I have the integral $$ I= \int_0^\infty\frac{w^2dw}{(1+w^2)^4}$$ and I think I can solve it using the fourier transform and Plancherels formula. I can write Plancherels as: $$\int_{-\infty}^\infty |\hat{f}(w)|^2dw = 2\pi\int_{-\infty}^\infty |f(t)|^2dt $$ Which mean that I can set $\hat{f}(w) = \frac{w}{(1+w^2)^2}$
If I look at the table with known transformations I think that I can use:
$$ tf(t) \rightarrow i\frac{d\hat{f}}{dw}$$ And $$ e^{-|t|} \rightarrow \frac{2}{1+w^2}$$ Those two together will be $$ te^{-|t|} \rightarrow -4i\frac{w}{(1+w^2)^2}$$ Because of this i know that the reverse transform of $\frac{w}{(1+w^2)}$ would be $\frac{te^{-|t|}}{-4i}$
Then I could rewrite I using Plancherels formula:
$$ 2\pi\int_{-\infty}^\infty |\frac{te^{-|t|}}{-4i}|^2 dt$$
I don't think that this is right, according to Wolfram Aplha. Is there some abvious mistake that I have done? How do I change the lower $\infty$ to 0, could I just do that and then take the whole integral times two?