Although I am not sure, but I think the improper integral $\int_0^\infty \sin x \:dx$ is divergent. I have done a proof for that, which looks like-
Firstly I take $\int_{0}^{n \pi}\sin x\:dx=\int_{0}^{\pi}\sin x\:dx+\int_{\pi}^{2 \pi}\sin x\:dx+...+\int_{(n-1)\pi}^{n \pi}\sin x\:dx$
Then after a little calculation, I get $\int_{0}^{\pi}\sin x\:dx=2,\: \int_{\pi}^{2 \pi}\sin x\:dx=-2,\: ...,\: \int_{(n-1)\pi}^{n \pi}\sin x\:dx=(-1)^{n-1}2$
Therefore, $\int_{0}^{n \pi}\sin x\:dx=\sum_{r=1}^n (-1)^{r-1}2$
Now, $\int_0^\infty \sin x \:dx=\lim_{n\to\infty} \int_{0}^{n \pi}\sin x\:dx=\lim_{n\to\infty}\sum_{r=1}^n (-1)^{r-1}2$, but this limit does not exist because the alternating series $\sum_{r=1}^\infty (-1)^{r-1}2$ is divergent.
Am I right? Can anybody tell me is there any mistake in my proof?If there is any, please mark it help me with a proper way out.
Thanks for your assistance in advance.