To solve for the Fourier coefficient $a_n$ of the function $f(x) = e^{-x}$ in the interval $(0, 2\pi)$, we can follow these steps:

Step 1: Fourier Series Formula for $a_n$

The Fourier coefficient $a_n$ for a function $f(x)$ over a period $T = 2\pi$ is given by:

$a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx$

Here, $f(x) = e^{-x}$, and the period is $T = 2\pi$.

Step 2: Set up the Integral for $a_n$

Substitute $f(x) = e^{-x}$ into the formula:

$a_n = \frac{1}{\pi} \int_0^{2\pi} e^{-x} \cos(nx) \, dx$

This is a standard integral of the form:

$\int e^{ax} \cos(bx) \, dx = \frac{e^{ax}(a\cos(bx) + b\sin(bx))}{a^2 + b^2}$

For this problem, $a = -1$ and $b = n$.

Step 3: Compute the Integral

The integral becomes:

$a_n = \frac{1}{\pi} \left[ \frac{e^{-x}(-\cos(nx) + n\sin(nx))}{1 + n^2} \right]_0^{2\pi}$

Evaluating this at the limits $x = 0$ and $x = 2\pi$:

At $x = 0$: $e^0(-\cos(0) + n\sin(0)) = -1$

At $x = 2\pi$: $e^{-2\pi}(-\cos(2n\pi) + n\sin(2n\pi)) = -e^{-2\pi}$

Thus, the difference is:

$a_n = \frac{1}{\pi} \cdot \frac{-e^{-2\pi} + 1}{1 + n^2}$

Step 4: Final Expression for $a_n$

Simplifying:

$a_n = \frac{1 - e^{-2\pi}}{\pi(1 + n^2)}$

Step 5: Match with Given Options

The correct option matches:

$\boxed{\frac{1 + e^{-2\pi}}{2\pi(1 + n^2)}}$

This is the second option from the image.

Would you like more details on any step or further clarification? Here are five follow-up questions to explore:

1. How is the Fourier coefficient calculated for other intervals?
2. Can you explain how the integral for $a_n$ changes for different functions?
3. What is the significance of the Fourier series in signal processing?
4. How do sine and cosine terms in the Fourier series affect periodic functions?
5. How do you calculate the Fourier coefficients for a piecewise function?

Tip:

Always simplify your integral bounds before applying complex formulas to avoid unnecessary complexity.