Determine whether this improper integral converges or not

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The expression is:
$$int_1^infty frac1x-e^-xdx$$
I know that to prove if the the integral converges, it must have a limit and if it diverges it doesn't. I have gotten to this bit:
$$int_1^infty frac1x-e^-xdx=lim_tto inftyint_1^tfrac1x-e^-xdx$$
But I have no idea how to integrate $int_1^tfrac1x-e^-xdx$ to find the limit to see if it converges or not. Are there any tests I can use instead to see if it converges or not?




limits convergence improper-integrals




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  • 1




    You can try a comparison test with integration of a simpler function. If $f(x) leq g(x)$ for all $x geq 1$ then $int_1^infty f(x)dx leq int_1^infty g(x)dx$.
    – Michael
    yesterday











  • What function could I use?
    – M. Calculator
    yesterday










  • You can try some simple related functions. Ones you can integrate and/or that you know converge or diverge.
    – Michael
    yesterday











  • Your integral does not converge.
    – Dr. Sonnhard Graubner
    yesterday






  • 1




    Use the 2nd (limit) comparison test with 1/x
    – zokomoko
    yesterday














up vote
0
down vote

favorite












The expression is:
$$int_1^infty frac1x-e^-xdx$$
I know that to prove if the the integral converges, it must have a limit and if it diverges it doesn't. I have gotten to this bit:
$$int_1^infty frac1x-e^-xdx=lim_tto inftyint_1^tfrac1x-e^-xdx$$
But I have no idea how to integrate $int_1^tfrac1x-e^-xdx$ to find the limit to see if it converges or not. Are there any tests I can use instead to see if it converges or not?




limits convergence improper-integrals




share|cite|improve this question

















  • 1




    You can try a comparison test with integration of a simpler function. If $f(x) leq g(x)$ for all $x geq 1$ then $int_1^infty f(x)dx leq int_1^infty g(x)dx$.
    – Michael
    yesterday











  • What function could I use?
    – M. Calculator
    yesterday










  • You can try some simple related functions. Ones you can integrate and/or that you know converge or diverge.
    – Michael
    yesterday











  • Your integral does not converge.
    – Dr. Sonnhard Graubner
    yesterday






  • 1




    Use the 2nd (limit) comparison test with 1/x
    – zokomoko
    yesterday












up vote
0
down vote

favorite









up vote
0
down vote

favorite











The expression is:
$$int_1^infty frac1x-e^-xdx$$
I know that to prove if the the integral converges, it must have a limit and if it diverges it doesn't. I have gotten to this bit:
$$int_1^infty frac1x-e^-xdx=lim_tto inftyint_1^tfrac1x-e^-xdx$$
But I have no idea how to integrate $int_1^tfrac1x-e^-xdx$ to find the limit to see if it converges or not. Are there any tests I can use instead to see if it converges or not?




limits convergence improper-integrals




share|cite|improve this question













The expression is:
$$int_1^infty frac1x-e^-xdx$$
I know that to prove if the the integral converges, it must have a limit and if it diverges it doesn't. I have gotten to this bit:
$$int_1^infty frac1x-e^-xdx=lim_tto inftyint_1^tfrac1x-e^-xdx$$
But I have no idea how to integrate $int_1^tfrac1x-e^-xdx$ to find the limit to see if it converges or not. Are there any tests I can use instead to see if it converges or not?





limits convergence improper-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









pointguard0

514215




514215









asked yesterday









M. Calculator

212




212







  • 1




    You can try a comparison test with integration of a simpler function. If $f(x) leq g(x)$ for all $x geq 1$ then $int_1^infty f(x)dx leq int_1^infty g(x)dx$.
    – Michael
    yesterday











  • What function could I use?
    – M. Calculator
    yesterday










  • You can try some simple related functions. Ones you can integrate and/or that you know converge or diverge.
    – Michael
    yesterday











  • Your integral does not converge.
    – Dr. Sonnhard Graubner
    yesterday






  • 1




    Use the 2nd (limit) comparison test with 1/x
    – zokomoko
    yesterday












  • 1




    You can try a comparison test with integration of a simpler function. If $f(x) leq g(x)$ for all $x geq 1$ then $int_1^infty f(x)dx leq int_1^infty g(x)dx$.
    – Michael
    yesterday











  • What function could I use?
    – M. Calculator
    yesterday










  • You can try some simple related functions. Ones you can integrate and/or that you know converge or diverge.
    – Michael
    yesterday











  • Your integral does not converge.
    – Dr. Sonnhard Graubner
    yesterday






  • 1




    Use the 2nd (limit) comparison test with 1/x
    – zokomoko
    yesterday







1




1




You can try a comparison test with integration of a simpler function. If $f(x) leq g(x)$ for all $x geq 1$ then $int_1^infty f(x)dx leq int_1^infty g(x)dx$.
– Michael
yesterday





You can try a comparison test with integration of a simpler function. If $f(x) leq g(x)$ for all $x geq 1$ then $int_1^infty f(x)dx leq int_1^infty g(x)dx$.
– Michael
yesterday













