# Computing limit of integral [duplicate]

We are supposed to calculate the limit of the integral $\lim_{n\to\infty}\int_0^\pi \frac{\sin nx}{nx}dx$.

What I am currently thinking is this. Since $|\frac{\sin nx}{nx}|<\frac{1}{n}$, we can prove that this converges uniformly and can then use the dominated convergence theorem to shift the limit inside the integral, as in $\int_0^\pi \lim_{n\to\infty}\frac{\sin nx}{nx}dx = \int_0^\pi \frac{1}{n} \,dx$.

However, I don't think that this is correct. Does someone know the correct way to solve this?

## marked as duplicate by Arnaud D., Ernie060, Cesareo, Yanior Weg, Henry LeeOct 16 at 21:58

• It is true that $\left| \dfrac{\sin(nx)} n \right| \le \dfrac 1 n$ but it is not true that $\left| \dfrac{\sin(nx)} {nx} \right| \le \dfrac 1 n,$ since $x$ may be near $0$ so that $1/x$ is large. In particular $\dfrac{\sin(nx)}{nx} \to 1$ as $x\to0. \qquad$ – Michael Hardy Nov 19 '17 at 5:35
• The first part of your comment says that the same thing is both true and not true, could you clarify? – user485755 Nov 19 '17 at 5:37
• Typo fixed. $\qquad$ – Michael Hardy Nov 19 '17 at 5:38
• 1) It's not true that $\left\lvert\frac{\sin(nx)}{nx}\right \rvert<\frac1n$. 2) $\frac{\sin(nx)}{nx}$ converges pointwise but not uniformly on $(0,\pi)$. 3) $\lim_{n\to\infty}\frac{\sin(nx)}{nx}\ne \frac1n$. – user228113 Nov 19 '17 at 5:38
• Don't know if you noticed, but the answer you accepted is simply wrong. – David C. Ullrich Nov 19 '17 at 18:19

By enforcing the substitution $x=\frac{z}{n}$ we have $$\int_{0}^{\pi}\frac{\sin(nx)}{nx}\,dx = \frac{1}{n}\int_{0}^{\pi n}\frac{\sin z}{z}\,dz.$$ In order to prove that the wanted limit is zero it is enough to show that $\int_{0}^{\pi n}\frac{\sin z}{z}\,dz$ is $o(n)$.
Since $|\sin(x)|<\min(1,x)$ we have $$\left|\int_{0}^{\pi n}\frac{\sin z}{z}\right|\leq \int_{0}^{1}\frac{z}{z}\,dz + \int_{1}^{\pi n}\frac{dz}{z} = 1+\log(\pi n)$$ and we are done.
Use $|\sin u|\leq|u|$ to get $|\sin(nx)/nx|\leq 1$ for $x\in(0,\pi]$, then apply Lebesgue Dominated Convergence Theorem because $\displaystyle\int_0^{\pi}1 \, dx = \pi < \infty$.
Okay, sorry for my bad English, in fact, I was saying that the dominated function could be taken as the constant function $1$ on $[0,\pi]$. I was not saying that the limit function is $1$. And the limit function is $0$ (except for one point). So the integral is zero as well.
• How to show that $\frac{\sin nx}{nx}$ is measurable – Learnmore Nov 19 '17 at 5:39
• How is the function continuous on $[0,\pi]$ – Learnmore Nov 19 '17 at 5:42
• Yes, we can use DCT here. But $\sin(nx)/(nx)\to0$ as $n\to\\infty$, not $1$. – David C. Ullrich Nov 19 '17 at 16:17