Antidifferentiation involving natural log [closed]












-2














$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$



Sigma is not a constant.



Have I found the correct antiderivative?










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closed as off-topic by amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo Nov 28 '18 at 1:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Is the integration with respect to $dsigma $?Otherwise the answer is surely wrong.
    – Thomas Shelby
    Nov 27 '18 at 16:29










  • Yes, it is with respect to dσ
    – M Do
    Nov 27 '18 at 16:31










  • That's the anti-derivative with respect to $d(2pi sigma^2)$ but but not to $dsigma$. You're going to have to do integration by parts.
    – fleablood
    Nov 27 '18 at 16:35










  • .. D'oh! $sqrt{sigma^2} = |sigma|$ (as per Emilo's answer) so no need for integration be parts (except in the trivial sense of subbing a function for a constant times a function)
    – fleablood
    Nov 27 '18 at 16:44










  • If @EmilioNovati's answer works, click the check mark next to the voting button
    – clathratus
    Nov 27 '18 at 18:29
















-2














$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$



Sigma is not a constant.



Have I found the correct antiderivative?










share|cite|improve this question















closed as off-topic by amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo Nov 28 '18 at 1:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Is the integration with respect to $dsigma $?Otherwise the answer is surely wrong.
    – Thomas Shelby
    Nov 27 '18 at 16:29










  • Yes, it is with respect to dσ
    – M Do
    Nov 27 '18 at 16:31










  • That's the anti-derivative with respect to $d(2pi sigma^2)$ but but not to $dsigma$. You're going to have to do integration by parts.
    – fleablood
    Nov 27 '18 at 16:35










  • .. D'oh! $sqrt{sigma^2} = |sigma|$ (as per Emilo's answer) so no need for integration be parts (except in the trivial sense of subbing a function for a constant times a function)
    – fleablood
    Nov 27 '18 at 16:44










  • If @EmilioNovati's answer works, click the check mark next to the voting button
    – clathratus
    Nov 27 '18 at 18:29














-2












-2








-2







$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$



Sigma is not a constant.



Have I found the correct antiderivative?










share|cite|improve this question















$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$



Sigma is not a constant.



Have I found the correct antiderivative?







integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 16:59









Robert Howard

1,9161822




1,9161822










asked Nov 27 '18 at 16:27









M DoM Do

124




124




closed as off-topic by amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo Nov 28 '18 at 1:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo Nov 28 '18 at 1:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Davide Giraudo, KReiser, Shailesh, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Is the integration with respect to $dsigma $?Otherwise the answer is surely wrong.
    – Thomas Shelby
    Nov 27 '18 at 16:29










  • Yes, it is with respect to dσ
    – M Do
    Nov 27 '18 at 16:31










  • That's the anti-derivative with respect to $d(2pi sigma^2)$ but but not to $dsigma$. You're going to have to do integration by parts.
    – fleablood
    Nov 27 '18 at 16:35










  • .. D'oh! $sqrt{sigma^2} = |sigma|$ (as per Emilo's answer) so no need for integration be parts (except in the trivial sense of subbing a function for a constant times a function)
    – fleablood
    Nov 27 '18 at 16:44










  • If @EmilioNovati's answer works, click the check mark next to the voting button
    – clathratus
    Nov 27 '18 at 18:29


















  • Is the integration with respect to $dsigma $?Otherwise the answer is surely wrong.
    – Thomas Shelby
    Nov 27 '18 at 16:29










  • Yes, it is with respect to dσ
    – M Do
    Nov 27 '18 at 16:31










  • That's the anti-derivative with respect to $d(2pi sigma^2)$ but but not to $dsigma$. You're going to have to do integration by parts.
    – fleablood
    Nov 27 '18 at 16:35










  • .. D'oh! $sqrt{sigma^2} = |sigma|$ (as per Emilo's answer) so no need for integration be parts (except in the trivial sense of subbing a function for a constant times a function)
    – fleablood
    Nov 27 '18 at 16:44










  • If @EmilioNovati's answer works, click the check mark next to the voting button
    – clathratus
    Nov 27 '18 at 18:29
















Is the integration with respect to $dsigma $?Otherwise the answer is surely wrong.
– Thomas Shelby
Nov 27 '18 at 16:29




Is the integration with respect to $dsigma $?Otherwise the answer is surely wrong.
– Thomas Shelby
Nov 27 '18 at 16:29












Yes, it is with respect to dσ
– M Do
Nov 27 '18 at 16:31




Yes, it is with respect to dσ
– M Do
Nov 27 '18 at 16:31












That's the anti-derivative with respect to $d(2pi sigma^2)$ but but not to $dsigma$. You're going to have to do integration by parts.
– fleablood
Nov 27 '18 at 16:35




That's the anti-derivative with respect to $d(2pi sigma^2)$ but but not to $dsigma$. You're going to have to do integration by parts.
– fleablood
Nov 27 '18 at 16:35












