# Upper bound of $\int_{-\infty}^{\infty}\sin(x)dx$. [duplicate]

From another question, improprer integral $$\int_{-\infty}^{\infty}\sin(x)dx$$ is not $$\lim_{a \to \infty} \int_{-a}^a \sin x \, d x$$ but $$\lim_{a \to \infty}\lim_{b \to \infty}\int_{-a}^b \sin x \, d x.$$ This is also valid for $$\lim_{a \to \infty}\lim_{b \to \infty}\int_{-a}^b \cos x \, d x.$$ For this reason there is no limit. However, is it possible to obtain some estimations (from above) of the following? $$\int_{-a}^b \cos x \, d x, \ \ \int_{-a}^b \sin x \, d x$$ $$\left|\int_{-a}^b \cos x \, d x\right|, \ \ \left|\int_{-a}^b \sin x \, d x\right|$$ I think that the upper bound of these integrals is $4$.

## marked as duplicate by Elliot G, N. F. Taussig, colormegone, 6005, user223391 Jan 14 '16 at 5:51

Simply use $$\bigg|\int_a^b \cos (x)dx \bigg|=|\sin (b)-\sin (a)|\leq 2.$$ By choosing $b = \pi/2$ and $a=-\pi/2$, one can see that this bound cannot be improved. The calculation for $\sin$ is analogous.
The absolute value of the definite integral of $\sin$ or $\cos$ over any interval is at most the area under one arch of the sine curve. Any larger interval of integration can only introduce cancellation.