\( I = \int \limits_{0}^{\infty} \frac{\sin ^{2} x}{1+x^{2}} d x = \frac 12 \int \limits_{-\infty}^{\infty} \frac{\sin ^{2} x}{1+x^{2}} d x\)
Mit \(\sin^2 x = \frac 12(1-\cos 2x) \) bekommen wir
\(I = \underbrace{\frac 14 \int_{-\infty}^\infty \frac 1{1+x^2}dx}_{I_1} - \underbrace{\frac 14\int_{-\infty}^\infty \frac{\cos 2x}{1+x^2}dx}_{I_2}\)
Beachte nun \(\cos 2x = \Re \left(e^{2ix}\right)\). Außerdem ist \(1+x^2 = (x+i)(x-i)\).
\(I_1\) kannst du per \(\arctan x\) direkt berechnen oder auch per Residuum:
\(I_1 = \frac 14\cdot 2\pi i Res_i\frac 1{(z-i)(z+i)} = \frac 12 \pi i \frac 1{2i} = \frac \pi4\)
\(I_2 = \frac 14 \Re \left(2\pi i Res_i \frac{e^{2iz}}{(z-i)(z+i)}\right)= \frac 14 \Re \left(2\pi i \frac{e^{-2}}{2i}\right) = \frac \pi4 e^{-2}\)
Also
\(I = \int \limits_{0}^{\infty} \frac{\sin ^{2} x}{1+x^{2}} d x = I_1 - I_2 = \frac\pi4(1-e^{-2})\approx 0.6791\)
Probe 1
Probe 2
