the area between $e^{-x}sin xquad (xge0)$ and x-axis [closed]












-1












$begingroup$


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.










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closed as off-topic by Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos Dec 31 '18 at 12:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 3




    $begingroup$
    If you know when $sin, x$ is positive and when it is negative you can just split the integral accordingly and use integration by parts. Have you made any attempt so far?
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 6:43








  • 1




    $begingroup$
    In addition to what Kavi Rama Murthy said: the indefinite integral of $e^{-x}sin x$ is a standard table integral. Often given in the more general form $int e^{ax}sin bx,dx$. It is a bit tricky to find, as a standard way is to try and find that and $int e^{ax}cos bx,dx$ together, using a pair of equations that you can generate with integration by parts. Mind you, there is a simpler way relying on the properties of the complex exponential function, but calculus students are typically not expected to have heard about that at this point in their studies.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 6:53












  • $begingroup$
    @KaviRamaMurthy Thank you, I followed your hint and posted my answer below.
    $endgroup$
    – burg1ar
    Dec 31 '18 at 8:21
















-1












$begingroup$


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.










share|cite|improve this question











$endgroup$



closed as off-topic by Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos Dec 31 '18 at 12:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 3




    $begingroup$
    If you know when $sin, x$ is positive and when it is negative you can just split the integral accordingly and use integration by parts. Have you made any attempt so far?
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 6:43








  • 1




    $begingroup$
    In addition to what Kavi Rama Murthy said: the indefinite integral of $e^{-x}sin x$ is a standard table integral. Often given in the more general form $int e^{ax}sin bx,dx$. It is a bit tricky to find, as a standard way is to try and find that and $int e^{ax}cos bx,dx$ together, using a pair of equations that you can generate with integration by parts. Mind you, there is a simpler way relying on the properties of the complex exponential function, but calculus students are typically not expected to have heard about that at this point in their studies.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 6:53












  • $begingroup$
    @KaviRamaMurthy Thank you, I followed your hint and posted my answer below.
    $endgroup$
    – burg1ar
    Dec 31 '18 at 8:21














-1












-1








-1





$begingroup$


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.










share|cite|improve this question











$endgroup$




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.







calculus integration improper-integrals






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edited Dec 31 '18 at 15:22







burg1ar

















asked Dec 31 '18 at 6:40









burg1arburg1ar

487




487




closed as off-topic by Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos Dec 31 '18 at 12:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos Dec 31 '18 at 12:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, RRL, Kavi Rama Murthy, user91500, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    $begingroup$
    If you know when $sin, x$ is positive and when it is negative you can just split the integral accordingly and use integration by parts. Have you made any attempt so far?
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 6:43








  • 1




    $begingroup$
    In addition to what Kavi Rama Murthy said: the indefinite integral of $e^{-x}sin x$ is a standard table integral. Often given in the more general form $int e^{ax}sin bx,dx$. It is a bit tricky to find, as a standard way is to try and find that and $int e^{ax}cos bx,dx$ together, using a pair of equations that you can generate with integration by parts. Mind you, there is a simpler way relying on the properties of the complex exponential function, but calculus students are typically not expected to have heard about that at this point in their studies.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 6:53












  • $begingroup$
    @KaviRamaMurthy Thank you, I followed your hint and posted my answer below.
    $endgroup$
    – burg1ar
    Dec 31 '18 at 8:21














  • 3




    $begingroup$
    If you know when $sin, x$ is positive and when it is negative you can just split the integral accordingly and use integration by parts. Have you made any attempt so far?
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 6:43








  • 1




    $begingroup$
    In addition to what Kavi Rama Murthy said: the indefinite integral of $e^{-x}sin x$ is a standard table integral. Often given in the more general form $int e^{ax}sin bx,dx$. It is a bit tricky to find, as a standard way is to try and find that and $int e^{ax}cos bx,dx$ together, using a pair of equations that you can generate with integration by parts. Mind you, there is a simpler way relying on the properties of the complex exponential function, but calculus students are typically not expected to have heard about that at this point in their studies.
    $endgroup$
    – Jyrki Lahtonen
    Dec 31 '18 at 6:53












  • $begingroup$
    @KaviRamaMurthy Thank you, I followed your hint and posted my answer below.
    $endgroup$
    – burg1ar
    Dec 31 '18 at 8:21








3




3




$begingroup$
If you know when $sin, x$ is positive and when it is negative you can just split the integral accordingly and use integration by parts. Have you made any attempt so far?
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 6:43






$begingroup$
If you know when $sin, x$ is positive and when it is negative you can just split the integral accordingly and use integration by parts. Have you made any attempt so far?
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 6:43






1




1




$begingroup$
In addition to what Kavi Rama Murthy said: the indefinite integral of $e^{-x}sin x$ is a standard table integral. Often given in the more general form $int e^{ax}sin bx,dx$. It is a bit tricky to find, as a standard way is to try and find that and $int e^{ax}cos bx,dx$ together, using a pair of equations that you can generate with integration by parts. Mind you, there is a simpler way relying on the properties of the complex exponential function, but calculus students are typically not expected to have heard about that at this point in their studies.
$endgroup$
– Jyrki Lahtonen
Dec 31 '18 at 6:53






$begingroup$
In addition to what Kavi Rama Murthy said: the indefinite integral of $e^{-x}sin x$ is a standard table integral. Often given in the more general form $int e^{ax}sin bx,dx$. It is a bit tricky to find, as a standard way is to try and find that and $int e^{ax}cos bx,dx$ together, using a pair of equations that you can generate with integration by parts. Mind you, there is a simpler way relying on the properties of the complex exponential function, but calculus students are typically not expected to have heard about that at this point in their studies.
$endgroup$
– Jyrki Lahtonen
Dec 31 '18 at 6:53














$begingroup$
@KaviRamaMurthy Thank you, I followed your hint and posted my answer below.
$endgroup$
– burg1ar
Dec 31 '18 at 8:21




$begingroup$
@KaviRamaMurthy Thank you, I followed your hint and posted my answer below.
$endgroup$
– burg1ar
Dec 31 '18 at 8:21










3 Answers
3






active

oldest

votes


















3












$begingroup$

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.



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






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$endgroup$





















    2












    $begingroup$

    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
    $$S=sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{2kpi}^{(2k+1)pi}-sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{(2k+1)pi}^{(2k+2)pi}\
    =sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+1)pi}+e^{-2kpi}right]+sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+2)pi}+e^{-(2k+1)pi}right]\
    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-(2k+1)pi}\
    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=1}^{infty}e^{-2kpi}\
    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
    =sum_{k=0}^{infty}e^{-(2k+1)pi}+sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
    =sum_{k=0}^{infty}left[e^{-2kpi}+e^{-(2k+1)pi}right]-frac{1}{2}\
    =sum_{n=0}^{infty}e^{-npi}-frac{1}{2}\
    =frac{1}{1-e^{-pi}}-frac{1}{2}$$

    $$left(sum_{n=0}^{infty}x^n=frac{1}{1-x}Rightarrow sum_{n=0}^{infty}(e^{-pi})^n=frac{1}{1-e^{-pi}}right)$$






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      1












      $begingroup$

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






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        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.



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






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          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.



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






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            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.



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






            share|cite|improve this answer











            $endgroup$



            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.



