Antidifferentiation involving natural log [closed]
$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$
Sigma is not a constant.
Have I found the correct antiderivative?
integration
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
asked Nov 27 '18 at 16:27
M Do
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.
add a comment |
$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$
Sigma is not a constant.
Have I found the correct antiderivative?
integration
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
asked Nov 27 '18 at 16:27
M Do
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.
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
add a comment |
$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$
Sigma is not a constant.
Have I found the correct antiderivative?
integration
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
asked Nov 27 '18 at 16:27
M Do
124
$$int frac 1{sqrt{2pisigma^2}}dsigma = lnleft(sqrt{2pi sigma^2}right)$$
Sigma is not a constant.
Have I found the correct antiderivative?
integration
integration
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
asked Nov 27 '18 at 16:27
M Do
124
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
asked Nov 27 '18 at 16:27
M Do
124
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
edited Nov 27 '18 at 16:59
Robert Howard
1,9161822
1,9161822
asked Nov 27 '18 at 16:27
M Do
124
asked Nov 27 '18 at 16:27
M Do
124
asked Nov 27 '18 at 16:27
M 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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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:
$$
frac{1}{sqrt{ 2 pi }}ln |sigma| + C
$$
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
Thank you so much!
– M Do
Nov 27 '18 at 16:45
add a comment |
$begin{align}
int frac{1}{sqrt{2pisigma^2}} dsigma &=
frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
& =
frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
end{align}$
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
$$
frac{1}{sqrt{ 2 pi }}ln |sigma| + C
$$
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
Thank you so much!
– M Do
Nov 27 '18 at 16:45
add a comment |
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:
$$
frac{1}{sqrt{ 2 pi }}ln |sigma| + C
$$
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
Thank you so much!
– M Do
Nov 27 '18 at 16:45
add a comment |
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:
$$
frac{1}{sqrt{ 2 pi }}ln |sigma| + C
$$
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
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:
$$
frac{1}{sqrt{ 2 pi }}ln |sigma| + C
$$
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
answered Nov 27 '18 at 16:36
Emilio Novati
51.5k43472
51.5k43472
Thank you so much!
– M Do
Nov 27 '18 at 16:45
add a comment |
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
add a comment |
$begin{align}
int frac{1}{sqrt{2pisigma^2}} dsigma &=
frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
& =
frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
end{align}$
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
add a comment |
$begin{align}
int frac{1}{sqrt{2pisigma^2}} dsigma &=
frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
& =
frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
end{align}$
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
add a comment |
$begin{align}
int frac{1}{sqrt{2pisigma^2}} dsigma &=
frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
& =
frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
end{align}$
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
$begin{align}
int frac{1}{sqrt{2pisigma^2}} dsigma &=
frac{1}{sqrt{2pi}} int frac{1}{|sigma|} dsigma\
& =
frac{1}{sqrt{2pi}} operatorname{ln}(|sigma|)+c
end{align}$
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
edited Nov 28 '18 at 6:51
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
answered Nov 27 '18 at 16:45
Thomas Shelby
1,445216
1,445216
add a comment |
add a comment |
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