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Use of big mysterious shortcuts in academic papers, example integration by parts

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1

1

$begingroup$

I read a paper where the author did a very strange but valid integration by parts:

What I thought was unusual is the repeating occurrence of $(q_T - q_t)$ in essentially all the terms (ignore $h$). Less unusually (sadly), I had absolutely no clue how the author obtained the expression.

After (embarrassingly) many hours, I figured out the author used a very unusual "definite integration by parts" shortcut:

$$
int_a^b udv = -[(v(b)-v)u]Big{|}_a^b + int_a^b (v(b) - v) du
$$

What I would like to do is not be so troubled by such shortcuts in the future, and would greatly appreciate your advice and knowledge regarding the use of such shortcuts in academic papers, etc.

• Is this a well-known expression somewhere or in some field?




• Or is this just a bit of hidden manipulation to get nice aesthetic
  properties?




• Is there an expectation that the reader won't be confused by the use
  of such a shortcut?




• Is it a bad sign that it confused me (a grad student) so much and took me a few hours to get around it?

Thanks for your kind responses.

integration soft-question problem-solving

share|cite|improve this question

edited Mar 15 at 10:38

OrangeSherbet

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

$endgroup$

add a comment | 

1

1

$begingroup$

I read a paper where the author did a very strange but valid integration by parts:

What I thought was unusual is the repeating occurrence of $(q_T - q_t)$ in essentially all the terms (ignore $h$). Less unusually (sadly), I had absolutely no clue how the author obtained the expression.

After (embarrassingly) many hours, I figured out the author used a very unusual "definite integration by parts" shortcut:

$$
int_a^b udv = -[(v(b)-v)u]Big{|}_a^b + int_a^b (v(b) - v) du
$$

What I would like to do is not be so troubled by such shortcuts in the future, and would greatly appreciate your advice and knowledge regarding the use of such shortcuts in academic papers, etc.

• Is this a well-known expression somewhere or in some field?




• Or is this just a bit of hidden manipulation to get nice aesthetic
  properties?




• Is there an expectation that the reader won't be confused by the use
  of such a shortcut?




• Is it a bad sign that it confused me (a grad student) so much and took me a few hours to get around it?

Thanks for your kind responses.

integration soft-question problem-solving

share|cite|improve this question

edited Mar 15 at 10:38

OrangeSherbet

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

$endgroup$

add a comment | 

$begingroup$

I read a paper where the author did a very strange but valid integration by parts:

What I thought was unusual is the repeating occurrence of $(q_T - q_t)$ in essentially all the terms (ignore $h$). Less unusually (sadly), I had absolutely no clue how the author obtained the expression.

After (embarrassingly) many hours, I figured out the author used a very unusual "definite integration by parts" shortcut:

$$
int_a^b udv = -[(v(b)-v)u]Big{|}_a^b + int_a^b (v(b) - v) du
$$

What I would like to do is not be so troubled by such shortcuts in the future, and would greatly appreciate your advice and knowledge regarding the use of such shortcuts in academic papers, etc.

• Is this a well-known expression somewhere or in some field?




• Or is this just a bit of hidden manipulation to get nice aesthetic
  properties?




• Is there an expectation that the reader won't be confused by the use
  of such a shortcut?




• Is it a bad sign that it confused me (a grad student) so much and took me a few hours to get around it?

Thanks for your kind responses.

integration soft-question problem-solving

share|cite|improve this question

edited Mar 15 at 10:38

OrangeSherbet

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

$endgroup$

share|cite|improve this question

edited Mar 15 at 10:38

OrangeSherbet

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

share|cite|improve this question

edited Mar 15 at 10:38

OrangeSherbet

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

asked Mar 15 at 9:26

OrangeSherbetOrangeSherbet

22718

2 Answers
2

active

oldest

votes

3

$begingroup$

Sadly, there is no one recipe for addressing this issue. In my own personal experience, the depth of explanation is generally not the fault of the author but the restrictions of the Journal in how many pages/words are permitted. As such, you will very often see compressed working to accomodate such restrictions.

The best approach is to email the authors directly. I know that when I've received questions I'm more than happy to respond. And when I've asked questions, I've had nothing but positive experiences in the responses I receive.

share|cite|improve this answer

answered Mar 15 at 9:53

user619699

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great advice and thanks for mentioning your experiences. I haven't contacted many authors about questions in my short experience reading papers, so my fears are probably overinflated (fear they might find my question offensive or annoying).
  $endgroup$
  – OrangeSherbet
  Mar 15 at 10:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, not at all. If you email through with your work so far and are polite, I can not imagine you encountering any issues. Authors want their material to be read and understood, so you contacting means you've read it and want to learn more.
  $endgroup$
  – user619699
  Mar 15 at 10:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for last paragraph.
  $endgroup$
  – Paramanand Singh
  Mar 16 at 2:37
  

  
  
  
  

add a comment | 

1

$begingroup$

It's not a secret trick at all. If I ask you to choose $u,,v$ so that an integrand is $uv^prime$, after fixing $u$ your choice of $v$ isn't unique because a constant can be added to it. Let $v_0$ denote the choice for $v$ that was in your head; the author was thinking of $v_0-v_0(b)$ instead. In other words, they made $v$ unique with the convention $v(b)=0$.

