I am getting undefined as the answer of this integral problem $intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$....












2












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Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$





My Attempt:



Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?




Thank you in advance.






integration definite-integrals







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  • $begingroup$
    Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
    $endgroup$
    – Dbchatto67
    20 hours ago


















2












$begingroup$



Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$





My Attempt:



Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?




Thank you in advance.






integration definite-integrals







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$endgroup$












  • $begingroup$
    Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
    $endgroup$
    – Dbchatto67
    20 hours ago
















2












2








2





$begingroup$



Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$





My Attempt:



Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?




Thank you in advance.






integration definite-integrals







share|cite|improve this question











$endgroup$





Find $$intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)}$$





My Attempt:



Let $$begin{align}frac{1}{(n-2)(3-n)}&=frac{A}{n-2}+frac{B}{3-n} \ &= frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$begin{align}therefore intlimits_{2}^{3}frac{mathrm dn}{(n-2)(3-n)} &= intlimits_{2}^{3}bigg(frac{1}{n-2}+frac{1}{3-n}bigg),mathrm dn \ &= intlimits_{2}^{3}frac{1}{n-2},mathrm dn+intlimits_{2}^{3}frac{1}{3-n},mathrm dn \ &=left[log(n-2)right]_{small 2}^{small 3}+left[log(3-n)right]_{small 2}^{small 3} \ &= left[log(3-2)-log(2-2)right]+left[log(3-3)+log(3-2)right] \ &= left[log 1-log 0right]+left[log 0 - log 1right]end{align}$$ but $log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?




Thank you in advance.






integration definite-integrals




integration definite-integrals






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edited 15 hours ago









YuiTo Cheng

2,41341037




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asked 20 hours ago









arandomguyarandomguy

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  • $begingroup$
    Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
    $endgroup$
    – Dbchatto67
    20 hours ago




















  • $begingroup$
    Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
    $endgroup$
    – Dbchatto67
    20 hours ago


















$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
20 hours ago






$begingroup$
Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand.
$endgroup$
– Dbchatto67
20 hours ago












2 Answers
2






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No.



There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.






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  • $begingroup$
    Whats quite strong mean
    $endgroup$
    – Mikey Spivak
    20 hours ago










  • $begingroup$
    @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
    $endgroup$
    – The_Sympathizer
    20 hours ago










  • $begingroup$
    More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
    $endgroup$
    – The_Sympathizer
    20 hours ago





















2












$begingroup$

No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    No.



    There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Whats quite strong mean
      $endgroup$
      – Mikey Spivak
      20 hours ago










    • $begingroup$
      @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
      $endgroup$
      – The_Sympathizer
      20 hours ago










    • $begingroup$
      More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
      $endgroup$
      – The_Sympathizer
      20 hours ago


















    3












    $begingroup$

    No.



    There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Whats quite strong mean
      $endgroup$
      – Mikey Spivak
      20 hours ago










    • $begingroup$
      @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
      $endgroup$
      – The_Sympathizer
      20 hours ago










    • $begingroup$
      More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
      $endgroup$
      – The_Sympathizer
      20 hours ago
















    3












    3








    3





    $begingroup$

    No.



    There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.






    share|cite|improve this answer









    $endgroup$



    No.



    There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 20 hours ago









    The_SympathizerThe_Sympathizer

    7,8752246




    7,8752246












    • $begingroup$
      Whats quite strong mean
      $endgroup$
      – Mikey Spivak
      20 hours ago










    • $begingroup$
      @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
      $endgroup$
      – The_Sympathizer
      20 hours ago










    • $begingroup$
      More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
      $endgroup$
      – The_Sympathizer
      20 hours ago




















    • $begingroup$
      Whats quite strong mean
      $endgroup$
      – Mikey Spivak
      20 hours ago










    • $begingroup$
      @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
      $endgroup$
      – The_Sympathizer
      20 hours ago










    • $begingroup$
      More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
      $endgroup$
      – The_Sympathizer
      20 hours ago


















    $begingroup$
    Whats quite strong mean
    $endgroup$
    – Mikey Spivak
    20 hours ago




    $begingroup$
    Whats quite strong mean
    $endgroup$
    – Mikey Spivak
    20 hours ago












    $begingroup$
    @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
    $endgroup$
    – The_Sympathizer
    20 hours ago




    $begingroup$
    @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is.
    $endgroup$
    – The_Sympathizer
    20 hours ago












    $begingroup$
    More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
    $endgroup$
    – The_Sympathizer
    20 hours ago






    $begingroup$
    More quantitatively, a singular term $x^{-alpha}$ with $alpha ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 le alpha < 1$, it actually is weak enough that it can converge.
    $endgroup$
    – The_Sympathizer
    20 hours ago













    2












    $begingroup$

    No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.






        share|cite|improve this answer









        $endgroup$



        No, there is nothing wrong. You havebegin{align}int_2^3frac1{(x-2)(x-3)},mathrm dx&=int_2^{5/2}frac1{(x-2)(x-3)},mathrm dx+int_{5/2}^3frac1{(x-2)(x-3)},mathrm dx\&=lim_{tto2^+}int_t^{5/2}frac1{(x-2)(x-3)},mathrm dx+lim_{tto3^-}int_{5/2}^tfrac1{(x-2)(x-3)},mathrm dx.end{align}None of these limits exist, since the function that's being integrated behaves as $frac1{x-3}$ near $3$ and as $frac1{x-2}$ near $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 20 hours ago









        José Carlos SantosJosé Carlos Santos

        175k23134243




        175k23134243






























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