Calculate
It is "good" integral
at x=0 =0
at x=inf decreases as $\displaystyle 1/x^2$
and its value is
integrate sin x dx/ (1+x^2) from x=0 to +inf - Wolfram|Alpha
Thank you I thought in the following way :
By integration by parts, we know that $\displaystyle 2\int_{0}^{\infty}a e^{-a^{2}(1+x^{2})}da =\frac{1}{1+x^{2}}$
We then get:
$\displaystyle I =\int_{0}^{\infty}\frac{\sin x}{1+x^{2}}dx = 2\int_{0}^{\infty}ae^{-a^{2}}\boxed{\int_{0}^{\infty}e^{-a^{2}x^{2}}\sin x dx}da$
by working out the boxed part using integration by parts we get the answer
Who hasn't at disposal Wolfram's tools but only a table of Laplace Tranforms can remember that is...
$\displaystyle \displaystyle \varphi(s)= \mathcal{L} \{\frac{1}{1+t^{2}}\} = \int_{0}^{\infty} \frac{e^{-s\ t}}{1+t^{2}}\ dt = \cos s \ \{\frac{\pi}{2} - Si (s) \} - \sin s \ Ci(s) $ (1)
... so that is...
$\displaystyle \displaystyle \int_{0}^{\infty} \frac{\sin t}{1+t^{2}} \ dt = Im \{ \varphi(-i)\} $ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$