Talk:Integration by parts

(Untitled section)

An alternative notation has the advantage that the factors of the original expression are identified as f and g

Why is that an advantage? It seems arbitrary to say it's better to call them f and g than to call them u and v. Michael Hardy 00:27, 15 Nov 2003 (UTC)

That's not what I meant. The "classic" notation tells you what to do when you are integrating "f(x)g'(x)". The "alternative" integrates "f(x)g(x)". See? -- Tarquin 18:58, 15 Nov 2003 (UTC)

Justification

Could we chop of the justification section, since the rule is now justified at the beginning? Also I put the definite integral notation first, because the semantics of the indefinite integral form are a lot less clear. It is still in there since it may still be part of the calculus curriculum. Need to explain bound and free variables.

The distribution comment at the end is informative. Maybe more can be said. The distribution article I think should also be rewritten, having two parts

• a 1-dimensional dsitribution theory and
• a general theory for open sets and manifolds.

CSTAR 21:15, 11 May 2004 (UTC)

A practical note

I have removed this section from the end of the article. I think it probably belongs on Wikibooks, being a suggestion for students in second-semester calculus. I also did my best to clean up the writing style somewhat, to eliminate use of second person ("you") and incorrect word usage (e.g. the word "derivate", which may or may not actually mean "derivative", but if it does, it's not standard by any means) - please see my changes here: [1]. Aerion//talk 22:07, 23 Dec 2004 (UTC)

In a general way, when an exponential or trigonometric function appears in the expression, it should be chosen as $v'$:

$\int f(x)e^{2x}\,dx={\frac {1}{2}}f(x)e^{2x}-{\frac {1}{2}}\int f'(x)e^{2x}\,dx\ \dots$

$\int f(x)\cos(4x)\,dx={\frac {1}{4}}f(x)\sin(4x)-{\frac {1}{4}}\int f'(x)\sin(4x)\,dx\ \dots$

A good way to choose U and dV is the mnemonic "DETAIL". D stands for differential, which is what is to be chosen (dV). The rest of the letters give the order of functions to consider. E stands for exponential. T is for trigonometric function. A stands for algebraic. I is for inverse trigonometric, and finally L is for logarithmic, which is typically a poor choice.

This is especially useful when $f(x)$ is a polynomial, since each consecutive derivative of f(x) is simpler, and eventually, is constant. In general, integration by parts if $f(x)$ is easy to differentiate. Otherwise, the substitution rule may need to be employed.

By contrast, logarithmic or inverse trigonometric functions should be chosen as $u$:

$\int f(x)\log(3x)\,dx=F(x)\log(3x)-\int F(x){\frac {3}{3x}}\,dx\ \dots$

$\int f(x)\arcsin(6x)\,dx=F(x)\arcsin(6x)-\int F(x){\frac {6}{\sqrt {1-(6x)^{2}}}}\,dx\ \dots$

The objective is to reduce the inverse trigonometric function to a fraction inside the integral. If the derivative contains a radical, trigonometric substitution may be useful.

ILATE rule

I have removed the text below, which is similar to the section I previously removed, seen above. It doesn't really fit the style of the rest of the article, and ought to be on Wikibooks instead, especially since it is clearly targeted for calculus students. :(Aerion//talk 23:03, 5 Feb 2005 (UTC)'

Alternately a bit handy rule is ILATE rule. This rule helps to decide which function must be used as a substitute for f and which for g. This rule works fine in most cases, making the calculations easier.
KEY:
I = Inverse functions
L = Logarithamic functions
A = Algebric functions
T = Trigonometric functions
E = Exponential functions
So if you get cos(x) and log(x) in the product then, according to the rule take "log(x)" as equivalent to "f" in the equation, while "cos(x)" takes the position of second function.
P.S. Sometimes you need to fiddle around and use the LIATE instead in some cases.

Let's keep the LIATE section (or ILATE, but a quick web search reveals that LIATE is more popular). It's extremely handy for someone looking up integration by parts (as I just did). --Ben Kovitz 12:37, 1 October 2005 (UTC)

That section was written by an anon. I did some formatting on it. I would incline on keeping it, as it is at the very bottom of the article and therefore not interfering with anything else, and it seems that it might be indeed useful. Oleg Alexandrov 01:53, 3 October 2005 (UTC)

It's not at the bottom any longer... Melchoir 08:45, 8 December 2005 (UTC)

ILATE - WTF?!? LIATE is not only "more popular" on the web; it's in every current calculus text I've seen. Also, the example about "nontrivial polynomial splits" is self-serving and unnecessary. Use u-substitution and then parts as ever. But, of course, LUNCH knows best, so no further edits are warranted. "Concensus"...right. 209.129.254.58 22:47, 5 September 2007 (UTC)

"It's in every current calculus text I've seen." Can you name one? Or, better yet, several?

