Proof $\text{Si}(n) $ is convergent

Question

I am trying to prove that the sequence formed by the Si function, $\text{Si}(n) = \int_0^n \frac{\sin(u)}{u} \mathrm{d}u$, is convergent as $n\rightarrow \infty$. The only twist is the lower bound of the integral is 1 instead of zero. At first I looked at this as a sort of alternating series, and the sequence formed was that of the partial sums. I attempted to change the upper bound of the integral so that I was looking at integer multiples of $\pi$. I was then going to bound each "bump" with a rectangle with base pi, and height equal to the function value at the middle each interval. Shortly in I realized that the "bumps" were not symmetric about their centers, so I did not have a way to calculate the height of the function at these midpoints. I cannot think of another way to attack this problem. Any help appreciated.

Answer 1

Hint: Prove that $$ \int_0^{2n\pi}\frac{\sin(x)}{x}\,\mathrm{d}x $$ converges using the estimates $$ \int_0^{2\pi}\left|\,\frac{\sin(x)}{x}\,\right|\,\mathrm{d}x\le2\pi $$ and, for $n\ge2$, $$ \begin{align} \left|\,\int_{2(n-1)\pi}^{2n\pi}\frac{\sin(x)}{x}\,\mathrm{d}x\,\right| &=\left|\,\int_{2(n-1)\pi}^{2n\pi}\left(\frac{\sin(x)}{x}-\frac{\sin(x)}{2n\pi}\right)\,\mathrm{d}x\,\right|\\ &\le\int_{2(n-1)\pi}^{2n\pi}\left|\,\frac{\sin(x)}{x}-\frac{\sin(x)}{2n\pi}\,\right|\,\mathrm{d}x\\ &\le\int_{2(n-1)\pi}^{2n\pi}\left|\,\frac1{2n(n-1)\pi}\,\right|\,\mathrm{d}x\\ &=\frac1{n-1}-\frac1n \end{align} $$ Then prove that $$ \int_{2n\pi}^{2n\pi+s}\left|\,\frac{\sin(x)}{x}\,\right|\,\mathrm{d}x\le\frac1n $$ for $0\le s\le2\pi$

Answer 2

We can write the $Si$ function as

$$\begin{align} Si(n)&=\int_0^n \frac{\sin x}{x}dx\\ &=\int_0^1 \frac{\sin x}{x}dx+\int_1^n \frac{\sin x}{x}dx \end{align}$$

The first integral on the right-hand side is obviously convergent. Let's examine the second one more carefully.

Integrating by parts reveals

$$\begin{align} \lim_{n\to \infty} \int_1^n \frac{\sin x}{x}dx&=\cos(1)+\lim_{n\to \infty}\left(-\int_1^n \frac{\cos x}{x^2}dx\right) \end{align}$$

Since the integral on the right-hand side satisfies the inequality

$$\begin{align} \left|-\int_1^n \frac{\cos x}{x^2}dx\right|&\le \int_1^n \left|\frac{\cos x}{x^2}\right|dx\\\\ &\le \int_1^n \frac{1}{x^2}dx\\\\ &=1-\frac1{n} \end{align}$$

it is absolutely convergent. Thus, the original integral, $Si(n)$ converges (not absolutely), as was to be shown!

Answer 3

For $n \ge 0$, let $I_n = \int_{2\pi n}^{2\pi (n+1)}\frac{\sin(x)}{x}dx $.

Then

$\begin{array}\\ I_n &= \int_{2\pi n}^{2\pi (n+1)}\frac{\sin(x)}{x}dx\\ &= \int_{2\pi n}^{(2n+1)\pi }\frac{\sin(x)}{x}dx + \int_{(2 n+1)\pi}^{2(n+1)\pi }\frac{\sin(x)}{x}dx\\ &= \int_{0}^{\pi }\frac{\sin(x+2\pi n)}{x+2\pi n}dx + \int_{0}^{\pi }\frac{\sin(x+(2n+1)\pi)}{x+(2n+1)\pi}dx\\ &= \int_{0}^{\pi }\frac{\sin(x)}{x+2\pi n}dx + \int_{0}^{\pi }\frac{-\sin(x)}{x+(2n+1)\pi}dx\\ &= \int_{0}^{\pi }\sin(x)\left(\frac{1}{x+2\pi n}-\frac1{x+(2n+1)\pi}\right)dx\\ &= \int_{0}^{\pi }\sin(x)\frac{(x+(2n+1)\pi)-(x+2\pi n)}{(x+(2n+1)\pi)(x+2\pi n)}dx\\ &= \int_{0}^{\pi }\sin(x)\frac{\pi}{(x+(2n+1)\pi)(x+2\pi n)}dx\\ \end{array} $

so

$\begin{array}\\ |I_n| &=\big|\int_{0}^{\pi }\sin(x)\frac{\pi}{(x+(2n+1)\pi)(x+2\pi n)}dx\big|\\ &\le\big|\int_{0}^{\pi }\frac{\pi}{(x+(2n+1)\pi)(x+2\pi n)}dx\big|\\ &\le\big|\int_{0}^{\pi }\frac{\pi}{(2\pi n)^2}dx\big|\\ &= \frac1{4n^2} \end{array} $

Therefore $\sum_{n=0}^{\infty} I_n $ converges.

Note that this proof works for any bounded function $f$ such that $f(x)+f(x+\pi)=0$ such as $\cos(x)$.