If `f(theta) = [cos i theta - i sin i theta],i=sqrt(-1)` then the value of `int_(-3)^(-1) [f (theta) + f(2 theta) + f(3 theta) + .....oo] d theta `=

Options

A. `"log"(1 + e + e^2]`
B. `"log" (1 + e^(-1) + e^2]`
C. `"log" (1 + e + e^(-1))`
D. `"log" (1 + e^(-1) + e^(-2))`

Answer

To solve the given integral, we start by analyzing the function \( f(\theta) \) defined as:

\[ f(\theta) = \cos(i\theta) - i\sin(i\theta) \]

Using the properties of exponential functions, we can express \( f(\theta) \) in terms of the exponential function:

\[ f(\theta) = \cos(i\theta) - i\sin(i\theta) = e^{i(i\theta)} = e^{-\theta} \]

Now, we need to compute the integral:

\[ \int_{-3}^{-1} [f(\theta) + f(2\theta) + f(3\theta) + \ldots] \, d\theta \]

Substituting for \( f(n\theta) \):

\[ f(n\theta) = e^{-n\theta} \]

Thus, we have:

\[ f(\theta) + f(2\theta) + f(3\theta) + \ldots = e^{-\theta} + e^{-2\theta} + e^{-3\theta} + \ldots \]

This series is a geometric series with the first term \( a = e^{-\theta} \) and common ratio \( r = e^{-\theta} \). The sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} = \frac{e^{-\theta}}{1 - e^{-\theta}} \quad \text{for } |r| < 1 \]

Now, we can rewrite the integral:

\[ \int_{-3}^{-1} \frac{e^{-\theta}}{1 - e^{-\theta}} \, d\theta \]

Next, we simplify the integral. We can use the substitution \( u = e^{-\theta} \), which gives \( du = -e^{-\theta} d\theta \) or \( d\theta = -\frac{du}{u} \). The limits change as follows:

- When \( \theta = -3 \), \( u = e^{3} \)
- When \( \theta = -1 \), \( u = e^{1} \)

Thus, the integral becomes:

\[ -\int_{e^{3}}^{e^{1}} \frac{u}{1 - u} \left(-\frac{du}{u}\right) = \int_{e^{3}}^{e^{1}} \frac{1}{1 - u} \, du \]

Now we can compute this integral:

\[ \int \frac{1}{1 - u} \, du = -\log|1 - u| \]

Evaluating the definite integral:

\[ \left[-\log|1 - u|\right]_{e^{3}}^{e^{1}} = -\log|1 - e| + \log|1 - e^{3}| \]

This simplifies to:

\[ \log\left(\frac{|1 - e^{3}|}{|1 - e|}\right) \]

Thus, the final result of the integral is:

\[ \log\left(\frac{1 - e^{3}}{1 - e}\right) \]