Area under Integral Calculation

[azatkgz]

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}

 

[NateTG]

Because sine and cosine cross the x-axis, you can't just take the absolute value thorough the integral like that.

 

[azatkgz]

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

 

[azatkgz]

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

 

[azatkgz]

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

 

[azatkgz]

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



Is this answer true?Please,check.

 

[Gib Z]

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.

 

[azatkgz]

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]

Your last post makes no sense to me :( Maybe I'm doing the mistake, so here's 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})}

 

[azatkgz]

Hey,Gib Z
Should't be your answer be same as mine.

 

[Gib Z]

lol well i don't know if your making the mistake or me, but i can't see mine, so unless u do, check ur answer again