You are meant to (somehow) use the inversion formula:

We’ve been writing the inversion formula since we first talked about Fourier transform 30-Jan. What’s new ( and theorem ) is that you can actually trust the inversion formula to tell the truth.

For question 2B, partial fraction clearly won’t work on this. And if I complete the square, the equation looks horrible. How should I tackle this question?

You have the right idea – you just need to stick with it.

You can factor the denominator as:
,
with constants and not too horrible. Then use partial fractions.

Question from a student:

You are meant to (somehow) use the inversion formula:

We’ve been writing the inversion formula since we first talked about Fourier transform 30-Jan. What’s new ( and

theorem) is that you can actually trust the inversion formula to tell the truth.See 06-Feb, slide 5

Question from a student:

You have the right idea – you just need to stick with it.

You can factor the denominator as:

,

with constants and not too horrible. Then use partial fractions.

I still don’t get it. If coefficients of both “iw” are 1 then one of the terms we get is but we want . So far I have done this

You need to factor out the 2. eg: