Integrate $e^{y-2e^y}$












3












$begingroup$


I would like to compute:



$$
int_{0}^{infty} e^{y-2e^y} dy
$$



I thought about using integration by parts but it didn't lead me anywhere. I also tried changing variables but it didn't work as well.



Any suggestion is appreciated :)






calculus integration







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I would like to compute:



    $$
    int_{0}^{infty} e^{y-2e^y} dy
    $$



    I thought about using integration by parts but it didn't lead me anywhere. I also tried changing variables but it didn't work as well.



    Any suggestion is appreciated :)






    calculus integration







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I would like to compute:



      $$
      int_{0}^{infty} e^{y-2e^y} dy
      $$



      I thought about using integration by parts but it didn't lead me anywhere. I also tried changing variables but it didn't work as well.



      Any suggestion is appreciated :)






      calculus integration







      share|cite|improve this question









      $endgroup$




      I would like to compute:



      $$
      int_{0}^{infty} e^{y-2e^y} dy
      $$



      I thought about using integration by parts but it didn't lead me anywhere. I also tried changing variables but it didn't work as well.



      Any suggestion is appreciated :)






      calculus integration




      calculus integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 31 at 15:00









      superuser123superuser123

      48628




      48628






















          3 Answers
          3






          active

          oldest

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          4












          $begingroup$

          Hint



          Write $displaystyle e^{y-2e^y}$ as $dfrac{e^y}{e^{2e^y}}$ and let $u=e^y$. This leaves you with a nicer expression to integrate. Then it's just a matter of setting the limit for the upper bound of the integral as characteristic of improper integrals and evaluate the lower bound as the corresponding $u$ evaluated at $y=0$.
          $$int_{0}^{infty} dfrac{e^y}{e^{2e^y}}mathrm dy=lim_{b to infty}int_{1}^{b} dfrac{mathrm du}{e^{2u}}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If $u=e^y$ then $du=e^y dy$, so how does that work?
            $endgroup$
            – superuser123
            Jan 31 at 15:11










          • $begingroup$
            The differential $mathrm du$ substitutes for $e^y mathrm dy$.
            $endgroup$
            – Paras Khosla
            Jan 31 at 15:13










          • $begingroup$
            oh right, thanks
            $endgroup$
            – superuser123
            Jan 31 at 15:15



















          1












          $begingroup$

          The integral itself is a u-substitution problem:



          $$
          begin{align}
          int e^{y-2e^y} dy
          &=int e^ye^{-2e^y} dy\
          &=int e^{-2e^y}frac{d}{dy}left(e^yright)dy\
          &=int e^{-2e^y}dleft(e^yright) (u=e^y)\
          &=int e^{-2u}du\
          &=-frac{1}{2}int e^{-2u}frac{d}{du}left(-2uright)du\
          &=-frac{1}{2}int e^{-2u}dleft(-2uright) (w=-2u)\
          &=-frac{1}{2}int e^wdw\
          &=-frac{1}{2}e^w+C\
          &=-frac{1}{2}e^{-2e^y}+C.
          end{align}
          $$



          And here's how you should do your improper integral:



          $$
          begin{align}
          int_{0}^{infty} e^{y-2e^y} dy
          &=lim_{b to infty}int_{0}^{b} e^{y-2e^y} dy\
          &=lim_{b to infty}left(-frac{1}{2}e^{-2e^y}Big|_{0}^{b}right)\
          &=lim_{b to infty}left(-frac{1}{2}e^{-2e^b}+frac{1}{2}e^{-2e^0}right)\
          &=0+frac{1}{2}e^{-2}\
          &=frac{1}{2e^2}.
          end{align}
          $$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Rewrite the integrand as $$e^ye^{-2e^y}=(e^y)'e^{-2e^y}$$ and by the chain rule the antiderivative must be



            $$-frac12e^{-2e^y}.$$






            share|cite|improve this answer









            $endgroup$














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              3 Answers
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              active

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              3 Answers
              3






              active

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              active

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              votes






              active

              oldest

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              4












              $begingroup$

              Hint



              Write $displaystyle e^{y-2e^y}$ as $dfrac{e^y}{e^{2e^y}}$ and let $u=e^y$. This leaves you with a nicer expression to integrate. Then it's just a matter of setting the limit for the upper bound of the integral as characteristic of improper integrals and evaluate the lower bound as the corresponding $u$ evaluated at $y=0$.
              $$int_{0}^{infty} dfrac{e^y}{e^{2e^y}}mathrm dy=lim_{b to infty}int_{1}^{b} dfrac{mathrm du}{e^{2u}}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If $u=e^y$ then $du=e^y dy$, so how does that work?
                $endgroup$
                – superuser123
                Jan 31 at 15:11










              • $begingroup$
                The differential $mathrm du$ substitutes for $e^y mathrm dy$.
                $endgroup$
                – Paras Khosla
                Jan 31 at 15:13










              • $begingroup$
                oh right, thanks
                $endgroup$
                – superuser123
                Jan 31 at 15:15
















              4












              $begingroup$

              Hint



              Write $displaystyle e^{y-2e^y}$ as $dfrac{e^y}{e^{2e^y}}$ and let $u=e^y$. This leaves you with a nicer expression to integrate. Then it's just a matter of setting the limit for the upper bound of the integral as characteristic of improper integrals and evaluate the lower bound as the corresponding $u$ evaluated at $y=0$.
              $$int_{0}^{infty} dfrac{e^y}{e^{2e^y}}mathrm dy=lim_{b to infty}int_{1}^{b} dfrac{mathrm du}{e^{2u}}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If $u=e^y$ then $du=e^y dy$, so how does that work?
                $endgroup$
                – superuser123
                Jan 31 at 15:11










              • $begingroup$
                The differential $mathrm du$ substitutes for $e^y mathrm dy$.
                $endgroup$
                – Paras Khosla
                Jan 31 at 15:13










              • $begingroup$
                oh right, thanks
                $endgroup$
                – superuser123
                Jan 31 at 15:15














              4












              4








              4





              $begingroup$

              Hint



              Write $displaystyle e^{y-2e^y}$ as $dfrac{e^y}{e^{2e^y}}$ and let $u=e^y$. This leaves you with a nicer expression to integrate. Then it's just a matter of setting the limit for the upper bound of the integral as characteristic of improper integrals and evaluate the lower bound as the corresponding $u$ evaluated at $y=0$.
              $$int_{0}^{infty} dfrac{e^y}{e^{2e^y}}mathrm dy=lim_{b to infty}int_{1}^{b} dfrac{mathrm du}{e^{2u}}$$






              share|cite|improve this answer









              $endgroup$



              Hint



              Write $displaystyle e^{y-2e^y}$ as $dfrac{e^y}{e^{2e^y}}$ and let $u=e^y$. This leaves you with a nicer expression to integrate. Then it's just a matter of setting the limit for the upper bound of the integral as characteristic of improper integrals and evaluate the lower bound as the corresponding $u$ evaluated at $y=0$.
              $$int_{0}^{infty} dfrac{e^y}{e^{2e^y}}mathrm dy=lim_{b to infty}int_{1}^{b} dfrac{mathrm du}{e^{2u}}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 31 at 15:07









              Paras KhoslaParas Khosla

              2,883523




              2,883523












              • $begingroup$
                If $u=e^y$ then $du=e^y dy$, so how does that work?
                $endgroup$
                – superuser123
                Jan 31 at 15:11










              • $begingroup$
                The differential $mathrm du$ substitutes for $e^y mathrm dy$.
                $endgroup$
                – Paras Khosla
                Jan 31 at 15:13










              • $begingroup$
                oh right, thanks
                $endgroup$
                – superuser123
                Jan 31 at 15:15


















