Tricky Integral (fourier transforms)

[wolf party]

the question is on Fourier transforms and i am stuck on how to approch the integral


INT[ (1/(2*pi))*e^(ik(x-y)-(1+ik)ct) dk


i know that the integral INT [(1/(2*pi))*e^(ik(x-y-ct)) dk] yields the delta function

= DELTA(x-y-ct) but the terms in the exponential from the integral in question have stumped me

any ideas would be great, cheers

 

[Pere Callahan]

What don't you like about the delta function result?

The e^-ct is just a constant...

 

[wolf party]

would that mean we now have

(INT[e^-ct]) *DELTA(x-y-ct) ?

 

[Pere Callahan]

would that mean we now have

(INT[e^-ct]) *DELTA(x-y-ct) ?

No, the integral is gone, but the constant still there. What you end up with is

e^-ctDelta(x-y-ct)

 

[wolf party]

cheers

from that, the integral goes into the second integral

u(x,t)=[INT (u(y, 0)dy*e^-ct*DELTA(x-y-ct))] - integrated over real space

does this equal e^-ct*u(x-ct,0) ?

 

[Pere Callahan]

cheers

from that, the integral goes into the second integral

u(x,t)=[INT (u(y, 0)dy*e^-ct*DELTA(x-y-ct))] - integrated over real space

does this equal e^-ct*u(x-ct,0) ?

Yes.