Subbing infinity into limits of an Integral (1 Viewer)

[frenzal_dude]

How would you work this out?:
This is the answer to the integral, just need to sub in values for t.

(1/4)[(e^(j4PIct - j2PIft))/(j4PIc - j2PIf)]

from t = infinity to t = -infinity.

Hope you guys can help.

 

[cutemouse]

These are improper integral and are not in the scope of the current NSW syllabus for Extension 2 Mathematics I believe.

 

[kaz1]

(1/4)[(e^(j4PIct - j2PIft))/(j4PIc - j2PIf)]

wtf?

Use Latex

 

[frenzal_dude]

now with latex:

 

[shaon0]

now with latex:

Let u=j*4*pi*c-j*2*pi*f
y= t: (-inf,inf) (1/4) (e^tu/u)
= inf

 

[Affinity]

Let u=j*4*pi*c-j*2*pi*f
y= t: (-inf,inf) (1/4) (e^tu/u)
= inf

That j is probably sqrt(-1) so doesn't quite work that way

suspect the original poster made mistake somewhere. seems like it's some fourier transform. can you post the question?

 

[jet]

Yeah it looks like a fourier transform, which is beyond the HSC level. Moving it to extracurricular.