# Area under Integral.

This question is quite hard for me.

## Homework Statement

Find the area between the curve $$y=e^{-x}\left|sinx\right|$$ and the straight line y=0 for $$x\geq 0$$

## The Attempt at a Solution

$$\int_{0}^{\infty}e^{-x}\left|sinx\right|dx=-e^{-x}\left|sinx\right|+\int_{0}^{\infty}e^{-x}\left|cosx\right|dx$$
$$=-e^{-x}\left|sinx\right|-e^{-x}\left|cosx\right|-\int_{0}^{\infty}e^{-x}\left|sinx\right|dx$$
$$\int_{0}^{\infty}e^{-x}\left|sinx\right|dx=\frac{-e^{-x}\left|sinx\right|-e^{-x}\left|cosx\right|}{2}$$

Related Calculus and Beyond Homework Help News on Phys.org
NateTG
Homework Helper
Because sine and cosine cross the x-axis, you can't just take the absolute value thorough the integral like that.

Than ,actually, i should divide it to pieces.
$$\int_{0}^{\infty}e^{-x}\left|sinx\right|dx=\int_{0}^{\pi}e^{-x}sinxdx+\int_{\pi}^{2\pi}e^{-x}(-sinx)dx+\cdots$$
$$=\sum_{k=0}^{\infty}(-1)^k\int_{k\pi}^{(k+1)\pi}e^{-x}sinxdx$$

Last edited:
Is it equals to
$$(-1)^k\sum_{k=1}^{\infty}\frac{e^{-k\pi}}{2}$$ ?

Oh,sorry
$$\frac{1}{2}+\sum_{k=1}^{\infty}(-1)^ke^{-k\pi}$$

from this one I've found $$\frac{(1-e^{-\pi})}{2(1+e^{-\pi})}$$

Gib Z
Homework Helper
Posts 3 and 5 are definitely correct, though either you or I have made a sign error on the sum of the geometric series, check that again just to be safe.

I did it in this way
$$\frac{1}{2}-\frac{e^{-\pi}}{1-e^{-2\pi}}+\frac{e^{-2\pi}}{1-e^{-2\pi}}$$

Gib Z
Homework Helper
Your last post makes no sense to me :( Maybe I'm doing the mistake, so heres how I'm getting it

$$\frac{1}{2}+\sum_{k=1}^{\infty}(-1)^ke^{-k\pi} = \frac{1}{2} + \frac{-e^{-\pi}}{1+e^{-\pi}} = \frac{1-3e^{-\pi}}{2(1-e^{-\pi})}$$

Hey,Gib Z