Fourier series of (t+4)/(t²+8t+25)²

[dengulakungen]

1.
Find the Fourier series of :
$$g(t)=\frac{t+4}{(t^2+8t+25)^2}$$

2. I have been trying to write the function to match the formula $$\mathcal{F} [\frac{1}{1+t^2}] = \pi e^{-\mid(\omega)\mid}$$


3.
I have simplified the function to
$$(t+4)(\frac{1}{9}(\frac{1}{1+\frac{(t+4)^2}{9}})^2)$$
Cant really see where to go from here.

 

[Khashishi]

Try doing a partial fraction decomposition on g.

 

[Incand]

I may be misleading you here but if you complete the square and then do a variable substitution you get what I would say is an easier problem.

 

[Ray Vickson]

1.
Find the Fourier series of :
$$g(t)=\frac{t+4}{(t^2+8t+25)^2}$$

2. I have been trying to write the function to match the formula $$\mathcal{F} [\frac{1}{1+t^2}] = \pi e^{-\mid(\omega)\mid}$$3.
I have simplified the function to
$$(t+4)(\frac{1}{9}(\frac{1}{1+\frac{(t+4)^2}{9}})^2)$$
Cant really see where to go from here.

You can write the Fourier transform as ${\cal F}(g) (w) = \int_{-\infty}^{\infty} e^{-i w t} g(t) \, dt$, then apply integration by parts: $\int u \, dv = uv - \int v \, du$, with $u = e^{-iwt}$ and $dv = g(t) \, dt$. By a change-of-variables, the integral $\int g\, dt$ is do-able, and afterwards you will be left with a form in which your previous FT formula can now be used.

BTW: you are NOT doing Fourier series, you are doing the Fourier transform. These two concepts are very different and should never be mixed up.