Theorem: Half-Range Fourier Sine Series/Sine of Non-Integer Multiple of x over 0 to Pi

Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = \sin \lambda x$
Then its half-range Fourier sine series can be expressed as:
{{begin-eqn}}
{{eqn | l = \map f x
      | o = \sim
      | r = \frac {2 \sin \lambda \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n x} {\lambda^2 - n^2} }
      | c = 
}}
{{eqn | r = \frac {2 \sin \lambda \pi} \pi \paren {-\frac {\sin x} {\lambda^2 - 1} + \frac {2 \sin 2 x} {\lambda^2 - 4} - \frac {3 \sin 3 x} {\lambda^2 - 9}  + \frac {4 \sin 4 x} {\lambda^2 - 16} - \dotsb}
      | c = 
}}
{{end-eqn}}

Proof:
By definition of half-range Fourier sine series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \sin n x$
where for all $n \in \Z_{> 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$
Because $\lambda \notin \Z$ we have that $\lambda \ne n$ for all $n$.
Thus for $n > 0$:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x
      | c = 
}}
{{eqn | r = \frac 2 \pi \int_0^\pi \sin \lambda x \sin n x \rd x
      | c = Definition of $f$
}}
{{eqn | r = \frac 2 \pi \intlimits {\frac {\map \sin {\lambda - n} x} {2 \paren {\lambda - n} } - \frac {\map \sin {\lambda + n} x} {2 \paren {\lambda + n} } } 0 \pi
      | c = Primitive of $\sin \lambda x \sin n x$
}}
{{eqn | r = \frac 2 \pi \paren {\paren {\frac {\sin \paren {\lambda - n} \pi} {2 \paren {\lambda - n} } - \frac {\sin \paren {\lambda + n} \pi} {2 \paren {\lambda + n} } } - \paren {\frac {\sin 0} {2 \paren {\lambda - n} } - \frac {\sin 0} {2 \paren {\lambda + n} } } }
      | c = 
}}
{{eqn | r = \frac 1 \pi \paren {\frac {\sin \paren {\lambda - n} \pi} {\lambda - n} - \frac {\sin \paren {\lambda + n} \pi} {\lambda + n} }
      | c = Sine of Multiple of Pi and simplification
}}
{{eqn | r = \frac 1 \pi \paren {\frac {\sin \lambda \pi \cos n \pi - \cos \lambda \pi \sin n \pi} {\lambda - n} - \frac {\sin \lambda \pi \cos n \pi + \cos \lambda \pi \sin n \pi} {\lambda + n} }
      | c = Sine of Sum
}}
{{eqn | r = \frac {\sin \lambda \pi \cos n \pi} \pi \paren {\frac 1 {\lambda - n} - \frac 1 {\lambda + n} }
      | c = Sine of Multiple of Pi and simplification
}}
{{eqn | r = \frac {\paren {-1}^n \sin \lambda \pi} \pi \frac {\lambda + n - \lambda + n} {\paren {\lambda - n} \paren {\lambda + n} }
      | c = Cosine of Multiple of Pi and manipulation
}}
{{eqn | r = \paren {-1}^n \frac {2 \sin \lambda \pi} \pi \frac n {\lambda^2 - n^2}
      | c = Difference of Two Squares
}}
{{end-eqn}}
{{qed|lemma}}
Finally:
{{begin-eqn}}
{{eqn | l = \map f x
      | o = \sim
      | r = \sum_{n \mathop = 1}^\infty b_n \sin n x
      | c = 
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \sin \pi} \pi \frac n {\lambda^2 - n^2} \sin n x
      | c = substituting for $b_n$ from above
}}
{{eqn | r = \frac {2 \sin \lambda \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac n {\lambda^2 - n^2} \sin n x}
      | c = simplifying
}}
{{eqn | r = \frac {2 \sin \lambda \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n x} {\lambda^2 - n^2} }
      | c = further manipulation
}}
{{end-eqn}}
{{qed}}