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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?

rablentain
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    Yes, you can just make that integral twice the integral from zero to $\infty$. – Ron Gordon Oct 19 '14 at 10:42
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    @theva Just another route to check your result, using $\displaystyle \left(\arctan (\alpha x)\right)'_x= \frac{\alpha}{1+\alpha^2x^2} $ and derivation under the $\displaystyle\int$ sign.

    You readily write $$ \int_0^{\infty } \frac{1}{\left(a+bx^2\right)} , dx=\frac{\pi }{2 \sqrt{a b}} \quad a>0,b>0, $$ giving

    $$ \int_0^{\infty } \frac{x^2}{\left(1+x^2\right)^4} , dx=-\left.\frac16\partial_a^2\partial_b\int_0^{\infty } \frac{1}{\left(a+bx^2\right)} , dx\right|{a=1,b=1}=-\left.\frac16\partial_a^2\partial_b\frac{\pi }{2 \sqrt{a b}}\right|{a=1,b=1}=\frac{\pi }{32}. $$

     – Olivier Oloa Oct 19 '14 at 11:00
  • I don't get it, am I on the right way or not? – rablentain Oct 19 '14 at 11:18

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