Aligning long equations to the left, when split is used

Question

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{equation}
\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}
\end{equation}

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{equation}
\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}
\end{equation}

Answer 1

You want to set an alignment point.

\documentclass{article}
\usepackage{amsmath}

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

\begin{document}

Your attempt
\begin{equation}
\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}
\end{equation}

With alignment at the equals sign and with fixes
\begin{equation}
\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}
\end{equation}

\end{document}