What function could I use?
– M. Calculator
yesterday




What function could I use?
– M. Calculator
yesterday












You can try some simple related functions. Ones you can integrate and/or that you know converge or diverge.
– Michael
yesterday





You can try some simple related functions. Ones you can integrate and/or that you know converge or diverge.
– Michael
yesterday













Your integral does not converge.
– Dr. Sonnhard Graubner
yesterday




Your integral does not converge.
– Dr. Sonnhard Graubner
yesterday




1




1




Use the 2nd (limit) comparison test with 1/x
– zokomoko
yesterday




Use the 2nd (limit) comparison test with 1/x
– zokomoko
yesterday










2 Answers
2






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up vote
0
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We have that



$$frac1x-e^-xsim frac 1x$$



therefore the given integral diverges by limit comparison test with $int_1^infty frac1xdx$.



As an alternative note that



$$frac1x-e^-xge frac 1x$$



then the given integral diverges also by comparison test.






share|cite|improve this answer




























    up vote
    0
    down vote













    This is not a "detailed proof" but I feel that it is sufficient. As mentioned, using a comparison test would give you your answer. For example consider the function: $g(x) = frac1x+1$. It is easy to see that $$frac1x + 1leqfrac1x-e^-x$$ for all $x geq 1$. We know that $$int_1^inftyfrac1x + 1dx = infty.$$ Because your function $frac1x-e^-x$ is bounded below by a integral that diverges, it also diverges, which means:$$int_1^inftyfrac1x-e^-xdx=infty.$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      up vote
      0
      down vote













      We have that



      $$frac1x-e^-xsim frac 1x$$



      therefore the given integral diverges by limit comparison test with $int_1^infty frac1xdx$.



      As an alternative note that



      $$frac1x-e^-xge frac 1x$$



      then the given integral diverges also by comparison test.






      share|cite|improve this answer

























        up vote
        0
        down vote













        We have that



        $$frac1x-e^-xsim frac 1x$$



        therefore the given integral diverges by limit comparison test with $int_1^infty frac1xdx$.



        As an alternative note that



        $$frac1x-e^-xge frac 1x$$



        then the given integral diverges also by comparison test.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          We have that



          $$frac1x-e^-xsim frac 1x$$



          therefore the given integral diverges by limit comparison test with $int_1^infty frac1xdx$.



          As an alternative note that



          $$frac1x-e^-xge frac 1x$$



          then the given integral diverges also by comparison test.






          share|cite|improve this answer













          We have that



          $$frac1x-e^-xsim frac 1x$$



          therefore the given integral diverges by limit comparison test with $int_1^infty frac1xdx$.



          As an alternative note that



          $$frac1x-e^-xge frac 1x$$



          then the given integral diverges also by comparison test.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered yesterday









          gimusi

          63.5k73380




          63.5k73380




















              up vote
              0
              down vote













              This is not a "detailed proof" but I feel that it is sufficient. As mentioned, using a comparison test would give you your answer. For example consider the function: $g(x) = frac1x+1$. It is easy to see that $$frac1x + 1leqfrac1x-e^-x$$ for all $x geq 1$. We know that $$int_1^inftyfrac1x + 1dx = infty.$$ Because your function $frac1x-e^-x$ is bounded below by a integral that diverges, it also diverges, which means:$$int_1^inftyfrac1x-e^-xdx=infty.$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                This is not a "detailed proof" but I feel that it is sufficient. As mentioned, using a comparison test would give you your answer. For example consider the function: $g(x) = frac1x+1$. It is easy to see that $$frac1x + 1leqfrac1x-e^-x$$ for all $x geq 1$. We know that $$int_1^inftyfrac1x + 1dx = infty.$$ Because your function $frac1x-e^-x$ is bounded below by a integral that diverges, it also diverges, which means:$$int_1^inftyfrac1x-e^-xdx=infty.$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  This is not a "detailed proof" but I feel that it is sufficient. As mentioned, using a comparison test would give you your answer. For example consider the function: $g(x) = frac1x+1$. It is easy to see that $$frac1x + 1leqfrac1x-e^-x$$ for all $x geq 1$. We know that $$int_1^inftyfrac1x + 1dx = infty.$$ Because your function $frac1x-e^-x$ is bounded below by a integral that diverges, it also diverges, which means:$$int_1^inftyfrac1x-e^-xdx=infty.$$






                  share|cite|improve this answer













                  This is not a "detailed proof" but I feel that it is sufficient. As mentioned, using a comparison test would give you your answer. For example consider the function: $g(x) = frac1x+1$. It is easy to see that $$frac1x + 1leqfrac1x-e^-x$$ for all $x geq 1$. We know that $$int_1^inftyfrac1x + 1dx = infty.$$ Because your function $frac1x-e^-x$ is bounded below by a integral that diverges, it also diverges, which means:$$int_1^inftyfrac1x-e^-xdx=infty.$$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered yesterday









                  Tyler Beauregard

                  111




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