- #1

kbrijesh

- 2

- 0

i.e. ∫((1+at^2)^-n)*exp(-jωt)dt; limits of integration goes from -∞ to ∞

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- Thread starter kbrijesh
- Start date

- #1

kbrijesh

- 2

- 0

i.e. ∫((1+at^2)^-n)*exp(-jωt)dt; limits of integration goes from -∞ to ∞

- #2

jackmell

- 1,804

- 53

i.e. ∫((1+at^2)^-n)*exp(-jωt)dt; limits of integration goes from -∞ to ∞

You can do it via contour integration for specific small n but not sure for general n. But first, look at a simple case:

[tex]\int_{-\infty}^{\infty} \frac{e^{-i\omega t}}{(1+2t^2)^2}dt[/tex]

Now that can be solved by the Residue Theorem. First get that one straight, then go on to n=3, maybe 4, then try and come up with an expression for the general case.

- #3

kbrijesh

- 2

- 0

You can do it via contour integration for specific small n but not sure for general n. But first, look at a simple case:

[tex]\int_{-\infty}^{\infty} \frac{e^{-i\omega t}}{(1+2t^2)^2}dt[/tex]

Now that can be solved by the Residue Theorem. First get that one straight, then go on to n=3, maybe 4, then try and come up with an expression for the general case.

Yes, you are true. Even using contour integration we can do it upto n=3. But I am y\trying to get a generalized solution.

Anyway, thanks for your reply.

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