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\int_{0}^{\infty}\sin{t^{2}} dt=\sum_{n=0}^{\infty}\int_{\sqrt{nx}}^{\sqrt{(n+1)x}}\sin(t^{2}) dt \tea{gilt da:}\tag{*}\\
\int_{0}^{\infty}t^{2} dt=\int_{\sqrt{t}}^{\sqrt{t+1}}t^{2} dt +\int_{\sqrt{2t}}^{\sqrt{2(t+1)}}t^{2}dt\dots\\
*=\sum_{n=0}^{\infty} \int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}}(-1)^{n}\sin(t^{2})dt \tea{da gilt:}\\\sin(t^{2})\in [0,\infty ]=[0,\sqrt{\pi}]\cup [\sqrt{\pi},\sqrt{2\pi}]\cup\dots\\
\Rightarrow *\leq \sum_{n=0}^{\infty} (-1)^{n} \int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}}1dt\\
=\sqrt{\pi}\sum_{n=0}^{\infty}(-1)^{n}(\sqrt{n-1}-\sqrt{n})\\
\Rightarrow \tea{konvergiert nach Leipnetz da gilt:} \sqrt{n-1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}