              • $begingroup$
                If $u=e^y$ then $du=e^y dy$, so how does that work?
                $endgroup$
                – superuser123
                Jan 31 at 15:11










              • $begingroup$
                The differential $mathrm du$ substitutes for $e^y mathrm dy$.
                $endgroup$
                – Paras Khosla
                Jan 31 at 15:13










              • $begingroup$
                oh right, thanks
                $endgroup$
                – superuser123
                Jan 31 at 15:15
















              $begingroup$
              If $u=e^y$ then $du=e^y dy$, so how does that work?
              $endgroup$
              – superuser123
              Jan 31 at 15:11




              $begingroup$
              If $u=e^y$ then $du=e^y dy$, so how does that work?
              $endgroup$
              – superuser123
              Jan 31 at 15:11












              $begingroup$
              The differential $mathrm du$ substitutes for $e^y mathrm dy$.
              $endgroup$
              – Paras Khosla
              Jan 31 at 15:13




              $begingroup$
              The differential $mathrm du$ substitutes for $e^y mathrm dy$.
              $endgroup$
              – Paras Khosla
              Jan 31 at 15:13












              $begingroup$
              oh right, thanks
              $endgroup$
              – superuser123
              Jan 31 at 15:15




              $begingroup$
              oh right, thanks
              $endgroup$
              – superuser123
              Jan 31 at 15:15











              1












              $begingroup$

              The integral itself is a u-substitution problem:



              $$
              begin{align}
              int e^{y-2e^y} dy
              &=int e^ye^{-2e^y} dy\
              &=int e^{-2e^y}frac{d}{dy}left(e^yright)dy\
              &=int e^{-2e^y}dleft(e^yright) (u=e^y)\
              &=int e^{-2u}du\
              &=-frac{1}{2}int e^{-2u}frac{d}{du}left(-2uright)du\
              &=-frac{1}{2}int e^{-2u}dleft(-2uright) (w=-2u)\
              &=-frac{1}{2}int e^wdw\
              &=-frac{1}{2}e^w+C\
              &=-frac{1}{2}e^{-2e^y}+C.
              end{align}
              $$



              And here's how you should do your improper integral:



              $$
              begin{align}
              int_{0}^{infty} e^{y-2e^y} dy
              &=lim_{b to infty}int_{0}^{b} e^{y-2e^y} dy\
              &=lim_{b to infty}left(-frac{1}{2}e^{-2e^y}Big|_{0}^{b}right)\
              &=lim_{b to infty}left(-frac{1}{2}e^{-2e^b}+frac{1}{2}e^{-2e^0}right)\
              &=0+frac{1}{2}e^{-2}\
              &=frac{1}{2e^2}.
              end{align}
              $$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The integral itself is a u-substitution problem:



                $$
                begin{align}
                int e^{y-2e^y} dy
                &=int e^ye^{-2e^y} dy\
                &=int e^{-2e^y}frac{d}{dy}left(e^yright)dy\
                &=int e^{-2e^y}dleft(e^yright) (u=e^y)\
                &=int e^{-2u}du\
                &=-frac{1}{2}int e^{-2u}frac{d}{du}left(-2uright)du\
                &=-frac{1}{2}int e^{-2u}dleft(-2uright) (w=-2u)\
                &=-frac{1}{2}int e^wdw\
                &=-frac{1}{2}e^w+C\
                &=-frac{1}{2}e^{-2e^y}+C.
                end{align}
                $$



