Talk:Differentiation under the integral sign

message: Pretty sure second step of example 2 is wrong.  —Preceding unsigned comment added by 129.110.241.33 (talk) 23:00, 9 April 2010 (UTC)

message: Is there really a point to having the Feynman quote at the top of the page?  If this was not a reference work, but instead a more literary medium (like a math book), it would be acceptable, but as it stands, I believe it is horribly out of place.
:Not only that, but if it's the same quotation still in the article I don't think it belongs at all anywhere in the article, except perhaps as a footnote.  For example:
: This technique is not widely taught, and there is a long-winded anecdote that can be told that kind of relates[1].
:—DIV (138.194.11.244 (talk) 07:50, 6 January 2012 (UTC)
thread_title:  Beginning Quote 

message: Consider the function defined by  f(x,t) = 0 if  t = 0
and by 
:  f(x,t) =  \frac{1}{\sqrt{t}} sin \frac{x}{\sqrt{t}} 
if  0< t \le 1.
This is integrable with respect to  t for  t \in [0,1]
so the function
:  F(x) = \int_0^1 f(x,t) dt 
is at least defined. But
 \frac{\partial f}{\partial x}(x,t) = \frac{1}{t} cos \frac{x}{\sqrt{t}} 
isn't integrable, so one of the integrals appearing in the result is undefined. And
in fact F is not differentiable.
So at the least we must add the hypothesis that F be differentiable. 
But I doubt if this will be enough to make the result true. If we allow
t to vary over an infinite inteval then there is the following counter example. Let
:  f(x,t) = x^3 \exp (-x^2t) 
for all  x and for  t \in [0,\infty).
Then
:  F(x) = \int_0^\infty f(x,t) dt = x,
so  F is differentiable, with  F'(0) = 1. But
:  \int_0^\infty  \frac{\partial f}{\partial x}(0,t)  dt = \int_0^\infty 0 dt = 0 
so the derivative of F at 0 is not given by differentiating under the integral sign. 
 88.105.188.214 05:23, 1 December 2006 (UTC)
thread_title:  Some other concern 

message: This article treats as if x and t are two independent variables, and together x and t define a plane.
Does everything still hold if x is a function of t?  —Preceding unsigned comment added by 69.134.241.125 (talk) 09:56, 8 April 2010 (UTC)
thread_title:  Does everything still hold if x is a function of t? 

message: I think this might be an error. Shouldn't this text:
*"A form of the mean value theorem, \int_a^bf(x)dx=(b-a)f'(\xi)\,, where a<\xi<b\,,"
actually be:
*"A form of the mean value theorem, \int_a^bf(x)dx=(b-a)f(\xi)\,, where a<\xi<b\,," ?
ssepp(talk) 14:10, 1 October 2007 (UTC)
replies: text: I made the change. ssepp(talk) 18:31, 4 October 2007 (UTC)
thread_title: mean value theorem error?

message: I would suggest that the example of integration with limits not depending on 'x'(for fixed real numbers) be treated in the beginning. The present theorem is too general I believe for the beginning section. Perhaps the article can describe the different formulations in increasing generality? Ulner 20:28, 10 November 2007 (UTC)
thread_title:  Simple example earlier in the article 

message: I have a degree in Physics, minor in math and I'm trying to learn this technique.  I'm trying to do Example 1, and I follow it up to the integral sign where it suddenly integrates it in 1 step.  I'm not seeing how that step follows, sorry maybe I'm just slow, but I really don't get how that step actually worked.  Did the person use a table? Mathematica?  Is there some technique being used that I'm not seeing?  Is there a lightning bolt or a ninja sneaking in and doing the work to finish the integral?  —Preceding unsigned comment added by 24.117.47.160 (talk) 22:43, 26 December 2007 (UTC)
thread_title:  Lightning Bolt or Ninjas Missing for the Intregal Step in Example 1 ? 

message: I am not familiar with the subject, but I was trying to make a careful read of the definition and I don't understand the constant use of x_0\leq x\leq x_1.  I suspect that is supposed to be a(x)\leq x \leq b(x)?  Is that right? 66.216.172.3 (talk) 15:22, 25 March 2008 (UTC)
thread_title: Lead Definition

message: I just edited this section to include the dx after the integral signs in the formulae. I'm pretty sure these are correct and that the statements were meaningless before - but just wanted to check.  —Preceding unsigned comment added by A good brew (talk • contribs) 09:44, 21 July 2008 (UTC)
thread_title:  Derivation - integral parameters 

message: I might not know what I'm talking about here, but it occurred to me that in none of the examples are the limits on the integrals non-constant functions.  Does someone want to add an example where you have non-trivial a(x) and b(x) as the integration limits?  Or does this rarely come up in practice?  Ceresly (talk) 23:00, 5 September 2009 (UTC)
thread_title:  Examples don't use the full power of the technique. 

