If we extend the domain from
to
, we can write the new Fourier coefficients
using the techniques from part (a) as
![{\displaystyle d_{n}={\frac {1}{2}}\int _{0}^{2}xe^{-2\pi {}inx/2}{\textrm {d}}x.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/aa0620866e3bfcf95b7fcc305438ad0d899a3eb4)
We could go and directly compute these integrals but we want to relate them to the coefficients
from part (a) computed on a domain
. To do this, we can use a substitution and let
. Then
. With this transformation we have
and
and so putting everything together,
![{\displaystyle d_{n}=2\int _{0}^{1}ue^{-2\pi {}inu}{\textrm {d}}u=2c_{n}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/30078d805a1eaf31f7c4a58d412a494ea8625eb2)
where we have recognized that the new integral is precisely what we computed to get the coefficients on a period
. Therefore on
the Fourier coefficients double and
![{\displaystyle d_{n}={\begin{cases}2{\frac {1}{2}}=1,\qquad {}n=0\\2{\frac {i}{2\pi {}n}}={\frac {i}{\pi {}n}},\qquad {}n\neq 0\end{cases}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/63e04a0ad431b3700ca30b93606fce04a16659db)
Therefore we have that
and the points to plot for the frequency amplitude are
.