The problem asks for the sum of the squared magnitudes of the Fourier series coefficients (\(|c_n|^2\)) for a given periodic function \(f(t)\). The function has a period \(T = 2\pi\), and in the interval \((-\pi, \pi)\), it is defined as \(f(t) = e^{-t}\). The Fourier series is given in the complex exponential form \(f(t) = \sum\nolimits_{ - \infty }^\infty {{c_n}{e^{{\mathop{\rm int}} }}}\).

To find the sum \({\sum\nolimits_{ - \infty }^\infty {\left| {{c_n}} \right|} ^2}\), we can use Parseval's Theorem (also known as Parseval's Identity) for Fourier series. Parseval's Theorem relates the sum of the squares of the magnitudes of the Fourier coefficients to the integral of the square of the magnitude of the function over one period.

Parseval's Theorem for Fourier Series

For a periodic function \(f(t)\) with period \(T\) and complex Fourier series coefficients \(c_n\), Parseval's Theorem states:

\(\frac{1}{T}\int_{t_0}^{t_0+T} |f(t)|^2 dt = \sum_{n=-\infty}^{\infty} |c_n|^2\)

where \(t_0\) is any convenient point to start the integration over one period \(T\).

Applying Parseval's Theorem

In this problem, the period is \(T = 2\pi\), and the function is given as \(f(t) = e^{-t}\) in the interval \((-\pi, \pi)\). We can choose the integration interval from \(-\pi\) to \(\pi\). So, according to Parseval's Theorem, the sum we need to find is:

\(\sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(t)|^2 dt\)

Substitute \(f(t) = e^{-t}\) into the integral:

\(\sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |e^{-t}|^2 dt\)

Since \(e^{-t}\) is a real function, \(|e^{-t}| = e^{-t}\). The square of the magnitude is \((e^{-t})^2 = e^{-2t}\).

\(\sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-2t} dt\)

Evaluating the Integral

Now, we need to evaluate the definite integral \(\int_{-\pi}^{\pi} e^{-2t} dt\).

The integral of \(e^{-2t}\) with respect to \(t\) is \(-\frac{1}{2} e^{-2t}\). So, the definite integral is:

\(\int_{-\pi}^{\pi} e^{-2t} dt = \left[ -\frac{1}{2} e^{-2t} \right]_{-\pi}^{\pi}\)

\(= -\frac{1}{2} \left( e^{-2(\pi)} - e^{-2(-\pi)} \right)\)

\(= -\frac{1}{2} \left( e^{-2\pi} - e^{2\pi} \right)\)

\(= \frac{1}{2} \left( e^{2\pi} - e^{-2\pi} \right)\)

Final Calculation

Now substitute the result of the integral back into the Parseval's Theorem equation:

\(\sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \cdot \frac{1}{2} \left( e^{2\pi} - e^{-2\pi} \right)\)

\(= \frac{1}{2\pi} \cdot \frac{e^{2\pi} - e^{-2\pi}}{2}\)

Recall the definition of the hyperbolic sine function: \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). Using this definition with \(x = 2\pi\), we have \(\sinh(2\pi) = \frac{e^{2\pi} - e^{-2\pi}}{2}\).

Therefore, the sum is:

\(\sum_{n=-\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \sinh(2\pi)\)

This matches one of the given options.