Is this integration finite?

Question

Let $I = \int_1^{\infty}\sin x dx$. Is $I < \infty$? I can get no clue to solve this. Using trivial bound for $\sin x$ gives me no conclusion. Any help will be appreciated. Thank you.

Answer 1

You can see by yourself that

$$\lim_{\lambda \to +\infty} \int_1^{\lambda} \sin(x)\ dx$$

Which turns out to be a cosine, in the limit of the argument to infinity.

Let's talk about the Sine. The same speech will hold for Cosine function.

If $\sin x$ had a limit $L$ for $x\to\infty$, then for every sequence $(x_n)$ such that $x_n\to\infty$ we would have $$\lim\limits_{n\to\infty} \sin x_n=L.$$ In particular, this limit would exist and would have the same value for every choice of such sequence $(x_n)$. (See e.g. here http://en.wikipedia.org/wiki/Limit_of_a_function#Sequential_limits)

If you choose $x_n= 2n\pi$, then this limit is equal $0$.

If you choose $x_n=\frac\pi2+2n\pi$, then this limit is equal to $1$.

The limit is not defined and surely it does not exist.

Hence the integral does not converge in the usual sense.

Answer 2

$$\lim_{M\to +\infty}\int_{1}^{M}\sin(x)\,dx $$ does not exist, hence your integral is simply not convergent. On the other hand, we may define a regularized value in the following way: $$ R_0=\lim_{\lambda\to 0^+}\int_{0}^{+\infty}\sin(x)e^{-\lambda x}\,dx $$ since for any $\lambda>0$ the function $\sin(x)e^{-\lambda x}$ is both Lebesgue-integrable and improperly Riemann integrable over $\mathbb{R}^+$. By integration by parts $$ R_0=\lim_{\lambda\to 0^+}\frac{1}{1+\lambda^2}=1.$$ Similarly, $$ R_1 = \lim_{\lambda\to 0^+}\int_{1}^{+\infty}\sin(x)\,e^{-\lambda x}\,dx = \cos(1).$$