Usually $v$ is chosen either to vanish at one end or the other, or to be "the obvious" option (e.g. $x^2$ instead of $x^2+5$, regardless of the integration limits.) When an author doesn't tell you what they did, check those three options until one makes sense. Usually the upper limit won't be used to "calibrate" $v$ in the way it was here, although I think financial analysis might be exceptional in that regard because of how often the at-$T$ behaviour matters.

share|cite|improve this answer

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Spot on analysis! I see now that only changes in $v$ matter, like you said. Intuitively for this finance context, $v$ is the number of shares being owned, and this is a calculation of transaction cost, so only the derivative matters. I guess adding that constant is the viewpoint of the "planner", where they have a future target $v(T)$ and are planning how to get there.
  $endgroup$
  – OrangeSherbet
  Mar 15 at 20:03
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

It's not a secret trick at all. If I ask you to choose $u,,v$ so that an integrand is $uv^prime$, after fixing $u$ your choice of $v$ isn't unique because a constant can be added to it. Let $v_0$ denote the choice for $v$ that was in your head; the author was thinking of $v_0-v_0(b)$ instead. In other words, they made $v$ unique with the convention $v(b)=0$.

Usually $v$ is chosen either to vanish at one end or the other, or to be "the obvious" option (e.g. $x^2$ instead of $x^2+5$, regardless of the integration limits.) When an author doesn't tell you what they did, check those three options until one makes sense. Usually the upper limit won't be used to "calibrate" $v$ in the way it was here, although I think financial analysis might be exceptional in that regard because of how often the at-$T$ behaviour matters.

share|cite|improve this answer

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Spot on analysis! I see now that only changes in $v$ matter, like you said. Intuitively for this finance context, $v$ is the number of shares being owned, and this is a calculation of transaction cost, so only the derivative matters. I guess adding that constant is the viewpoint of the "planner", where they have a future target $v(T)$ and are planning how to get there.
  $endgroup$
  – OrangeSherbet
  Mar 15 at 20:03
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

It's not a secret trick at all. If I ask you to choose $u,,v$ so that an integrand is $uv^prime$, after fixing $u$ your choice of $v$ isn't unique because a constant can be added to it. Let $v_0$ denote the choice for $v$ that was in your head; the author was thinking of $v_0-v_0(b)$ instead. In other words, they made $v$ unique with the convention $v(b)=0$.

Usually $v$ is chosen either to vanish at one end or the other, or to be "the obvious" option (e.g. $x^2$ instead of $x^2+5$, regardless of the integration limits.) When an author doesn't tell you what they did, check those three options until one makes sense. Usually the upper limit won't be used to "calibrate" $v$ in the way it was here, although I think financial analysis might be exceptional in that regard because of how often the at-$T$ behaviour matters.

share|cite|improve this answer

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Spot on analysis! I see now that only changes in $v$ matter, like you said. Intuitively for this finance context, $v$ is the number of shares being owned, and this is a calculation of transaction cost, so only the derivative matters. I guess adding that constant is the viewpoint of the "planner", where they have a future target $v(T)$ and are planning how to get there.
  $endgroup$
  – OrangeSherbet
  Mar 15 at 20:03
  
  
  

  
  
  
  

add a comment | 

$begingroup$

It's not a secret trick at all. If I ask you to choose $u,,v$ so that an integrand is $uv^prime$, after fixing $u$ your choice of $v$ isn't unique because a constant can be added to it. Let $v_0$ denote the choice for $v$ that was in your head; the author was thinking of $v_0-v_0(b)$ instead. In other words, they made $v$ unique with the convention $v(b)=0$.

Usually $v$ is chosen either to vanish at one end or the other, or to be "the obvious" option (e.g. $x^2$ instead of $x^2+5$, regardless of the integration limits.) When an author doesn't tell you what they did, check those three options until one makes sense. Usually the upper limit won't be used to "calibrate" $v$ in the way it was here, although I think financial analysis might be exceptional in that regard because of how often the at-$T$ behaviour matters.

share|cite|improve this answer

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250

$endgroup$

share|cite|improve this answer

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250

answered Mar 15 at 10:47

J.G.J.G.

32.1k23250