I have three undergrad calculus texts sitting right in front of me: Stewart's "Calculus", Simmons' "Calculus with Analytic Geometry", and Leithold's "Calculus with Analytic Geometry". These are all popular texts. None of them have any rule of thumb listed (neither LIATE nor ILATE). They all give general advice on how to split the integrand into parts, but do not codify that advice into a hard rule. IHMO, that's how it should be.

However, I recognize that many people want a pat rule; many editors here recognize that need, too. Thus the section hasn't been rewritten to eliminate the rules of thumb. In fairness, both ILATE and LIATE are mentioned. However, there isn't a rational way of favoring one over the other -- other than a popularity contest. If you want to start a systematic search of textbooks, you're welcome to post the results here. Otherwise, your unconstructive potty talk and name calling isn't welcome, and you should just go away. Lunch 00:41, 6 September 2007 (UTC)

I stopped by the library on the way home to look at more calculus textbooks. I looked through Apostol; Salas and Hille (now with Etgen); Thomas and Finney; Edwards and Penney; and Spivak. None of them make any mention of the LIATE, ILATE, or LIPET rules of thumb; all they do is give general advice on picking parts. All of these are popular books.

While we're at it, of the eight textbooks I have thumbed through, only one (Thomas and Finney) mentions tabular integration. Lunch 03:16, 6 September 2007 (UTC)

"Go away"? C'mon, I was having such a fine time provoking you! Oh well.

Actually, I was beginning to wonder if you'd respond (other than on the history log)...but I knew you would, which is why I chose you in the first place. I must say that you exceeded my expectations by a remarkable margin; congratulations on wasting time in "winning" such an important (and ultimately one-sided) argument. "I stopped by the library"...are you kidding me? Plus a potshot at tabular integration? Whoa. I don't care what you call the rule, what books it is (or isn't) in, how humble your opinion is...whatever. I'll go away, indeed - as far from your type as I can get. Try to enjoy life, and thanks for your cooperation. 207.69.139.134 06:23, 7 September 2007 (UTC)

Wowee, zing! You won that one, you anonymous IP! In your sick little mind, someone actually spending time to improve the Internet and respond to your stupid ass complaints is actually a loser. Gee... I wonder how long you can go through life with that mentality.

FWIW, I appreciate Lunch's efforts in researching for this article and finding good sources to justify both inclusion and exclusion of content. Dcoetzee 14:05, 13 November 2007 (UTC)

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Multidimensional case

I don't have a proof at the moment, but the textbook I'm looking at has this (using $f_{i,j}$ to mean the value of the ith component of the derivative in the j direction of f and the Einstein summation convention).

$\int _{\Omega }w_{i}\sigma _{ij,j}\,d\Omega =-\int _{\Omega }w_{i,j}\sigma _{ij}\,d\Omega +\int _{\Gamma }w_{i}\sigma _{ij}n_{j}\,d\Gamma$

Where Γ is the perimeter of Ω and n is the outward normal. This should be added (with a clearer notation). I'm not sure if this is related to the divergence theorem. —BenFrantzDale 20:22, 18 October 2005 (UTC)

Something like this definitely must be in the article. I don't recall for now how this is called, but I think that the divergence theorem is a particular case of this. From what I know, the most general version of this formula is

$\int _{\Omega }{\frac {\partial u}{\partial x_{i}}}v\,dV=\int _{\partial \Omega }uv\,\mathbf {n} _{i}\,dS-\int _{\Omega }u{\frac {\partial v}{\partial x_{i}}}\,dV$

from where one gets for example

$\int _{\Omega }\nabla u\cdot \mathbf {v} +u\nabla \cdot \mathbf {v} \,dV=\int _{\partial \Omega }u\mathbf {v} \cdot \mathbf {n} \,dS$

which is a very useful integration by parts formula in higher dimensions, which implies for example the Ostrogradsky-Gauss theorem and probably other cases of the Stokes theorem. Oleg Alexandrov (talk) 00:59, 19 October 2005 (UTC)

I agree that especially the last formula above should be in text. It should also be noted, that the

surface term on the right commonly goes to zero when integrating over all space and u, v "behave well". This is very often used in physics calculations, and so common than some textbooks even gloss over it. gbrandt 13:33, 26 January 2006 (UTC)

Please revise the "recursive formulation" section

I suggest the author of section 4 “recursive formulation” http://en.wikipedia.org/w/index.php?title=Integration_by_parts&action=edit&section=4 of the article Integration_by_parts to revise the recursive formula that is suspected to be mistaken or notation needs to be further clarified. It seems to differ from the corresponding formula in Mathworld: http://mathworld.wolfram.com/IntegrationbyParts.html. Consider inclusion of another term on RHS, integral of product of some functions.

