I have a problem that says: find the Fourier Transform by definition of the function $f(t)$, being $$ f(t)= e^{-2 \vert t\vert } $$ Then, use this result to solve $$ \int_{-\infty} ^{-\infty} \frac{1}{(x^2+2^2)^2}dx $$
I found that $$\mathscr{F}\{f(t)\}= \frac{2^2}{2^2+w^2}$$ But now I don't know how to relate this result with the integral
$$\mathscr{F}\{e^{-2|t|}\}= \frac{4}{w^2+4}$$ $$\mathscr{F}\left\{\dfrac {e^{-2|t|}}4 \right\}= \frac{1}{w^2+4}$$ Now by Plancherel's theorem you have $$\int_{-\infty}^\infty\hat f^2 (\xi)d\xi=2\pi\int_{-\infty}^\infty\ f^2 (t)dt $$ $$\int_{-\infty}^\infty\frac{1}{(\xi^2+4)^2}d\xi=\frac {\pi}8\int_{-\infty}^\infty\ e^{-4|t|}dt $$ You need to evaluate this integral: $$ I=\frac {\pi}8\int_{-\infty}^\infty e^{-4|t|}dt $$ $$I=\frac {\pi}8\left (\int_{0}^\infty e^{-4t}dt +\int_{-\infty}^0\ e^{4t}dt \right ) $$ $$I=\frac {\pi}8\left (\left | \dfrac {e^{-4t}}{-4} \right |_{0}^\infty+ \left |\dfrac {e^{4t}}{4} \right |_{-\infty}^0 \right ) $$ $$ I=\frac {\pi}8\left ( \dfrac 14+\dfrac 14 \right ) =\frac {\pi} {16}$$ Therefore: $$\int_{-\infty} ^{\infty} \frac{dx}{(x^2+4)^2}=\frac {\pi} {16}$$ As you see the integral dosen't diverge.