The curves \(C_1\) and \(C_2\) are defined by \[y=e^{-x}\qquad (x>0)\qquad \text{and}\qquad y=e^{-x}\sin x \qquad (x>0),\] respectively. Sketch roughly \(C_1\) and \(C_2\) on the same diagram.
When does \(e^{-x}\sin x\) equal \(e^{-x}\)?
Let \(x_n\) denote the \(x\)-coordinate of the \(n\)th point of contact between the two curves, where \(0<x_1< x_2<\dotsb\), and let \(A_n\) denote the area of the region enclosed by the two curves between \(x_n\) and \(x_{n+1}\). Show that \[A_n=\frac{1}{2}\left(e^{2\pi}-1\right)e^{-(4n+1)\pi/2}\]
Where do the points \(x_n\) occur?
If we look at the area that forms \(A_1\) on your graph, how can we write that in terms of integrals? What about \(A_n\)?
How could we find \(I = \int e^{-x}\sin x \,dx\)? Could we use integration by parts twice?
…and hence find \(\displaystyle\sum\limits_{n=1}^{\infty} A_n\).
Could we write out the first few terms of the infinite sum \(\sum\limits_{n=1}^{\infty} A_n\)?
What’s the expression for the infinite sum in this case? Can we spot a geometric series anywhere? What’s the sum to infinity for this?
