I encounter the following equation $$\int_0^\infty \frac{\sin(x)}{x}dx = \lim_{\theta\to 0}\sum_{n=0}^\infty\left( \theta \cdot \frac{\sin(n\theta)}{n\theta} \right).$$
Intuitively, I think the limit on the RHS is related to Riemann sum. However, I am not able to understand the reason behind it. Any hint is appreciated.
I don't know if this answers your question or not. The LHS is $\frac{\pi}{2}$ which is well-known. Now you work on the RHS like this: \begin{eqnarray} \sum_{n=0}^\infty\frac{\sin(n\theta)}{n}&=&\Im\sum_{n=0}^\infty\frac{e^{n\theta}i}{n}\\ &=&-\Im\ln(1-e^{\theta i})\\ &=&-\Im\ln[(1-\cos(\theta))-i\sin\theta]\\ &=&-\Im\ln\left\{2\sin\left(\frac{\theta}{2}\right)\left[\sin\left(\frac{\theta}{2}\right)-i\cos\left(\frac{\theta}{2}\right)\right]\right\}\\ &=&-\Im\ln\left\{2\sin\left(\frac{\theta}{2}\right)\left[\cos\left(\frac{\theta}{2}-\frac{\pi}{2}\right)+i\sin\left(\frac{\theta}{2}-\frac{\pi}{2}\right)\right]\right\}\\ &=&-\left(\frac{\theta}{2}-\frac{\pi}{2}\right)\\ &=&\frac{\pi}{2}-\theta \end{eqnarray} and hence $$ \lim_{\theta\to 0}\sum_{n=0}^\infty\left( \theta \cdot \frac{\sin(n\theta)}{n\theta} \right)=\frac{\pi}{2}. $$