.. D'oh! $sqrt{sigma^2} = |sigma|$ (as per Emilo's answer) so no need for integration be parts (except in the trivial sense of subbing a function for a constant times a function)
– fleablood
Nov 27 '18 at 16:44




.. D'oh! $sqrt{sigma^2} = |sigma|$ (as per Emilo's answer) so no need for integration be parts (except in the trivial sense of subbing a function for a constant times a function)
– fleablood
Nov 27 '18 at 16:44












If @EmilioNovati's answer works, click the check mark next to the voting button
– clathratus
Nov 27 '18 at 18:29




If @EmilioNovati's answer works, click the check mark next to the voting button
– clathratus
Nov 27 '18 at 18:29










2 Answers
2






active

oldest

votes


















3














If your integral is:
$$
int frac{1}{sqrt{ 2 pi sigma^2}}d sigma
$$

then you can write it as:
$$
int frac{1}{sqrt{ 2 pi };|sigma|}d sigma=frac{1}{sqrt{ 2 pi };} int frac{1}{|sigma|}d sigma
$$

so the antiderivative is:
$$
frac{1}{sqrt{ 2 pi }}ln |sigma| + C
$$






share|cite|improve this answer





















  • Thank you so much!
    – M Do
    Nov 27 '18 at 16:45



















1














$begin{align}
int frac{1}{sqrt{2pisigma^2}} dsigma &=
frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
& =
frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
end{align}$






share|cite|improve this answer






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    If your integral is:
    $$
    int frac{1}{sqrt{ 2 pi sigma^2}}d sigma
    $$

    then you can write it as:
    $$
    int frac{1}{sqrt{ 2 pi };|sigma|}d sigma=frac{1}{sqrt{ 2 pi };} int frac{1}{|sigma|}d sigma
    $$

    so the antiderivative is:
    $$
    frac{1}{sqrt{ 2 pi }}ln |sigma| + C
    $$






    share|cite|improve this answer





















    • Thank you so much!
      – M Do
      Nov 27 '18 at 16:45
















    3














    If your integral is:
    $$
    int frac{1}{sqrt{ 2 pi sigma^2}}d sigma
    $$

    then you can write it as:
    $$
    int frac{1}{sqrt{ 2 pi };|sigma|}d sigma=frac{1}{sqrt{ 2 pi };} int frac{1}{|sigma|}d sigma
    $$

    so the antiderivative is:
    $$
    frac{1}{sqrt{ 2 pi }}ln |sigma| + C
    $$






    share|cite|improve this answer





















    • Thank you so much!
      – M Do
      Nov 27 '18 at 16:45














    3












    3








    3






    If your integral is:
    $$
    int frac{1}{sqrt{ 2 pi sigma^2}}d sigma
    $$

    then you can write it as:
    $$
    int frac{1}{sqrt{ 2 pi };|sigma|}d sigma=frac{1}{sqrt{ 2 pi };} int frac{1}{|sigma|}d sigma
    $$

    so the antiderivative is:
    $$
    frac{1}{sqrt{ 2 pi }}ln |sigma| + C
    $$






    share|cite|improve this answer












    If your integral is:
    $$
    int frac{1}{sqrt{ 2 pi sigma^2}}d sigma
    $$

    then you can write it as:
    $$
    int frac{1}{sqrt{ 2 pi };|sigma|}d sigma=frac{1}{sqrt{ 2 pi };} int frac{1}{|sigma|}d sigma
    $$

    so the antiderivative is:
    $$
    frac{1}{sqrt{ 2 pi }}ln |sigma| + C
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 16:36









    Emilio NovatiEmilio Novati

    51.6k43473




    51.6k43473












    • Thank you so much!
      – M Do
      Nov 27 '18 at 16:45


















    • Thank you so much!
      – M Do
      Nov 27 '18 at 16:45
















    Thank you so much!
    – M Do
    Nov 27 '18 at 16:45




    Thank you so much!
    – M Do
    Nov 27 '18 at 16:45











    1














    $begin{align}
    int frac{1}{sqrt{2pisigma^2}} dsigma &=
    frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
    & =
    frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
    end{align}$






    share|cite|improve this answer




























      1














      $begin{align}
      int frac{1}{sqrt{2pisigma^2}} dsigma &=
      frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
      & =
      frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
      end{align}$






      share|cite|improve this answer


























        1












        1








        1






        $begin{align}
        int frac{1}{sqrt{2pisigma^2}} dsigma &=
        frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
        & =
        frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
        end{align}$






        share|cite|improve this answer














        $begin{align}
        int frac{1}{sqrt{2pisigma^2}} dsigma &=
        frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
        & =
        frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
        end{align}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 6:51

























        answered Nov 27 '18 at 16:45









        Thomas ShelbyThomas Shelby

        2,007219




        2,007219















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