I am trying to show that $\int_\pi^\infty \mid\frac{sin(x)}{x}\mid dx$ does not converge.
My approach: I noticed that $\frac{sin(k\pi)}{k\pi}=0$ for $k \in \mathbb{N}$\{0}
Now i "split up" the function like this $\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi} \mid\frac{sin(x)}{x} \mid dx$
My Problem is that I can't find a estimate (that diverges) and is less or equal to the term above.
trying to split the sum is the right idea :) $$\begin{align*} \int_{\pi}^{n\pi}\left|\frac{\sin(x)}{x}\right|\,\mathrm{d}x&=\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}\left|\frac{1}{x}\right|\cdot|\sin(x)|\,\mathrm{d}x\\ &\geq \sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{1}{(k+1)\pi}\cdot|\sin(x)|\,\mathrm{d}x\\ &\geq \sum_{k=1}^{n-1}\frac{1}{(k+1)\pi}\cdot\int_{k\pi}^{(k+1)\pi}|\sin(x)|\,\mathrm{d}x\\ &=\sum_{k=1}^{n-1}\frac{1}{(k+1)\pi}\cdot 2\xrightarrow{n\to\infty}\infty \end{align*} $$
Hint:
$$\int_{k\pi}^{(k+1)\pi} \left| \frac{\sin x}{x}\right|\,dx \ge \frac{1}{(k+1)\pi} \int_{k\pi}^{(k+1)\pi} |\sin x|\,dx.$$
