Dtyjlui

I encountered this problem in a test,and I didn’t work it out back then.
what I know is the area
$$S=int_{0}^{infty}e^{-x}|sin x|dx$$
and I suppose the indefinite integral $$int e^{-x}sin x dx= -frac{1}{2}e^{-x}(sin x+cos x)+C $$
should help to solve this problem,but that’s all I got,No further progress for now.

One approach is relate the integral to a simpler one $int_0^infty e^{-x}sin x dx$ which
can be evaluated in many ways. The following uses a complex exponential function first mentioned by Jykri Lahtonen in comment.

$$begin{align}
int_0^infty e^{-x}|sin x| dx
&= sum_{k=0}^infty int_{kpi}^{(k+1)pi} e^{-x}|sin(x)|dx\
&= sum_{k=0}^infty e^{-kpi} int_0^{pi} e^{-x} sin x dx\
&= frac{1+e^{-pi}}{1-e^{-pi}}sum_{k=0}^infty (-1)^k e^{-kpi} int_0^{pi} e^{-x} sin x dx\
&= frac{1+e^{-pi}}{1-e^{-pi}}sum_{k=0}^infty int_{kpi}^{(k+1)pi} e^{-x} sin x dx\
&= frac{1+e^{-pi}}{1-e^{-pi}}int_0^infty e^{-x}sin x dx\
&= cothfrac{pi}{2}cdotImleft[int_0^infty e^{-(1-i)x} dxright]\
&= cothfrac{pi}{2}cdotImleft[frac{1}{1-i}right]\
&= frac12 cothfrac{pi}{2}
end{align}
$$

we know that
$$sin xge0,xin(2kpi,(2k+1)pi)$$
$$sin xle0,xin((2k+1)pi,(2k+2)pi)$$
$$k=0,1,2,dots$$
So
$$S=int_{0}^{infty}e^{-x}vertsin xvert dx\
=sum_{k=0}^{infty}int_{2kpi}^{(2k+1)pi}e^{-x}sin x dx-sum_{k=0}^{infty}int_{(2k+1)pi}^{(2k+2)pi}e^{-x}sin x dx$$
Let
$$I=int e^{-x}sin x dx\
=-int sin x de^{-x}\
=-e^{-x}sin x+int e^{-x} dsin x\
=-e^{-x}sin x+int e^{-x}cos x dx\
=-e^{-x}sin x-int cos x de^{-x}\
=-e^{-x}sin x-e^{-x}cos x+int e^{-x} dcos x\
=-e^{-x}(sin x+cos x)-int e^{-x}sin x dx\
=-e^{-x}(sin x+cos x)-I$$
Hence
$$I=-frac{1}{2}e^{-x}(sin x+cos x)+C$$
Then
$$S=sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{2kpi}^{(2k+1)pi}-sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{(2k+1)pi}^{(2k+2)pi}\
=sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+1)pi}+e^{-2kpi}right]+sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+2)pi}+e^{-(2k+1)pi}right]\
=sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-(2k+1)pi}\
=sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=1}^{infty}e^{-2kpi}\
=sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
=sum_{k=0}^{infty}e^{-(2k+1)pi}+sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
=sum_{k=0}^{infty}left[e^{-2kpi}+e^{-(2k+1)pi}right]-frac{1}{2}\
=sum_{n=0}^{infty}e^{-npi}-frac{1}{2}\
=frac{1}{1-e^{-pi}}-frac{1}{2}$$
$$left(sum_{n=0}^{infty}x^n=frac{1}{1-x}Rightarrow sum_{n=0}^{infty}(e^{-pi})^n=frac{1}{1-e^{-pi}}right)$$

Hint:

Let $f(x)=e^{-x}$, $g(x)= sin(x)$ and $h(x)=cos(x)$. Note that $f'=-f$, $f''=f$, $g'=h$ and $g''=-g$

With this,

$$(fg)'=f'g+fg'=-fg+fh \(fh)'=f'h+fh'=-fh-fg$$

Adding, $$(fg+fh)'=-2fg$$

Integrating, $$ fg+fh+C=int (fg+fh)' = -2int fg$$

Finally, we have $$int e^{-x}sin(x)dx = -frac{1}{2}e^{-x}(sin(x)+cos(x))+C$$