By definition of half-range Fourier cosine series: :$\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$ where for all $n \in \Z_{> 0}$: :$a_n = \displaystyle \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x$ Thus by definition of $f$: {{begin-eqn}} {{eqn | l = a_0       | r = \frac 2 \pi \int_0^\pi \map f x \rd x       | c = Cosine of Zero is One }} {{eqn | r = \frac 2 \pi \int_0^\pi \cos \lambda x \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 2 \pi \intlimits {\frac {\sin \lambda x} \lambda} 0 \pi       | c = Primitive of $\cos a x$ }} {{eqn | r = \frac 2 \pi \paren {\frac {\sin \lambda \pi} \lambda - \frac {\sin 0} \lambda}       | c =  }} {{eqn | r = \frac {2 \sin \lambda \pi} {\pi \lambda}       | c = Sine of Zero is Zero }} {{end-eqn}} {{qed|lemma}} 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 \cos n x \rd x       | c =  }} {{eqn | r = \frac 2 \pi \int_0^\pi \cos \lambda x \cos n x \rd x       | c = Definition of $f$ }} {{eqn | r = \frac 2 \pi \intlimits {\frac {\sin \paren {\lambda - n} x} {2 \paren {\lambda - n} } + \frac {\sin \paren {\lambda + n} x} {2 \paren {\lambda + n} } } 0 \pi       | c = Primitive of $\cos \lambda x \cos n x$ }} {{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 \lambda {\lambda^2 - n^2}       | c = Difference of Two Squares }} {{end-eqn}} {{qed|lemma}} Finally: {{begin-eqn}} {{eqn | l = \map f x       | o = \sim       | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x       | c =  }} {{eqn | r = \frac 1 2 \frac {2 \sin \lambda \pi} {\pi \lambda} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \sin \lambda \pi} \pi \frac \lambda {\lambda^2 - n^2} \cos n x       | c = substituting for $a_0$ and $a_n$ from above }} {{eqn | r = \frac {2 \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac \lambda {\lambda^2 - n^2} \cos n x}       | c = simplifying }} {{eqn | r = \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }       | c = further manipulation }} {{end-eqn}} {{qed}}