As the title states: **Could we solve $\int\_{0}^{\infty}\sin(x)dx$ and what does it say about $\lim\_{x\to\infty}\cos(x)$?**

It is clear we can't solve this using the fundamental theorem of Calculus, but I thought it would be solvable as a summation.

$$\int\_{0}^{\infty}\sin(x)dx=\lim\_{u\to\infty}\lim\_{v\to\infty}\sum\_{i=1}^{u}\frac{v\sin(\frac{vi}u)}{u}$$

I figured this would be hard to solve, so I went a different path:

Noting that $$\int\_{0}^{\pi}\sin(x)dx=2$$$$\int\_{n}^{n+\pi}\sin(x)dx=-\int\_{n-\pi}^{n}\sin(x)dx$$For $n=m\pi$, $m=0,1,2,3,\cdots$

Trying to put this all together, I arrive at 
$$\int\_{0}^{\infty}\sin(x)dx=\sum\_{i=0}^{\infty}(-1)^i2=2-2+2-2+\cdots$$This summation evaluates to equal:
$$2-2+2-2+\cdots=1$$So, I argue that
$$\int\_{0}^{\infty}\sin(x)dx=1$$
Furthermore, going back to the fundamental theorem of Calculus, we have:
$$\int\_{0}^{\infty}\sin(x)dx=\lim\_{p\to\infty}-\cos(p)-[-\cos(0)]$$$$=-\lim\_{p\to\infty}\cos(p)+1=1$$
Therefore, we have
$$-\lim\_{p\to\infty}\cos(p)=0=-0=\lim\_{p\to\infty}\cos(p)$$

Am I wrong? If so, what does the indefinite integral evaluate to and what does it mean for $\lim\_{p\to\infty}\cos(p)$?