I am getting undefined as the answer of this integral problem $intlimits_2^3fracmathrm dn(n-2)(3-n)$. Am I doing something wrong? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$an intriguing integral $I=intlimits_0^4 fracdx4+2^x $Computing the integral $intlimits_-infty^infty(t^2-1)delta(t):dt$Help to understand this property: $intlimits_ka^kbsleft(fracxkright)dx = kintlimits_a^bs(x)dx$Prove $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpia}mathrme^{-2sqrtab$Computing the integral $int fracdxxsqrt x^2+1$Why am I not getting the right answer for this integral?Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Calculating improper integral $int limits_0^inftyfracmathrme^-xsqrtx,mathrmdx$Finding the value of $limlimits_nrightarrow inftysqrtnint^fracpi4_0cos^2n-2(x)mathrm dx$

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I am getting undefined as the answer of this integral problem $intlimits_2^3fracmathrm dn(n-2)(3-n)$. Am I doing something wrong?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integral $int^pi/2_0 (operatornamecsc x - frac1x),mathrm dx$an intriguing integral $I=intlimits_0^4 fracdx4+2^x $Computing the integral $intlimits_-infty^infty(t^2-1)delta(t):dt$Help to understand this property: $intlimits_ka^kbsleft(fracxkright)dx = kintlimits_a^bs(x)dx$Prove $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpiamathrme^-2sqrtab$Computing the integral $int fracdxxsqrt x^2+1$Why am I not getting the right answer for this integral?Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Calculating improper integral $int limits_0^inftyfracmathrme^-xsqrtx,mathrmdx$Finding the value of $limlimits_nrightarrow inftysqrtnint^fracpi4_0cos^2n-2(x)mathrm dx$










2












$begingroup$



Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$




My Attempt:



Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ 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
    21 hours ago
















2












$begingroup$



Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$




My Attempt:



Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ 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











$endgroup$











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














2












2








2





$begingroup$



Find $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$




My Attempt:



Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ 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 $$intlimits_2^3fracmathrm dn(n-2)(3-n)$$




My Attempt:



Let $$beginalignfrac1(n-2)(3-n)&=fracAn-2+fracB3-n \ &= fracA(3-n)+B(n-2)(n-2)(3-n)\ Rightarrow 1 &= A(3-n)+B(n-2) \ &= 3A - An+Bn-2B \ &= n(B-A)+3A-2B.endalign$$ Equating coefficients on both sides, we obtain $$B-A=0 qquad and qquad 3A-2B=1.$$ $$therefore A=B=1.$$ $$$$ $$beginaligntherefore intlimits_2^3fracmathrm dn(n-2)(3-n) &= intlimits_2^3bigg(frac1n-2+frac13-nbigg),mathrm dn \ &= intlimits_2^3frac1n-2,mathrm dn+intlimits_2^3frac13-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]endalign$$ 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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share|cite|improve this question













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share|cite|improve this question








edited 16 hours ago









YuiTo Cheng

2,41341037




2,41341037










asked 21 hours ago









arandomguyarandomguy

17418




17418











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

















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
















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





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











2 Answers
2






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
    21 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
    21 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
    21 hours ago



















2












$begingroup$

No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    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
      21 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
      21 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
      21 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
      21 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
      21 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
      21 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 21 hours ago









    The_SympathizerThe_Sympathizer

    7,8752246




    7,8752246











    • $begingroup$
      Whats quite strong mean
      $endgroup$
      – Mikey Spivak
      21 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
      21 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
      21 hours ago

















    • $begingroup$
      Whats quite strong mean
      $endgroup$
      – Mikey Spivak
      21 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
      21 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
      21 hours ago
















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




    $begingroup$
    Whats quite strong mean
    $endgroup$
    – Mikey Spivak
    21 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
    21 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
    21 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
    21 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
    21 hours ago












    2












    $begingroup$

    No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.






        share|cite|improve this answer









        $endgroup$



        No, there is nothing wrong. You havebeginalignint_2^3frac1(x-2)(x-3),mathrm dx&=int_2^5/2frac1(x-2)(x-3),mathrm dx+int_5/2^3frac1(x-2)(x-3),mathrm dx\&=lim_tto2^+int_t^5/2frac1(x-2)(x-3),mathrm dx+lim_tto3^-int_5/2^tfrac1(x-2)(x-3),mathrm dx.endalignNone of these limits exist, since the function that's being integrated behaves as $frac1x-3$ near $3$ and as $frac1x-2$ near $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 21 hours ago









        José Carlos SantosJosé Carlos Santos

        175k24134243




        175k24134243



























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