Infinite summation of area of region For $j=0,1,2,\ldots,n$. Let $S_j$ be the area of region bounded by the $x$-axis and the curve $ye^x=\sin x$ for $j\pi\leq x\leq(j+1)\pi$.
The value of $\sum\limits_{j=0}^\infty S_j$ equals to
(A)$ \dfrac{e^\pi(1+e^\pi)}{2(e^\pi-1)}$
(B)$ \dfrac{1+e^\pi}{2(e^\pi-1)}$
(C) $\dfrac{1+e^\pi}{(e^\pi-1)}$
(D) $\dfrac{e^\pi(1+e^\pi)}{(e^\pi-1)}$
I tried to find out. I found $S_0,S_1,S_2,\ldots,S_\infty$. All these summations form a GP. $S_0=\frac{1}{2}(1+e^{-\pi}),\ S_1=\frac{-1}{2}e^{-\pi}(1+e^{-\pi}), S_2=\frac{-1}{2}e^{-2\pi}(1+e^{-\pi})$ and so on. Their summation comes out to be $\frac{1}{2}$. But it is not given in the choices. Where have i gone wrong?Is my approach not correct? What is correct way to solve it?
 A: I had a previous answer in which I think I made the same mistake as yours for the calculation of $S_j, j \ge 0$. What we want is the area of the region, so we can't say that:
$$S_j = \int_{j \pi}^{(j+1)\pi} \sin(x) e^{-x} dx$$
Because this assume that the curve of $y = \sin(x) e^{-x}$ is above the $x$-axis, which is not true. To correct that:
$$S_j = \int_{j\pi}^{(j+1)\pi} |\sin(x)| e^{-x}dx$$
And:
If $j$ is even, then $\sin \ge 0$ on $[j\pi, (j+1)\pi]$.
If $j$ is odd, then $\sin \le 0$ on $[j\pi, (j+1) \pi]$.
Edit:
To be sure that I'm working right, I'll do the calculations.
$$\int \sin(x) e^{-x} dx = -\frac12 (\cos(x) + \sin(x) ) e^{-x}$$
If $j$ is even:
$$S_j = \int_{j \pi}^{(j+1) \pi} \sin (x) e^{-x} dx = \frac{1 + e^{-\pi}}{2} (e^{-\pi})^j$$
(With $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$, $\forall$ $k \in \mathbb N$).
If $j$ is odd, we also get the same value of $S_j$ (by calculation or symmetry).
$$\sum_j S_j = \sum_{\text{$j$ is odd}} S_j + \sum_{\text{$j$ is even}} S_j = 2 \sum_{\text{$j$ is even}} S_j = (1 + e^{-\pi}) \sum_j (e^{-\pi})^j = \frac{1 + e^{-\pi}}{1- e^{-\pi}} = \frac{1 + e^{\pi}} {e^{\pi} - 1}$$
