'll be very glad if someone can explain to me why it's an even function.
Thanks!
I haven't worked a lot with the dirac delta function before, so i always thought that the shifting property would only work as:
[tex]\int\delta(x-h)\;f(x)\;dx=f(h)[/tex]
Now I've been reading some articles and I came across expressions like:
[tex]\int\delta(h-x)\;f(x)\;dx=f(h)[/tex]
which didn't make sense to me so I checked on the internet and saw that the delta function is supposed to be an even function.
Now I know it is not entirely a true function, but still the only description I know is that it's zero everywhere except for x. If I can give values to x other than zero, than
[tex]\delta(x)=\delta(-x)[/tex] means for example
[tex]\delta(5)=\delta(-5)[/tex] which doesn't make sense to me.
So I'll be very glad if someone can explain to me why it's an even function.
Thanks!