Could we solve $\int_{0}^{\infty}\sin(x)dx$ and what does it say about $\lim_{x\to\infty}\cos(x)$?

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 $\lim_{p\to\infty}\cos(p)$?

Solution 1:

No. The integral doesn't convergent and the limit doesn't exist given the periodicity of the functions involved.

Solution 2:

It is clear that neither $\int_0^{+\infty}\sin x\,dx$ nor $\lim\limits_{x\to+\infty}\cos x$ exist in the conventional meanings of these symbols. But similar to how one can assign more or less reasonable values to some divergent series, one can also assign more or less reasonable values to some divergent integrals.

Recall that a series $\sum_n a_n$ is Cesàro-summable if the sequence of arithmetic means of the partial sums converges,

$$(a_n)\quad\text{Cesàro-summable } :\mspace{-5mu}\iff \lim_{n\to\infty} \frac{1}{n+1}\sum_{k = 0}^n \sum_{m = 0}^k a_m\quad\text{exists.}$$

Analogously, we might call a locally integrable function $f\colon [0,+\infty) \to \mathbb{R}$ C-integrable if the limit of the arithmetic means of the partial integrals,

$$\frac{1}{x}\int_0^x \biggl(\int_0^t f(u)\,du\biggr)dt,$$

exists. If $L := \lim\limits_{x\to +\infty} \int_0^x f(t)\,dt$ exists, then $f$ is C-integrable, and the C-integral equals $L$, so C-integrability is compatible with integrability, like Cesàro-summability is compatible with summability.

Since $\int_0^t \sin u\,du = 1 - \cos t$, we have

$$\frac{1}{x}\int_0^x \biggl(\int_0^t \sin u\,du\biggr)dt = \frac{1}{x}\int_0^x 1 - \cos t\,dt = 1 - \frac{\sin x}{x} \to 1,$$

so $\sin$ is C-integrable to $1$.

There are other ways than Cesàro-summation to assign values to (some) divergent series, and we can also use other means to assign values to divergent integrals. For example, if $f$ is (locally integrable and) of at most polynomial growth, then

$$I_{\varepsilon}(f) := \int_0^{+\infty} e^{-\varepsilon t}f(t)\,dt$$

exists for every $\varepsilon > 0$, and if $\lim\limits_{\varepsilon \searrow 0} I_{\varepsilon}(f)$ exists, we can assign that value to the integral. If $f$ is Lebesgue-integrable (improperly Riemann-integrable), the limit exists and equals the Lebesgue integral by the dominated convergence theorem (resp. it equals the improper Riemann integral).

For $\sin$, we find

\begin{align} \int_0^{+\infty} e^{-\varepsilon x}\sin x\,dx &= \frac{1}{2i}\int_0^{+\infty} e^{(i -\varepsilon)x} - e^{-(i+\varepsilon)x}\,dx\\ &= \frac{1}{2i}\biggl[\frac{1}{\varepsilon - i} - \frac{1}{\varepsilon + i}\biggr]\\ &= \frac{1}{\varepsilon^2 + 1}, \end{align}

so also this method leads to the value $1$.