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|Proof:
:<math>\begin{align}
&\int_{-\infty}^\infty h(\tau)\cdot x_{_T}(t - \tau)\,d\tau \\
&={} &\sum_{k=-\infty}^\infty \left[\int_{t_o+kT}^{t_o+(k+1)T} h(\tau)\cdot x_{_T}(t - \tau)\ d\tau\right] \quad t_0 \text{ is an arbitrary parameter}\\
&=\sum_{k=-\infty}^\infty \left[\int_{t_o}^{t_o+T} h(\tau + kT)\cdot \underbrace{x_{_T}(t - \tau-kT)}_{x_{_T}(t - \tau), \text{ by periodicity}}\ d\tau\right] \quad \text{substituting } \tau \rightarrow \tau+kT\\
\stackrel{\tau \rightarrow \tau+kT}{=}{}
&=\int_{t_o}^{t_o+T} \left[\sum_{k=-\infty}^\infty \left[\int_{t_o}^{t_o+T} h(\tau + kT)\cdot x_{_T}(t - \tau -kT)\right]\ d\tau\right] \\
&={} &\int_{t_o}^{t_o+T} \underbrace{\left[\sum_{k=-\infty}^\infty h(\tau + kT)\cdotright]}_{\triangleq \underbrace{x_ h_{_T}(t - \tau-kT)}_{\cdot x_{_T}(t - \tau), \text{ by periodicity}}\right]\ d\tau\\
={} &\int_{t_o}^{t_o+T} \underbrace{\left[\sum_{k=-\infty}^\infty h(\tau + kT)\right]}_{\triangleq \ h_{_T}(\tau)}\cdot x_{_T}(t - \tau)\ d\tau
\end{align}</math>
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