Evaluating integral with delta function (Fourier Transform)

[grandpa2390]

Homework Statement

Evaluate the Following integrals

1. http://www4b.wolframalpha.com/Calculate/MSP/MSP10141fif9b428c5bab0b00005dc489hi851d28h7?MSPStoreType=image/gif&s=37&w=164.&h=35.

Homework Equations

http://www4b.wolframalpha.com/Calculate/MSP/MSP23802098254f7bhefdhd00002e1e8be006773e70?MSPStoreType=image/gif&s=64&w=184.&h=35.

The Attempt at a Solution

I don't know. please help. I don't know if the -3 is correct. it is what wolfram gave me when I typed it into paste it here. If it is correct I don't know why. I would think the x being cubed would affect the answer so that it isn't just -a[/B]

 

[grandpa2390]

and http://www4f.wolframalpha.com/Calculate/MSP/MSP18161hebeg6i2g9e2cbi0000626gc3eg86i0g2ab?MSPStoreType=image/gif&s=16&w=290.&h=35. but replace the y's with x'
once again I don't know if the result given by wolfrahm is correct. I don't understand how to break this down.

 

[PeroK]

All you have to do is to work out when the argument for the ##\delta## function is ##0##. Can you do that for the two examples?

 

[grandpa2390]

All you have to do is to work out when the argument for the ##\delta## function is ##0##. Can you do that for the two examples?

well that would be when x=0 right? δ(0)=0

so I would plug that into the f(x)? and get f(0^3-3)= -3
and -3^3= -27

for the next one

δ(y-x) = 0 when y=x
so: (x-3)^3

that's pretty simple. thanks!

 

[Ray Vickson]

Homework Statement

Evaluate the Following integrals

1. http://www4b.wolframalpha.com/Calculate/MSP/MSP10141fif9b428c5bab0b00005dc489hi851d28h7?MSPStoreType=image/gif&s=37&w=164.&h=35.

Homework Equations

http://www4b.wolframalpha.com/Calculate/MSP/MSP23802098254f7bhefdhd00002e1e8be006773e70?MSPStoreType=image/gif&s=64&w=184.&h=35.

The Attempt at a Solution

I don't know. please help. I don't know if the -3 is correct. it is what wolfram gave me when I typed it into paste it here. If it is correct I don't know why. I would think the x being cubed would affect the answer so that it isn't just -a[/B]

Why would you think that? ##\int_{-\infty}^{\infty} f(x) \delta(x-a) \, dx = f(a)## [Note: NOT what you wrote under Heading 2.!] So, for ##a = 0## and ##f(x) = x^3 -3## we get the answer ##f(0) = (0^3 - 3)##.

 

[grandpa2390]

Why would you think that? ##\int_{-\infty}^{\infty} f(x) \delta(x-a) \, dx = f(a)## [Note: NOT what you wrote under Heading 2.!] So, for ##a = 0## and ##f(x) = x^3 -3## we get the answer ##f(0) = (0^3 - 3)##.

They are both valid. My professor gave us both. to make my question less confusing and wordy, I wrote the one that I thought applied. but they are both different ways of expressing the same thing.

find where δ(x) = δ(0)

at x=0
f(0-a) = f(-a)

or in the equation you posted, x=a and f(x)=f(a)

it's the same thing. apply it to my problem. the same way.

so we get 0^3-3=-3

 

[Ray Vickson]

They are both valid. My professor gave us both. to make my question less confusing and wordy, I wrote the one that I thought applied. but they are both different ways of expressing the same thing.

find where δ(x) = δ(0)

at x=0
f(0-a) = f(-a)

or in the equation you posted, x=a and f(x)=f(a)

it's the same thing. apply it to my problem. the same way.

so we get 0^3-3=-3

They are both valid. My professor gave us both. to make my question less confusing and wordy, I wrote the one that I thought applied. but they are both different ways of expressing the same thing.

find where δ(x) = δ(0)

at x=0
f(0-a) = f(-a)

or in the equation you posted, x=a and f(x)=f(a)

it's the same thing. apply it to my problem. the same way.

so we get 0^3-3=-3

Sorry: I mis-typed it. What I meant was ##\int_{-\infty}^{\infty} f(x) \delta (x-a) \, dx = f(a)##, not what I wrote. Your equation was correct, and I had made a stupid mistake.