I am having trouble computing the sum of the integral:
$$ J(t)=\int_{-\infty}^{\infty} \frac{\cos(tx)}{(1+x^2)^2} \ dx $$
To compute the sum, I have to use the fact that the following integral:
$$ F(t)=\int_{-\infty}^{\infty} \frac{\cos(tx)}{1+x^2} \ dx = \pi e^{-|t|} $$
How could I compute the first integral (named $J(t)$) using the fact that the second integral (named $F(t)$) is computed using $F(t)$ = $F''(t)$ and has the form $F(t) = ae^t + be^{-t}$ with ($a = 0$) and ($b = \pi$)?
I tried differentiating to obtain an equation that resembles the form $J''(t) - J(t) = -\pi e^{-|t|}$. However, I am unsure if I am proceeding correctly. How can I properly perform this differentiation?
It's not true that $F^{\prime\prime}=F$. In fact $t\mapsto \pi e^{-t}$, when differentiated twice, yields a Dirac delta at $t=0$. And differentiating under the integral sign leads to a diverging integral.
Instead, observe that $J$ and $F$ are related by $$J^{\prime\prime}(t) = -\int_{-\infty}^{+\infty}\frac{x^2\cos(xt)}{(1+x^2)^2}dx=-F(t)+\int_{-\infty}^{+\infty}\frac{\cos(xt)}{(1+x^2)}dx=J(t)-F(t)$$
You already arrived at that conclusion. So if you have the expression for $F$ (and you do), then you can compute the expression for $J$ using ordinary differential equations. Can you finish?