How to find the Fourier transform of $\frac{(t^2+2)}{(t^4+4)}$ and use it to evaluate $\frac{(t^2+2)^2}{(t^4+4)^2}$?

Question

I'm asked to evaluate the Fourier transform of $\dfrac{(t^2+2)}{(t^4+4)}$, and then use it to evaluate the integral from minus infinity to plus infinity of $\dfrac{(t^2+2)^2}{(t^4+4)^2}$.

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Part 1:

So using the residue theorem and integrating along the upper half circle, I get residues at $(1+i)$ and $(-1+i)$.

Evaluating the residues, I get $$\frac{-ie^{\omega(1-i)}}{4}$$ and $$\frac{-ie^{\omega(1+i)}}{4}$$

I use the ML inequality to prove that the arc length goes to zero as $R$ (the radius) tends toward infinity.

Using the Residue Theorem, I get $$\frac{\pi * e^{\omega} *(e^{i \omega}+ e^{i \omega})}{2}$$ which could be turned into $$\pi*e^{\omega}*cos(\omega)$$

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Part 2: For the second part, Plancherel's theorem is obviously of use. The problem is that the integral from minus infinity to plus infinity of $$\pi*e^{\omega}*cos(\omega)$$ doesn't seem to converge, so I don't see if I made a mistake or forgot something somewhere, and if that's not the case, how I should proceed further.

Thanks for your help !

Edit: I forgot to mention that it's the Fourier transform of $\frac{(t^2+2)}{(t^4+4)}$ that I need to find, not "just" the integral from minus to plus infinity. Sorry !

Edit 2: corrected an error in the calculation of the fourier transform. There was an extra 2

Answer 1

First: note that the approach you have taken is partial. Since this expression scales like $e^{-i\omega t}/t^2,$ whether you can add a semicircle contour at infinity that integrates to zero depends on the sign of $\omega.$ If it is positive then we need to add a contour on the negative imaginary half-plane to get $(-i)^2<0$ to avoid exponential blowup; shrinking the curve we have loops about $t=\pm1-i.$ This analysis switches for negative $\omega$ so that you shrink to loops around $t=\pm1+i.$

Second: note that this will solve your convergence concerns as the expression you have is only valid for negative $\omega$ and thus describes only an exponential decay; the expression for positive $\omega$ has an $e^{-\omega}.$

Third: if the above is still unclear try to start from an easier place: show that the Lorentzian $$\int_{\mathbb R} dt~e^{-i\omega t}~\frac1{1+t^2}=\pi~e^{-|\omega|},$$ and practice from there.