Fourier transform missing 2pi?

[philip041]

I keep doing questions on Fourier transforms where the 1/2pi isn't there.

Example:

$$<br /> F\left[\frac{\partial^2u\left(x,y\right)}{\partial x^2}\right]<br />$$

for which I thought the next step would be

$$<br /> \[<br /> \frac{1}{2\pi}\int^{\infty}_{-\infty}\frac{\partial^2u\left(x,y\right)}{\partial x^2} e^{-isx} dx<br /> \[<br />$$

but I keep finding it written in my answer booklet thing

$$<br /> \[<br /> \int^{\infty}_{-\infty}\frac{\partial^2u\left(x,y\right)}{\partial x^2} e^{-isx} dx<br /> \[<br />$$

I have found one example in my book where the 1/2pi appears, why does it not feature in these examples?

Cheers!

 

[the1ceman]

I keep doing questions on Fourier transforms where the 1/2pi isn't there.

Example:

$$<br /> F\left[\frac{\partial^2u\left(x,y\right)}{\partial x^2}\right]<br />$$

for which I thought the next step would be

$$<br /> \[<br /> \frac{1}{2\pi}\int^{\infty}_{-\infty}\frac{\partial^2u\left(x,y\right)}{\partial x^2} e^{-isx} dx<br /> \[<br />$$

but I keep finding it written in my answer booklet thing

$$<br /> \[<br /> \int^{\infty}_{-\infty}\frac{\partial^2u\left(x,y\right)}{\partial x^2} e^{-isx} dx<br /> \[<br />$$

I have found one example in my book where the 1/2pi appears, why does it not feature in these examples?

Cheers!

Sometimes the 2pi is written on the outside like you did or sometimes it is incorporated into the exponential as $$e^{-2i\pi sx}$$ (substitution in the integrand then gets you from the 2pi in the expo to the outside). Make sure that it is of the same form in your book.

 

[philip041]

ok, i checked my book and there was no indication that 2pi was incorporated into the s but there was also nothing to say against it so thank you!

 

[Mute]

Typically the Fourier transform over "real space" variables (space and time, say) are defined without the factor of 1/2pi in them. It is the inverse transforms from frequency space (k and omega) that have the 1/2pi factors in them. (At least in physics - and engineering? Mathematicians may define the transforms symmetricly so that both the transform and inverse transform have a factor of $1/\sqrt{2\pi}$ out front.)

 

[philip041]

just looked at the question that i found with a 2pi in and what you said, (mute), makes a lot of sense cheers!