I am trying to calculate
$$Z = \int\limits_{\phi(\beta) = \phi(0) =0} D \phi\ e^{-\frac{1}{2} \int_0^{\beta} d\tau \dot{\phi}^2}$$
without transforming it to the Matsubara frequency space, I can show that $Z = \sqrt{\frac{1}{2\pi \beta}}$. However, I have a problem in obtaining the same result in the Matsubara frequency space:
\begin{equation}
\phi (\tau) = \frac{1}{\sqrt{\beta}} \left( \sum_{n} \phi_n \ e^{i\omega_n\tau} \right),
\end{equation}
with $\sum_n \phi_n =0, \omega_n = \frac{2\pi n}{\beta}$. And
\begin{equation}
Z = \int \prod_n D\phi_n\ \delta\left(\sum_n \phi_n\right)\ e^{-\frac{1}{2} \sum_n \phi_n \phi_{-n} \omega_n^2 }
\end{equation}
which, I think, vanishes.
I guess the problem lies in the measure. Any comments?
Info: I write the Schulman's derivation in imaginary time here.
\begin{eqnarray}
Z &=& \int\limits_{\phi(0) =\phi(\beta) = 0} D\phi(\tau) e^{-\frac{1}{2}\int_0^{\beta}d\tau\dot{\phi}^2}\\
&=& \text{lim}_{N \rightarrow \infty} (\frac{1}{2\pi \epsilon})^{(N+1)/2} \int d\phi_1
\dots d\phi_N e^{-\frac{1}{2\epsilon} \sum_{i =0}^N (\phi_{i+1} -\phi_i)^2}
\end{eqnarray}
Then, we can use the identity
\begin{equation}
\int_{-\infty}^{\infty} du \sqrt{\frac{a}{\pi}} e^{-a(x-u)^2}\sqrt{\frac{b}{\pi}} e^{-b(u -y)^2} = \sqrt{\frac{ab}{\pi(a+b)}} e^{-\frac{ab}{a+b}(x-y)^2}
\end{equation}
to evaluate the sum to be
\begin{equation}
Z = \sqrt{\frac{1}{2\pi \beta}}.
\end{equation}
Best Answer
I) The Euclidean path integral with $\hbar=1$ reads
$$ Z~=~\int_{DBC} \!{\cal D}x ~e^{-S},\tag{1}$$
with Dirichlet boundary conditions (DBC)
$$ x(0)~=~0~=~x(T).\tag{2}$$
We expand the real periodic variable $x\in\mathbb{R}$ in Fourier series$^1$
$$ \begin{align} x(t) ~=~&\frac{a_0}{2}+\sum_{n\in\mathbb{N}} \left\{a_n \cos(\omega_n t) + b_n \sin(\omega_n t)\right\}\cr ~=~& \sum_{n\in\mathbb{Z}} c_n e^{i\omega_n t}, \cr \omega_n~:=~&\frac{2\pi n}{T},\qquad a_n,b_n~\in~ \mathbb{R},\cr c_n~\in~& \mathbb{C}, \qquad c_n^{\ast}~=~c_{-n}. \end{align} \tag{3} $$
The DBC (2) becomes
$$ \sum_{n\in\mathbb{Z}} c_n~=~0\qquad\Leftrightarrow\qquad c_0 ~=~ -2{\rm Re}\sum_{n\in\mathbb{N}}c_n .\tag{4}$$
The action for a free non-relativistic point particle with mass $m=1$ reads:
$$ S~=~\frac{1}{2} \int_0^T \!dt~ \dot{x}^2~=~T\sum_{n\in\mathbb{N}}\omega_n^2 |c_n|^2 . \tag{5} $$
II) We know that the proper normalization of the path integral (1) is
$$ Z~=~\frac{1}{\sqrt{2 \pi T}}.\tag{6} $$
This can e.g. be deduced (without introducing fudge factors!) from the (semi)group property of Feynman path integrals, cf. this Phys.SE post and links therein. Up till now we have basically just restated what OP wrote in his question.
III) Now we would like to repeat the same calculation using Fourier series, i.e. work with the Matsubara frequencies. In this answer, we will not explore the (semi)group property, but just do a quick and dirty calculation using various fudge factors, and see what we get. Since this is homework, the explanation will be somewhat brief.
To make heuristic sense of the path integral (1), we will use the following zeta function regularization rules:
$$ \prod_{n\in\mathbb{N}} a ~=~\frac{1}{\sqrt{a}} \quad\text{and}\quad \prod_{n\in\mathbb{N}} n ~=~\sqrt{2\pi}, \tag{7}$$
stemming from the zeta function values
$$ \zeta(0)~=~-\frac{1}{2} \quad\text{and}\quad \zeta^{\prime}(0)~=~-\ln\sqrt{2\pi} , \tag{8} $$ respectively. Now let the path integral measure be
$$ \begin{align} {\cal D}x~:=~&\delta\left(B\sum_{n\in\mathbb{Z}} c_n\right) A\mathrm{d} c_0 \prod_{n\in\mathbb{N}} A^2 \mathrm{d}^2c_n \cr ~\stackrel{(7)}{=}~&\frac{1}{B}\delta\left(\sum_{n\in\mathbb{Z}} c_n\right) \mathrm{d} c_0 \prod_{n\in\mathbb{N}} \mathrm{d}^2c_n , \end{align}\tag{9} $$
where $A$, $B$ are fudge factors. Interestingly, eq. (9) is independent of the $A$-fudge factor! After performing the delta function integration and the Gaussian integrals, we find
$$ \begin{align} Z~=~&\frac{1}{B} \prod_{n\in\mathbb{N}} \frac{\pi}{T\omega_n^2}\cr ~=~&\frac{1}{B} \prod_{n\in\mathbb{N}} \frac{T}{4\pi n^2}\cr ~\stackrel{(7)}{=}~&\frac{1}{B\sqrt{\pi T}}. \end{align}\tag{10}$$
Apparently we should chose the fudge factor $B=\sqrt{2}$ in order to achieve the correct normalization (6).
--
$^1$ Note that the sine (cosine) modes (3) correspond trivially (non-trivially) to the even (odd) modes in my Phys.SE answer here, respectively.