Finding the inverse of an integral transform

[shaiguy6]

Hi, I've defined an integral transform that I'll call T (obviously very similar to the Fourier transform):

$$T[f(x)](k)=\int_{-\infty}^\infty e^{- 2 \pi i k (x-a)}f(x)dx=\int_{-\infty}^\infty e^{- 2 \pi i k x}e^{ 2 \pi i k a}f(x)dx$$

where a is a given parameter. perhaps we can call this the "a centered Fourier transform". I'm having trouble finding the inverse $$T^{-1}$$

I've looked through the Fourier inversion theorem (though I'm having some trouble understanding it completely), and I can't really tell if that's useful for me.

Any help is appreciated.

 

[shaiguy6]

Hi, I've defined an integral transform that I'll call T (obviously very similar to the Fourier transform):

$$T[f(x)](k)=\int_{-\infty}^\infty e^{- 2 \pi i k (x-a)}f(x)dx=\int_{-\infty}^\infty e^{- 2 \pi i k x}e^{ 2 \pi i k a}f(x)dx$$

where a is a given parameter. perhaps we can call this the "a centered Fourier transform". I'm having trouble finding the inverse $$T^{-1}$$

I've looked through the Fourier inversion theorem (though I'm having some trouble understanding it completely), and I can't really tell if that's useful for me.

Any help is appreciated.

Oh I am dumb:$$T[f(x)](k)=\int_{-\infty}^\infty e^{- 2 \pi i k x}e^{ 2 \pi i k a}f(x)dx=e^{ 2 \pi i k a} \int_{-\infty}^\infty e^{- 2 \pi i k x}f(x)dx=e^{ 2 \pi i k a}\mathcal{F}[f(x)]$$

So uhh, actually I'm still not sure what that makes the inverse of T