Tablature Method (possible to add)

Can we add the talbulture method, a simplified way of using integration by parts. It only works for a repeating function multiplied by a non repeating function. By repeating, i mean a function that doesent settle in on zero after a fininte amount of derivatives. Can we add this to the main article? or is it somewhere else on wikipedia? Swerty 21:34, 27 March 2006 (UTC)

How about creating a new tablature method article? Oleg Alexandrov (talk) 16:25, 28 March 2006 (UTC)

Failed to parse

Can someone with more time on their hands than me please fix the code? Every equation on the page currently reads "Failed to parse" in big red letters!

They're all working fine now. Michael Hardy 19:21, 19 April 2006 (UTC)

Q: Recursive formulation

In the section Recursive formulation, what are the $v_{i}\,$? Are they distinct?

Thanks!

cultural references

I removed the cultural references becuase it is of no significance to anyone who reads this article, and its sole entry is exceedingly minute. It is not in keeping with an encyclopedia; if junk like that is allowed to remain, the content will become diluted. — Preceding unsigned comment added by 128.104.160.172 (talk)

I reverted your removal. I don't think it's "junk". Popular culture has a place in mathematics as does mathematics in popular culture. This bit of trivia answers the question, "What was on the chalkboard in the movie? Was it just a bunch of nonsense or was it "real" math?" See the "Trivia" section of the snake lemma article for another example. Lunch 02:34, 23 February 2007 (UTC)

It may be useful to provide more details and context within the movie. Right now, it bears no significance for people who have not seen the movie. How was the method of integration featured in the movie? Was it on the chalkboard? As fodder, or was there an actual reason? This information is not listed in Stand and Deliver, so any viewers who wondered what was on the chalkboard has no idea by looking at that article. Pomte 05:15, 23 February 2007 (UTC)

Wow, in that case my criticism was understated. It's not even featured, it was on a chalkboard at some point. Ask yourself: if this is the standard of inclusion, what will become of this encyclopedia? Popular culture (which this does not nearly reach the threshold of) has no place in a mathematics encyclopedia article. BTW, it would need a citation anyway. —The preceding unsigned comment was added by 128.104.34.126 (talk) 23:26, 8 March 2007 (UTC).

,dx in nested notation

I have four issues with the arguments in the nested notation integration formula, http://upload.wikimedia.org/math/e/c/7/ec780362cd9a4a26c6a6302d83263d04.png

1. Shouldn't be integrating with respect to an argument since functions aren't of an argument
2. If arguments are implied, what is the point of adding the argument you're integrating w.r.t? Isn't it also implied?
3. This is the only formula on the whole page where the functions are not of an argument. Why not make it consistent?
4. If the dx's were removed, the formula would be more readable. If the function arguments were added, the formula would be less confusing. In either case, the formula would be more consistent. In the prior case, it would at least be internally consistent; in the latter case, it would be consistent with the rest of the article as well.

So why not do it? Arrenlex 03:43, 15 March 2007 (UTC)

The formulas above and below are of the same form, without arguments. All the other integrals in that section show the differential. I think the dx on the last integral makes the nesting less confusing. I don't know what the usual notation is, and I don't have a strong opinion on this. –Pomte 03:50, 15 March 2007 (UTC)

ILATE or LIPET?

This article suggests the ILATE rule of thumb. However, anyplace i've ever heard it, it is called LIPET, for Log, Inverse Trig, Powers of x, e, Trig functions. I personally have never heard of ILATE, and seeing as it lacks sources, i'd like to get some sort of consensus then change it. If anybody wants a citation for my information, it can be found at a page on UCSD's website. Firestorm 18:53, 5 June 2007 (UTC)

Mnemotechnics

I deleted the section entitled Mnemotechnics for a number of reasons:

• It was not referenced to a reliable English-language source. For most mathematical content, verifiability is rather easily met provided the reference list is a good one. However, these mnemonics seem unlikely and obscure. (See my next comments
• The mnemonic is more difficult than the rule itself. In fact, if I found out that my own students were using such a mnemonic to remember integration by parts, I would strongly discourage them. The reason is that the "correct" way to remember integration by parts is by the fundamental theorem of calculus:

$\int _{a}^{b}u\,dv+v\,du=\int _{a}^{b}d(uv)=u(b)v(b)-u(a)v(a)$

a formula which really should be in the article, instead of the silly mnemonic.

• The final issue is that of WP:WEIGHT. The section was way up towards the top of the article, but in comparison to the rest of the content is barely deserving of a footnote. The prominent placement of such peripheral material in the article assigns undue weight to it.

