Improper integrals with polynomial in the denominator - how does p-series apply?











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For example,
$$int_2^{infty} frac1{x^2+2x-3} dx$$



The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?










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    up vote
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    down vote

    favorite












    For example,
    $$int_2^{infty} frac1{x^2+2x-3} dx$$



    The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?










    share|cite|improve this question
























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      For example,
      $$int_2^{infty} frac1{x^2+2x-3} dx$$



      The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?










      share|cite|improve this question













      For example,
      $$int_2^{infty} frac1{x^2+2x-3} dx$$



      The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?







      calculus integration improper-integrals






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      asked Nov 27 at 0:38









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          begin{align}
          int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
          & = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
          & = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
          & = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
          & = -frac 14 ln Big(frac 15Big) \
          & = frac 14 ln5
          end{align}



          Basically, there are many subtle things going on here.



          For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".



          This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing



          $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$



          We know that



          $$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$



          Hence



          $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$



          which is where the "infinities cancel out".



          This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that



          $$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$



          However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.






          share|cite|improve this answer






























            up vote
            0
            down vote













            When you use partial fractions you may loose convergence on each single fraction.



            To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.






            share|cite|improve this answer





















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              2 Answers
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              active

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              2 Answers
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              up vote
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              down vote













              begin{align}
              int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
              & = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
              & = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
              & = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
              & = -frac 14 ln Big(frac 15Big) \
              & = frac 14 ln5
              end{align}



              Basically, there are many subtle things going on here.



              For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".



              This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing



              $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$



              We know that



              $$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$



              Hence



              $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$



              which is where the "infinities cancel out".



              This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that



              $$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$



              However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.






              share|cite|improve this answer



























                up vote
                3
                down vote













                begin{align}
                int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
                & = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
                & = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
                & = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
                & = -frac 14 ln Big(frac 15Big) \
                & = frac 14 ln5
                end{align}



                Basically, there are many subtle things going on here.



                For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".



                This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing



                $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$



                We know that



                $$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$



                Hence



                $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$



                which is where the "infinities cancel out".



                This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that



                $$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$



                However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.






                share|cite|improve this answer

























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  begin{align}
                  int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
                  & = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
                  & = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
                  & = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
                  & = -frac 14 ln Big(frac 15Big) \
                  & = frac 14 ln5
                  end{align}



                  Basically, there are many subtle things going on here.



                  For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".



                  This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing



                  $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$



                  We know that



                  $$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$



                  Hence



                  $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$



                  which is where the "infinities cancel out".



                  This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that



                  $$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$



                  However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.






                  share|cite|improve this answer














                  begin{align}
                  int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
                  & = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
                  & = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
                  & = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
                  & = -frac 14 ln Big(frac 15Big) \
                  & = frac 14 ln5
                  end{align}



                  Basically, there are many subtle things going on here.



                  For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".



                  This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing



                  $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$



                  We know that



                  $$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$



                  Hence



                  $$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$



                  which is where the "infinities cancel out".



                  This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that



                  $$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$



                  However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 27 at 0:56

























                  answered Nov 27 at 0:50









                  glowstonetrees

                  2,195317




                  2,195317






















                      up vote
                      0
                      down vote













                      When you use partial fractions you may loose convergence on each single fraction.



                      To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        When you use partial fractions you may loose convergence on each single fraction.



                        To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          When you use partial fractions you may loose convergence on each single fraction.



                          To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.






                          share|cite|improve this answer












                          When you use partial fractions you may loose convergence on each single fraction.



                          To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 27 at 0:55









                          G Cab

                          17.4k31237




                          17.4k31237






























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