91Ó°ÊÓ

Here, we can apply Dirichlet's test since we have the product of two functions where one is decreasing and bounded (the \(e^{-x}\) function), and the other one, the cosine function, is indefinetely differentiable and has a bounded primitive. With these conditions, the test ensures that our integral indeed converges.

We can apply the formula for integration by parts, which is \(\int u dv=uv-\int v du\). Here, we will choose \(u=e^{-x}\) and \(dv=\cos{x}dx\). Then \(du=-e^{-x}dx\) and \(v=\sin{x}\). Then the integral now becomes \[ \lim_{b \to \infty} [\int_{0}^{b} e^{-x} \sin x d x - \int_{0}^{b} e^{-x} \cos x d x] \]

Now we should perform a recursive application of integration by parts, until we reach our initial integral again. So we apply integration by parts again to the second integral in the expression:Set \(u=e^{-x}\) and \(dv=\sin{x}dx\). Then \(du=-e^{-x}dx\) and \(v=-\cos{x}\).The expression now becomes \[ \lim_{b \to \infty} [\int_{0}^{b} e^{-x} \sin x d x + \int_{0}^{b} e^{-x} \cos x d x] \]Subtract the initial integral from both sides, we get:\(\int_{0}^{b} e^{-x} \cos x d x = \frac{1}{2} \int_{0}^{b} e^{-x} \sin x d x\) This integral can be calculated fairly easily to become \( \frac{1}{2} [e^{-x}(-\cos(x) - \sin(x))]_0^b \). After taking the limit as \( b \rightarrow \infty \), the boundaries values give \( \frac{1}{2} \).

Since the result of the integral is a finite number, the initial improper integral is convergent and its value is \( \frac{1}{2} \).