message: Great article, but the lede does not actually define what the part of the initial definition is differentiation under the integral sign (or how it gets its name); it is mostly obvious from context, but if one reads closely, the term is just defined as a useful technique; it could use a final recap at the end of the paragraph saying "This set of steps [? or perhaps better: the technique of expanding out partial derivatives with respect to b and to a] is called Differentiation under the integral sign".  Thanks!  -- Michael Scott Cuthbert (talk) 16:11, 8 July 2010 (UTC)
thread_title:  Great article, but lede needs definition 

message: Firstly, nowhere (that I could find) in this article are the restrictions on whether you can or whether you can't integrate under the integral sign.  The article does talk about conditions under which you can differentiate under the integral sign, but I don't see those conditions written down.  As far as I know there are very specific conditions on whether you are allowed to integrate under the integral sign and they must all be verified before you can do it (cf. Section 8.2 in Lang (1993) "Real and functional analysis," Graduate Texts in Mathematics, volume 142, Third Edition, Springer-Verlag, New York), namely the function must be measurable, the partial derivative must exist and also there must exist an L^1 integrable majorant, which in general can be hard to find. HowiAuckland (talk) 02:01, 6 September 2010 (UTC)
thread_title:  Worried about the rigour of this article 

message: I am correcting the sign on the  \frac{\partial F}{\partial a} term of the formula near the beginning. It should be a plus, due to the chain rule. The final result picks up a minus sign on calculating the actual value of this derivative (note the "Derivation of the principle of blah blah..." section).  — Preceding unsigned comment added by 84.38.9.75 (talk) 04:43, 14 July 2011 (UTC)
replies: text: I disagree.  The minus sign comes from evaluating F at the lower limit of the integral.  I have changed it back. Garethvaughan (talk) 11:11, 16 July 2011 (UTC)
replies: text:  Right.  I've thought about it and I see I'm wrong, but I think there is some confusion regarding notation.  As it is currently defined, F(x) is a function solely of x: not a nor b.   And, moreover, the current definition is contrary to the common notation as the anti-derivative (a convention that IS also used in this article - see the derivation section).  I think clarifying this confusion would complicate things, and anyway, what the equation in question provides is already adequately provided in the derivation section.  So I've removed the first line of the equation.Garethvaughan (talk) 02:24, 17 July 2011 (UTC)
thread_title:  Stop misusing the chain rule 

message: The Reynolds theorem given here is correct for a material volume, which has Lagrangian boundaries.  It is not correct for a volume with arbitrary boundary velocity.  What Reynolds actually wrote had the boundary velocity in the surface integral.  See the talk page on Reynolds transport theorem.  Garethvaughan (talk) 02:32, 17 July 2011 (UTC)
thread_title:  On Higher Dimensions 

message: The prime (Lagrange's) notation, i.e. b'(x) etc. is inconsistent with the notation used in the rest of the article, as is the dot (Newton's) notation signifying the time derivative. At least in the case of using the dot notation its use was given as defined in the discussion.
I would suggest updating the page to change the notation to make it more in line with the rest of the article, or at the very least add a "Where b'(x) is the total derivative of b with respect to x." to the definition.
I know use of the prime notation is standard, but as a self contained article, such additions are a nicety. 
http://en.wikipedia.org/wiki/Derivative
--Tardyon (talk) 16:57, 24 July 2011 (UTC)
thread_title:  Inconsistent notation for the total derivative 

message: There appears to be a lot of duplication with that article.Dfeuer (talk) 18:18, 7 October 2012 (UTC)
replies: text: Totally agree. Both pages discuss the same exact topic.  — Preceding unsigned comment added by 99.241.86.114 (talk) 13:26, 1 January 2013 (UTC)
replies: text: I agree. The only theorem discussed in this is the a generalised version of the Leibniz integral rule, which doesn't contain variable limits, but the article proves the theorem with variable limits. thedoctar (talk) 07:16, 21 May 2013 (UTC)
replies: text: As far I am concerned, whoever has the initiative should go ahead and do it.  We seem to have consensus here.KlappCK (talk) 21:44, 14 March 2014 (UTC)
thread_title:  Merge with [[Leibniz integral rule]]? 

message: The main result (theorem) at the beginning is incomplete.  — Preceding unsigned comment added by Prof. Globi (talk • contribs) 13:53, 26 December 2014 (UTC)
thread_title:  Incomplete main result