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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 31 '18 at 9:48

























            answered Dec 31 '18 at 7:32









            achille huiachille hui

            95.6k5131258




            95.6k5131258























                2












                $begingroup$

                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
                $$S=sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{2kpi}^{(2k+1)pi}-sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{(2k+1)pi}^{(2k+2)pi}\
                =sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+1)pi}+e^{-2kpi}right]+sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+2)pi}+e^{-(2k+1)pi}right]\
                =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-(2k+1)pi}\
                =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=1}^{infty}e^{-2kpi}\
                =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                =sum_{k=0}^{infty}e^{-(2k+1)pi}+sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                =sum_{k=0}^{infty}left[e^{-2kpi}+e^{-(2k+1)pi}right]-frac{1}{2}\
                =sum_{n=0}^{infty}e^{-npi}-frac{1}{2}\
                =frac{1}{1-e^{-pi}}-frac{1}{2}$$

                $$left(sum_{n=0}^{infty}x^n=frac{1}{1-x}Rightarrow sum_{n=0}^{infty}(e^{-pi})^n=frac{1}{1-e^{-pi}}right)$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  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
                  $$S=sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{2kpi}^{(2k+1)pi}-sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{(2k+1)pi}^{(2k+2)pi}\
                  =sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+1)pi}+e^{-2kpi}right]+sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+2)pi}+e^{-(2k+1)pi}right]\
                  =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-(2k+1)pi}\
                  =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=1}^{infty}e^{-2kpi}\
                  =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                  =sum_{k=0}^{infty}e^{-(2k+1)pi}+sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                  =sum_{k=0}^{infty}left[e^{-2kpi}+e^{-(2k+1)pi}right]-frac{1}{2}\
                  =sum_{n=0}^{infty}e^{-npi}-frac{1}{2}\
                  =frac{1}{1-e^{-pi}}-frac{1}{2}$$

                  $$left(sum_{n=0}^{infty}x^n=frac{1}{1-x}Rightarrow sum_{n=0}^{infty}(e^{-pi})^n=frac{1}{1-e^{-pi}}right)$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    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
                    $$S=sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{2kpi}^{(2k+1)pi}-sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{(2k+1)pi}^{(2k+2)pi}\
                    =sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+1)pi}+e^{-2kpi}right]+sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+2)pi}+e^{-(2k+1)pi}right]\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-(2k+1)pi}\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=1}^{infty}e^{-2kpi}\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                    =sum_{k=0}^{infty}left[e^{-2kpi}+e^{-(2k+1)pi}right]-frac{1}{2}\
                    =sum_{n=0}^{infty}e^{-npi}-frac{1}{2}\
                    =frac{1}{1-e^{-pi}}-frac{1}{2}$$

                    $$left(sum_{n=0}^{infty}x^n=frac{1}{1-x}Rightarrow sum_{n=0}^{infty}(e^{-pi})^n=frac{1}{1-e^{-pi}}right)$$






                    share|cite|improve this answer









                    $endgroup$



                    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
                    $$S=sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{2kpi}^{(2k+1)pi}-sum_{k=0}^{infty}-frac{1}{2}e^{-x}(sin x+cos x)Big |_{(2k+1)pi}^{(2k+2)pi}\
                    =sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+1)pi}+e^{-2kpi}right]+sum_{k=0}^{infty}frac{1}{2}left[e^{-(2k+2)pi}+e^{-(2k+1)pi}right]\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-(2k+1)pi}\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=1}^{infty}e^{-2kpi}\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}+frac{1}{2}sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                    =sum_{k=0}^{infty}e^{-(2k+1)pi}+sum_{k=0}^{infty}e^{-2kpi}-frac{1}{2}\
                    =sum_{k=0}^{infty}left[e^{-2kpi}+e^{-(2k+1)pi}right]-frac{1}{2}\
                    =sum_{n=0}^{infty}e^{-npi}-frac{1}{2}\
                    =frac{1}{1-e^{-pi}}-frac{1}{2}$$

                    $$left(sum_{n=0}^{infty}x^n=frac{1}{1-x}Rightarrow sum_{n=0}^{infty}(e^{-pi})^n=frac{1}{1-e^{-pi}}right)$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 31 '18 at 8:18









                    burg1arburg1ar

                    487




                    487























                        1












                        $begingroup$

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






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

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






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

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






                            share|cite|improve this answer









                            $endgroup$



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







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 31 '18 at 7:02









                            Martín Vacas VignoloMartín Vacas Vignolo

                            3,811623




                            3,811623















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