Proof $\text{Si}(n) $ is convergent 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.
 A: 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!
A: 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 $0\le s\le2\pi$
A: 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 $\cos(x)$.