                And here's how you should do your improper integral:



                $$
                begin{align}
                int_{0}^{infty} e^{y-2e^y} dy
                &=lim_{b to infty}int_{0}^{b} e^{y-2e^y} dy\
                &=lim_{b to infty}left(-frac{1}{2}e^{-2e^y}Big|_{0}^{b}right)\
                &=lim_{b to infty}left(-frac{1}{2}e^{-2e^b}+frac{1}{2}e^{-2e^0}right)\
                &=0+frac{1}{2}e^{-2}\
                &=frac{1}{2e^2}.
                end{align}
                $$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The integral itself is a u-substitution problem:



                  $$
                  begin{align}
                  int e^{y-2e^y} dy
                  &=int e^ye^{-2e^y} dy\
                  &=int e^{-2e^y}frac{d}{dy}left(e^yright)dy\
                  &=int e^{-2e^y}dleft(e^yright) (u=e^y)\
                  &=int e^{-2u}du\
                  &=-frac{1}{2}int e^{-2u}frac{d}{du}left(-2uright)du\
                  &=-frac{1}{2}int e^{-2u}dleft(-2uright) (w=-2u)\
                  &=-frac{1}{2}int e^wdw\
                  &=-frac{1}{2}e^w+C\
                  &=-frac{1}{2}e^{-2e^y}+C.
                  end{align}
                  $$



                  And here's how you should do your improper integral:



                  $$
                  begin{align}
                  int_{0}^{infty} e^{y-2e^y} dy
                  &=lim_{b to infty}int_{0}^{b} e^{y-2e^y} dy\
                  &=lim_{b to infty}left(-frac{1}{2}e^{-2e^y}Big|_{0}^{b}right)\
                  &=lim_{b to infty}left(-frac{1}{2}e^{-2e^b}+frac{1}{2}e^{-2e^0}right)\
                  &=0+frac{1}{2}e^{-2}\
                  &=frac{1}{2e^2}.
                  end{align}
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  The integral itself is a u-substitution problem:



                  $$
                  begin{align}
                  int e^{y-2e^y} dy
                  &=int e^ye^{-2e^y} dy\
                  &=int e^{-2e^y}frac{d}{dy}left(e^yright)dy\
                  &=int e^{-2e^y}dleft(e^yright) (u=e^y)\
                  &=int e^{-2u}du\
                  &=-frac{1}{2}int e^{-2u}frac{d}{du}left(-2uright)du\
                  &=-frac{1}{2}int e^{-2u}dleft(-2uright) (w=-2u)\
                  &=-frac{1}{2}int e^wdw\
                  &=-frac{1}{2}e^w+C\
                  &=-frac{1}{2}e^{-2e^y}+C.
                  end{align}
                  $$



                  And here's how you should do your improper integral:



                  $$
                  begin{align}
                  int_{0}^{infty} e^{y-2e^y} dy
                  &=lim_{b to infty}int_{0}^{b} e^{y-2e^y} dy\
                  &=lim_{b to infty}left(-frac{1}{2}e^{-2e^y}Big|_{0}^{b}right)\
                  &=lim_{b to infty}left(-frac{1}{2}e^{-2e^b}+frac{1}{2}e^{-2e^0}right)\
                  &=0+frac{1}{2}e^{-2}\
                  &=frac{1}{2e^2}.
                  end{align}
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 31 at 15:28

























                  answered Jan 31 at 15:15









                  Michael RybkinMichael Rybkin

                  4,164422




                  4,164422























                      0












                      $begingroup$

                      Rewrite the integrand as $$e^ye^{-2e^y}=(e^y)'e^{-2e^y}$$ and by the chain rule the antiderivative must be



                      $$-frac12e^{-2e^y}.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Rewrite the integrand as $$e^ye^{-2e^y}=(e^y)'e^{-2e^y}$$ and by the chain rule the antiderivative must be



                        $$-frac12e^{-2e^y}.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Rewrite the integrand as $$e^ye^{-2e^y}=(e^y)'e^{-2e^y}$$ and by the chain rule the antiderivative must be



                          $$-frac12e^{-2e^y}.$$






                          share|cite|improve this answer









                          $endgroup$



                          Rewrite the integrand as $$e^ye^{-2e^y}=(e^y)'e^{-2e^y}$$ and by the chain rule the antiderivative must be



                          $$-frac12e^{-2e^y}.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 31 at 15:48









                          Yves DaoustYves Daoust

                          132k676230




                          132k676230






























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