# Aligning long equations to the left, when split is used

I haven't found any clear indication on how to align the given eqn to the left. It goes automatically to the right, and I have tried {ll} next to split, but to no use:

$$\begin{split} \alpha_k=\frac{1}{\pi}\sum_{i=0}^n\eta_i\int_{\xi_i}^{\xi_{i+1}} \cos kt \text{d}t=-\frac{1}{\pi}\sum_{i=0}^n\eta_i\bigg( \frac{\sin (k\xi_{i+1})}{k}-\frac{\sin (k\xi_{i})}{k}\bigg)=\\ \frac{1}{\pi}\sum_{i=0}^n\frac{\eta_i}{k}\big[\sin(\xi_i)-\sin(\xi_{i+1})\big]\rightarrow \text{let \ }\delta=\xi_i\\ \alpha_k=\frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k}\big[\sin\delta-\sin\gamma\big]\rightarrow \\ \alpha_k=\frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k} \end{split}$$


any hints on how to align this to the left?

Thanks

UPDATE, with the below given suggestion I get:

I changed back to the original version and added some more text, but the format is still the same thus getting the same problem:

$$\begin{split} \alpha_k=\frac{1}{\pi}\sum_{i=0}^n\eta_i\int_{\xi_i}^{\xi_{i+1}} \cos kt \text{d}t=-\frac{1}{\pi}\sum_{i=0}^n\eta_i\bigg( \frac{\sin (k\xi_{i+1})}{k}-\frac{\sin (k\xi_{i})}{k}\bigg)=\\ \frac{1}{\pi}\sum_{i=0}^n\frac{\eta_i}{k}\big[\sin(k\xi_i)-\sin(k\xi_{i+1})\big]\rightarrow \text{let \ }\delta=\xi_i,\text{and \ }\gamma=\xi{_i+1}\\ \alpha_k=\frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k}\big[\sin k\delta-\sin k\gamma\big]\rightarrow \text{apply the relation} \sin(\alpha+\beta)-\sin(\alpha-\beta)=2\sin\beta\cos\alpha\\ \text{let\ }\alpha=\frac{\delta+\gamma}{2}\text{\ and \ } \frac{\delta-\gamma}{2}=2\sin\frac{\delta+\gamma}{2}\cos\frac{\delta-\gamma}{2} \text{\, then obtain:}\\ \alpha_k=\frac{2}{\pi}\sum_{i=1}^n\frac{\eta_i}{k}\sin k\bigg(\frac{\xi_{i+1}-\xi_i}{2}\bigg)\cos k\bigg(\frac{\xi_i+\xi_{i+1}}{2}\bigg) \end{split}$$


You want to set an alignment point.

\documentclass{article}
\usepackage{amsmath}

\newcommand{\diff}{\mathop{}\!\mathrm{d}}% or just d, which I prefer

\begin{document}

$$\begin{split} \alpha_k=\frac{1}{\pi}\sum_{i=0}^n\eta_i\int_{\xi_i}^{\xi_{i+1}} \cos kt \text{d}t=-\frac{1}{\pi}\sum_{i=0}^n\eta_i\bigg( \frac{\sin (k\xi_{i+1})}{k}-\frac{\sin (k\xi_{i})}{k}\bigg)=\\ \frac{1}{\pi}\sum_{i=0}^n\frac{\eta_i}{k}\big[\sin(\xi_i)-\sin(\xi_{i+1})\big]\rightarrow \text{let \ }\delta=\xi_i\\ \alpha_k=\frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k}\big[\sin\delta-\sin\gamma\big]\rightarrow \\ \alpha_k=\frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k} \end{split}$$
$$\begin{split} \alpha_k &= \frac{1}{\pi}\sum_{i=0}^n\eta_i\int_{\xi_i}^{\xi_{i+1}}\cos kt \diff t = -\frac{1}{\pi}\sum_{i=0}^n\eta_i \biggl( \frac{\sin (k\xi_{i+1})}{k}-\frac{\sin (k\xi_{i})}{k} \biggr) \\ &= \frac{1}{\pi}\sum_{i=0}^n\frac{\eta_i}{k}\bigl[\sin(\xi_i)-\sin(\xi_{i+1})\bigr] \rightarrow \text{let \ } \delta=\xi_i \\ \alpha_k &= \frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k}\bigl[\sin\delta-\sin\gamma\bigr] \rightarrow \\ \alpha_k &= \frac{1}{\pi}\sum_{i=1}^n\frac{\eta_i}{k} \end{split}$$

• @vqngs Apparently you missed some &. Commented Mar 30, 2023 at 9:51