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

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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

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## 2 Answers

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.

You should be able to compute that area and find the best possible bound.