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This question already has an answer here:

I am asked to calculate $\displaystyle \lim_{n\rightarrow\infty} \int_n^{n+1}\frac{\sin x} x \, dx$.

Before letting the $\lim$ to confuse me I used integration by parts, but it didn't get me far.

Any hint?

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marked as duplicate by Arnaud D., Yanior Weg, RRL calculus 2 days ago

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    $\begingroup$ Start with $|\sin x|\le1$. $\endgroup$ – Gerry Myerson Oct 3 '15 at 2:51
  • $\begingroup$ And with that, look at the limit of the absolute value of your integral $\endgroup$ – Alan Oct 3 '15 at 2:51
  • $\begingroup$ So... I get the absolute value of the integral is between 0 and $\ln(\frac{n+1}n)$ which goes to 0 when n goes to $\infty$, so by the Sandwich theorem we get the limit of the integral is zero? $\endgroup$ – Whyka Oct 3 '15 at 3:08
  • $\begingroup$ You don't even need to go that far - just note that $|\sin x/x|<1/n$ on $(1/n,1/(n+1))$. $\endgroup$ – Thomas Andrews Oct 3 '15 at 3:14
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There are a couple of answers already but I'll provide the simple one that was alluded to in comments. For all $x\in[n,n+1]$, $|\frac{\sin x}x|\le\frac1n$ and so $|\int_n^{n+1}\frac{\sin x}x\,\mathrm dx|\le\frac1n\to0$.

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If you permit special functions, $\int \frac{\sin x}{x}dx = \text{Si}(x)$. Further, there is the well known identity $\displaystyle\lim_{x \to \infty} \text{Si}(x) = \frac{\pi}{2}$. From this, we get $$\lim_{n\rightarrow\infty} \int_n^{n+1}\frac{\sin x} x \, dx = \lim_{n \to \infty} [\text{Si}(n+1)-\text{Si}(n)]$$ We can see that the growth of $n$ in the first sine integral overtakes the addition by 1, so we get $$= \lim_{n \to \infty} [\text{Si}(n)-\text{Si}(n)] = \color{red}{0}$$

For proofs of the above identity, see here. While the proofs are not that hard to grasp, they are often long in length and are great in number; as such I leave the proof-checking to you.

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  • $\begingroup$ The "well known identity" you quote implies the result immediately, but requires a fair bit more to prove. For this reason I don't think using the function $\text{Si}(x)$ is appropriate here. $\endgroup$ – Jason Oct 3 '15 at 4:41
  • $\begingroup$ @Jason I have encountered the identity often, and I thus consider it fairly "well known", especially compared to other identities of trig integrals. Nevertheless, you can probably show this integral without needed show that Si$(\infty) = \frac{\pi}{2}$ as long as you can show that the integral converges to a finite value. I'll agree your solution is probably better as an elementary answer however, and is simpler to show a full proof of. $\endgroup$ – Brevan Ellefsen Oct 3 '15 at 4:48
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Let $y = x-n \Rightarrow dx = dy \Rightarrow \displaystyle \int_{n}^{n+1} \dfrac{\sin x}{x} dx = \displaystyle \int_{0}^1 \dfrac{\sin(y+n)}{y+n}dy$. We treat this as Lebesgue integral, and the function $\dfrac{\sin(y+n)}{y+n}$ is dominated by $1$ which is integrable over $[0,1]$ because $\sin(y+n) \leq 1 \leq y+n, y \geq 0 \Rightarrow \displaystyle \lim_{n\to \infty} \int_{n}^{n+1}\dfrac{\sin x}{x}dx = \displaystyle \lim_{n\to \infty} \int_{0}^1 \dfrac{\sin(y+n)}{y+n}dy=\displaystyle \int_{0}^1 \lim_{n\to \infty} \dfrac{\sin(y+n)}{y+n}dy=\displaystyle \int_{0}^1 0 dy = 0$

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