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








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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            2017 IndyCar Series Contents Series news Teams and drivers Schedule Season summary Footnotes References External links Navigation menu"INDYCAR: Initial 2018 bodywork concepts unveiled"the original"IndyCar confirms switch to Performance Friction brakes in 2017""AJ Foyt Racing will switch to Chevy"the original"Carlos Munoz, Conor Daly will drive for AJ Foyt Racing""Zach Veach's Indy 500 Debut Confirmed with Foyt""No mass exodus from Honda after Ganassi switch""Ex-F1 driver Sato joins Andretti Autosport for 2017 IndyCar season""IndyCar's Ryan Hunter-Reay, sponsor DHL paired through 2020""hhgregg and Andretti Autosport announce partnership for key races in 2016""INDYCAR: Rossi re-signs with Andretti"the original"McLaren Formula 1 - Fernando Alonso to race at Indy 500 with McLaren, Honda and Andretti Autosport""Shank will finally take part in Indy 500 with Harvey, Andretti | MotorSportsTalk""Andretti adds Jack Harvey to Indy 500 field""Ganassi switches to Honda power for 2017""INDYCAR: Chilton returns to Ganassi"the original"IndyCar silly season: Who's going where in 2017?""INDYCAR: Kanaan, NTT Data return to Ganassi"the original"Kimball to remain at Ganassi for 2017""Coyne confirms Bourdais for 2017 IndyCar season""Davison to sub for Bourdais in Indy 500"the original"Gutierrez confirmed for Detroit IndyCar debut""Gutierrez returns with Coyne for rest of 2017 season""Vautier to drive for Coyne at Texas"the original"INDYCAR: Coyne confirms Jones for 2017"the original"Pippa Mann returns to Coyne for Indy 500""Karam, Dreyer & Reinbold teaming up again for Indianapolis 500""Pigot to return to Ed Carpenter Racing""Hildebrand confirmed as full-time Ed Carpenter driver""Veach to replace injured Hildebrand at Barber"the originalNew Team Harding Racing Enters Chaves for 101st Indianapolis 500"Juncos Racing Announces Entry in 101st Running of the Indianapolis 500 :: Juncos Racing""Juncos confirms Pigot for Indy 500""Saavedra confirmed in Juncos' second 500 entry"the original"Lazier confirms Indy 500 run after son's USF2000 debut"the original"Claman DeMelo to race for RLLR at Sonoma"the original"Rahal signs Servia and ace engineer for 2017""IndyCar: Aleshin returns with Schmidt"the original"Aleshin replaced by Saavedra for Toronto""Jack Harvey will pilot SPM No. 7 car at Watkins Glen, Sonoma""Jay Howard confirmed in Tony Stewart's supported SPM Indy entry""INDYCAR: Newgarden to wave the flag at Penske"the original"Pagenaud opts for No. 1 in 2017"the original"Penske confirms Newgarden for 2017""Montoya to stay with Team Penske in 2017""Target leaving IndyCar after 27 seasons with Chip Ganassi""Cavin: IndyCar could see complete driver/team shakeup in 2017""End of the road for KV Racing?""KV Racing confirms closure, equipment sold to Juncos""Juncos confirms IndyCar Series entry"the original"Juncos readies IndyCar program, aims for '17 500"the original"Harding Racing to add Texas, Pocono to schedule"the original"Sato signs with Andretti Autosport for 2017""INDYCAR: Aleshin in Doubt at SPM"the original"Long Beach notebook: JR Hildebrand breaks hand""Hildebrand cleared to return at Phoenix"the original"Bourdais to undergo surgery on multiple fractures""Aleshin loses Schmidt Peterson IndyCar ride""Saavedra in at SPM for Pocono, Gateway"the original"Bourdais to make return at Gateway"the original"The IndyCar Grand Prix no longer is sponsored by Angie's List""2017 IndyCar Series rulebook""2017 Verizon IndyCar Series Official Rulebook"Official websiteeeeee

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            1905190919101911191219131914191519161917191819191920192119221923192419251926192719281929193019311932193319341935193619371938193919401941194619471948194919501951195219531954195519791980198119821983198419851986198719881989199019911992199319941995199619971998199920002001200220032004200520062007200819961996–971998199920002001200220032004200520062007200820092010201120122013201420152016201720182019Kenny BräckBuzz CalkinsScott DixonDario FranchittiSam Hornish Jr.Ryan Hunter-ReayTony KanaanBuddy LazierJosef NewgardenSimon PagenaudWill PowerGreg RayScott SharpTony StewartDan WheldonGatewayIndianapolisIowaPoconoTexasBarberCircuit of the AmericasIndianapolisLaguna SecaMid-OhioPortlandRoad AmericaDetroitLong BeachSt. PetersburgTorontoAtlantaBaltimoreCharlotteChicagolandDoverEdmontonFontanaHomesteadHoustonKansasKentuckyLas VegasLouisianaMichiganMilwaukeeMotegiNashvilleNazarethNew HampshirePhoenixPikes PeakRichmondSão PauloSonomaSurfers ParadiseWalt Disney WorldWatkins GlenBostonBras...
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