Finding value of $\displaystyle \int^{\infty}_{0}\sin(t)dt$
What i have tried yet
As we know that period of $\sin(x)$ is $2\pi$
So we can split the intehral as
$\displaystyle \lim_{n\rightarrow \infty}\bigg[\int^{2\pi}_{0}\sin(t)dt+\int^{4\pi}_{2\pi}\sin(t)dt+\int^{6\pi}_{4\pi}\sin(t)dt+\cdots\cdots +\int^{2n\pi}_{(2n-1)\pi}\sin(t)dt +\int^{(2n+2)\pi}_{2n\pi}\sin(t)dt\bigg)\bigg]=0$
Because $\displaystyle \int^{2\pi}_{0}\sin(t)dt=0$
What i have mention is right, if not then please explain me , thanks
Best Answer
The integral is undefined (aka. "divergent").
To see this, you need to consult the definition of a definite integral with upper bound $\infty$:
$$\int_{0}^{\infty} f(x)\ dx := \lim_{x_\mathrm{max} \rightarrow \infty} \int_{0}^{x_\mathrm{max}} f(x)\ dx$$
This limit is a limit over the real numbers, and hence your problem is equivalent to asking what the value of
$$\lim_{x_\mathrm{max} \rightarrow \infty} \int_{0}^{x_\mathrm{max}} \sin(x)\ dx$$
is. But since the right-hand integral is just $1 - \cos(x_\mathrm{max})$, then that means we are asking for
$$\lim_{x_\mathrm{max} \rightarrow \infty} [1 - \cos(x_\mathrm{max})]$$
And this limit does not exist since the trig always oscillates up and down periodically; hence also, the integral has no value.