= Solution
{c}
Set $u=x/L$ and $q=n\pi$. The required <integrals> follow from
$$
\int e^u\cos(qu)\,du=\frac{e^u(\cos(qu)+q\sin(qu))}{1+q^2},\qquad
\int e^u\sin(qu)\,du=\frac{e^u(\sin(qu)-q\cos(qu))}{1+q^2}.
$$
Consequently the <Fourier coefficients> are
$$
\boxed{a_0=2(e-1),\qquad
a_n=\frac{2[(-1)^ne-1]}{1+n^2\pi^2},\qquad
b_n=\frac{2n\pi[1-(-1)^ne]}{1+n^2\pi^2}\quad(n\geq1).}
$$
Thus the <Fourier cosine series> is
$$
\boxed{e-1+2\sum_{n=1}^\infty\frac{(-1)^ne-1}{1+n^2\pi^2}\cos\frac{n\pi x}{L},}
$$
and its sum is $e^{x/L}$ everywhere on $[0,L]$. The <Fourier sine series> is
$$
\boxed{2\sum_{n=1}^\infty\frac{n\pi[1-(-1)^ne]}{1+n^2\pi^2}\sin\frac{n\pi x}{L},}
$$
and its sum is $e^{x/L}$ on $(0,L)$ and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v-fourier-sketches.png]
{title=The sums and finite <partial sums> of the <Fourier cosine series> and <Fourier sine series> for $e^{x/L}$}
{description=The upper panels show the exact sums on $[0,L]$. The lower panels compare twelve-term <partial sums> with the <exponential function>: the <Fourier sine series> has zero endpoints and oscillations near its <periodic extension>'s jumps.}
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