Does this integral diverge? $int_{-pi/2}^{pi/2}csc{x}dx$












-1














According to my calculusbook the following integral diverges:
$$int_{-pi/2}^{pi/2}csc{x}dx$$



This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$



My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?



On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?






calculus integration limits improper-integrals







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    -1














    According to my calculusbook the following integral diverges:
    $$int_{-pi/2}^{pi/2}csc{x}dx$$



    This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$



    My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?



    On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?






    calculus integration limits improper-integrals







    share|cite|improve this question



























      -1












      -1








      -1







      According to my calculusbook the following integral diverges:
      $$int_{-pi/2}^{pi/2}csc{x}dx$$



      This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$



      My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?



      On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?






      calculus integration limits improper-integrals







      share|cite|improve this question















      According to my calculusbook the following integral diverges:
      $$int_{-pi/2}^{pi/2}csc{x}dx$$



      This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$



      My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?



      On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?






      calculus integration limits improper-integrals




      calculus integration limits improper-integrals






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      edited Nov 27 at 11:59









      José Carlos Santos

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      asked Nov 27 at 11:53









      GambitSquared

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          The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.






          share|cite|improve this answer





























            3














            No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).



            In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.






            share|cite|improve this answer





















            • But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
              – GambitSquared
              Nov 27 at 12:03










            • By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
              – José Carlos Santos
              Nov 27 at 12:04












            • @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
              – SinTan1729
              Nov 27 at 12:04












            • @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
              – GambitSquared
              Nov 27 at 12:22










            • @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
              – GambitSquared
              Nov 27 at 12:23











            Your Answer





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            2 Answers
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            active

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            2 Answers
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            active

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            active

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            2














            The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.






            share|cite|improve this answer


























              2














              The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.






              share|cite|improve this answer
























                2












                2








                2






                The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.






                share|cite|improve this answer












                The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 at 11:59









                Kavi Rama Murthy

                50k31854




                50k31854























                    3














                    No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).



                    In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.






                    share|cite|improve this answer





















                    • But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
                      – GambitSquared
                      Nov 27 at 12:03










                    • By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
                      – José Carlos Santos
                      Nov 27 at 12:04












                    • @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
                      – SinTan1729
                      Nov 27 at 12:04












                    • @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
                      – GambitSquared
                      Nov 27 at 12:22










                    • @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
                      – GambitSquared
                      Nov 27 at 12:23
















                    3














                    No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).



                    In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.






                    share|cite|improve this answer





















                    • But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
                      – GambitSquared
                      Nov 27 at 12:03










                    • By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
                      – José Carlos Santos
                      Nov 27 at 12:04












                    • @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
                      – SinTan1729
                      Nov 27 at 12:04












                    • @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
                      – GambitSquared
                      Nov 27 at 12:22










                    • @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
                      – GambitSquared
                      Nov 27 at 12:23














                    3












                    3








                    3






                    No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).



                    In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.






                    share|cite|improve this answer












                    No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).



                    In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 27 at 11:58









                    José Carlos Santos

                    150k22120221




                    150k22120221












                    • But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
                      – GambitSquared
                      Nov 27 at 12:03










                    • By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
                      – José Carlos Santos
                      Nov 27 at 12:04












                    • @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
                      – SinTan1729
                      Nov 27 at 12:04












                    • @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
                      – GambitSquared
                      Nov 27 at 12:22










                    • @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
                      – GambitSquared
                      Nov 27 at 12:23


















                    • But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
                      – GambitSquared
                      Nov 27 at 12:03










                    • By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
                      – José Carlos Santos
                      Nov 27 at 12:04












                    • @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
                      – SinTan1729
                      Nov 27 at 12:04












                    • @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
                      – GambitSquared
                      Nov 27 at 12:22










                    • @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
                      – GambitSquared
                      Nov 27 at 12:23
















                    But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
                    – GambitSquared
                    Nov 27 at 12:03




                    But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
                    – GambitSquared
                    Nov 27 at 12:03












                    By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
                    – José Carlos Santos
                    Nov 27 at 12:04






                    By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
                    – José Carlos Santos
                    Nov 27 at 12:04














                    @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
                    – SinTan1729
                    Nov 27 at 12:04






                    @GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
                    – SinTan1729
                    Nov 27 at 12:04














                    @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
                    – GambitSquared
                    Nov 27 at 12:22




                    @JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
                    – GambitSquared
                    Nov 27 at 12:22












                    @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
                    – GambitSquared
                    Nov 27 at 12:23




                    @SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
                    – GambitSquared
                    Nov 27 at 12:23


















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