Area Enclosed Between Exponential Curves and Intersection Lines

To find the required value, we first determine the points of intersection of the curves $C_1: y = e^{-x}$ and $C_2: y = e^{-x}(\sin x + \cos x)$ in the domain $x \in [0, 10\pi]$.

Step 1: Points of Intersection

Equating the two curves: $e^{-x} = e^{-x}(\sin x + \cos x)$

Since $e^{-x} \neq 0$ for all real $x$, we can divide by $e^{-x}$: $\sin x + \cos x = 1$

Dividing by $\sqrt{2}$: $\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}}$ $\sin\left(x + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)$

This gives the general solution: $x + \frac{\pi}{4} = 2k\pi + \frac{\pi}{4} \quad \text{or} \quad x + \frac{\pi}{4} = 2k\pi + \frac{3\pi}{4} \quad \text{for } k \in \mathbb{Z}$ $x = 2k\pi \quad \text{or} \quad x = 2k\pi + \frac{\pi}{2}$

Arranging the $x$-coordinates of the intersection points in $[0, 10\pi]$ in ascending order: $\alpha_1 = 0, \quad \alpha_2 = \frac{\pi}{2}, \quad \alpha_3 = 2\pi, \quad \alpha_4 = \frac{5\pi}{2}, \dots$

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Step 2: Set Up the Area Integral

The area $\beta$ enclosed between $C_1$, $C_2$, and the lines $x = \alpha_1 = 0$ and $x = \alpha_4 = \frac{5\pi}{2}$ is given by: $\beta = \int_{0}^{\frac{5\pi}{2}} |C_1(x) - C_2(x)| \, dx = \int_{0}^{\frac{5\pi}{2}} e^{-x} |1 - \sin x - \cos x| \, dx$

We split the domain of integration according to the sign of $(1 - \sin x - \cos x)$:

1. For $x \in \left[0, \frac{\pi}{2}\right]$: $\sin x + \cos x \ge 1 \implies |1 - \sin x - \cos x| = \sin x + \cos x - 1$
2. For $x \in \left[\frac{\pi}{2}, 2\pi\right]$: $\sin x + \cos x \le 1 \implies |1 - \sin x - \cos x| = 1 - \sin x - \cos x$
3. For $x \in \left[2\pi, \frac{5\pi}{2}\right]$: $\sin x + \cos x \ge 1 \implies |1 - \sin x - \cos x| = \sin x + \cos x - 1$

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Step 3: Evaluate the Indefinite Integral

Let $J(x) = \int e^{-x} (\sin x + \cos x - 1) \, dx$.

Using standard exponential-trigonometric integrals: $\int e^{-x} \sin x \, dx = \frac{e^{-x}}{2}(-\sin x - \cos x)$ $\int e^{-x} \cos x \, dx = \frac{e^{-x}}{2}(\sin x - \cos x)$

Adding these two: $\int e^{-x} (\sin x + \cos x) \, dx = -e^{-x} \cos x$

Thus: $J(x) = -e^{-x} \cos x - (-e^{-x}) = e^{-x}(1 - \cos x)$

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Step 4: Calculate Area $\beta$

$\beta = I_1 + I_2 + I_3$

1. First region: $I_1 = \int_0^{\frac{\pi}{2}} e^{-x}(\sin x + \cos x - 1) \, dx = [J(x)]_0^{\frac{\pi}{2}} = e^{-\frac{\pi}{2}}(1 - 0) - e^0(1 - 1) = e^{-\frac{\pi}{2}}$

2. Second region: $I_2 = \int_{\frac{\pi}{2}}^{2\pi} e^{-x}(1 - \sin x - \cos x) \, dx = -[J(x)]_{\frac{\pi}{2}}^{2\pi} = - \left( e^{-2\pi}(1 - 1) - e^{-\frac{\pi}{2}}(1 - 0) \right) = e^{-\frac{\pi}{2}}$

3. Third region: $I_3 = \int_{2\pi}^{\frac{5\pi}{2}} e^{-x}(\sin x + \cos x - 1) \, dx = [J(x)]_{2\pi}^{\frac{5\pi}{2}} = e^{-\frac{5\pi}{2}}(1 - 0) - e^{-2\pi}(1 - 1) = e^{-\frac{5\pi}{2}}$

Summing these up: $\beta = e^{-\frac{\pi}{2}} + e^{-\frac{\pi}{2}} + e^{-\frac{5\pi}{2}} = 2e^{-\frac{\pi}{2}} + e^{-\frac{5\pi}{2}}$

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Step 5: Evaluate the Final Expression

Substituting $\beta$ into the given expression: $-\frac{1}{\pi} \log_e \left( \beta - 2 e^{-\frac{\pi}{2}} \right) = -\frac{1}{\pi} \log_e \left( 2e^{-\frac{\pi}{2}} + e^{-\frac{5\pi}{2}} - 2 e^{-\frac{\pi}{2}} \right)$

$= -\frac{1}{\pi} \log_e \left( e^{-\frac{5\pi}{2}} \right) = -\frac{1}{\pi} \left( -\frac{5\pi}{2} \right) = \frac{5}{2} = 2.5$