--siℓℓy rabbit (talk) 13:05, 6 September 2008 (UTC)

Error

Moving this comment by User:204.153.78.195 from the article to here:

In this case in the last step it is necessary to integrate the product of the two bottom cells obtaining:

$\int e^{x}\cos x\,dx=e^{x}\sin x+e^{x}\cos x-\int e^{x}\cos x\,dx$

which is then solvable in the usual way. <<<<< this is not true because the integration left over is the same as the indefinate integral that you started with which would cause an indefinate number of integrations using tabular method. —Preceding unsigned comment added by 204.153.78.195 (talk • contribs)

The anonymous commentator clearly missed the point. "Solvable in the usual way" does not mean "keep going until you've evaluated the integral". That's not solving; that's evaluating. Michael Hardy (talk) 20:41, 12 October 2010 (UTC)

Visualisation

Would the following visualization be helpful for this article?

The area of the blue region is

$A_{1}=\int _{y_{1}}^{y_{2}}f(g^{-1}(y))dy=\int _{t_{1}}^{t_{2}}f(t)g'(t)dt$ (substitution $y=g(t)$)

The area of the red region is

$A_{2}=\int _{x_{1}}^{x_{2}}g(f^{-1}(x))dx=\int _{t_{1}}^{t_{2}}g(t)f'(t)dt$ (substitution $x=f(t)$)

The total area $A_{1}+A_{2}$ is equal to the area of the bigger rectangle, $f(t_{2})g(t_{2})$ minus the area of the smaller one, $f(t_{1})g(t_{1})$,

$\overbrace {\int _{t_{1}}^{t_{2}}f(t)g'(t)dt} ^{A_{1}}+\overbrace {\int _{t_{1}}^{t_{2}}g(t)f'(t)dt} ^{A_{2}}={\biggl .}f(t)g(t){\biggl |}_{t_{1}}^{t_{2}}$

This visualisation also explains why integration by parts is helpful to integrate an inverse, $f^{-1}(x)$, if you know antiderivative of the original function, $f(x)$. What do you think?(Igny (talk) 05:03, 4 June 2009 (UTC))

It is helpful - sorry to not include it earlier, just been busy and admittedly didn't understand it at a first read (and did not join wikipedia untill ~ september 2011)... thanks and well done! =) F = q(E+v×B) ⇄ ∑_ic_i 12:16, 18 May 2012 (UTC)

Thank you. I based this visualization on a spacific example of (imagine this picture here)

$\int _{1}^{x}\ln \xi d\xi +\int _{0}^{\ln(x)}e^{y}dy=x\ln x$

for x > 1. (Igny (talk) 23:22, 11 June 2012 (UTC))

A rule of thumb

While an engineering student at the University of Illinois in Chicago, 7/23/1980, I discovered a procedure for handling "integration by parts". It was frustrating having to guess which half to integrate. I was preparing for a calculus exam and was doing problem after problem and I started to sense a pattern in my solutions, but I couldn’t bring it to the fore of my mind. I knew if I continued to solve problems, the pattern would become obvious, after another hour it did. I will note the method briefly, anyone familiar with basic calculus should have no trouble following my argument. I presented this procedure to a number of the professors at the university and none could find a problem that couldn't be solved. My math professor was to have published it in the Ramanujan Journal; whether this was ever done I don't know. I would appreciate if anyone would try and find a problem that cannot be solved by this procedure and let me know! If not, it would be an honor to get recognition for this procedure. It might make solving these problems just a little easier. I still have my orginal notes dated and written in my Purcell Calculus book, third edition. 1. Choose the u value so that when you differentiate it will go to 0 or a constant. Do this only if you can integrate the other half of the problem. If you can not, then make it your u value. 2. If there is only one term in the integration, make it your u value and it will be dx. 3. If neither value will ever go to 0 when differentiated and you can integrate both, they will flip/flop on the second integration by parts and you will get the divided by two equation.

Michael Parish parishmp@aol.com —Preceding unsigned comment added by 24.15.98.91 (talk) 21:47, 9 September 2010 (UTC)

Very sloppy article

From the point of view of a pure mathematician (i.e. me!) this article has numerous problems. I'd clean them up myself, but it's too big a job for me take on right now.

The main problem is that the article is written from the point of view of someone who considers an integral such as $f(x)=\textstyle \int _{1}^{x}1/x\,\mathrm {d} x$ as "unsolved" and the same thing written as $f(x)=\ln x$ as "solved". To a mathematician neither is solved or unsolved, they are just two quantities that happen to be equal. Perhaps some properties of f are more easily proved by using the first form, and some more easily using the second. I realise that, in practice, many people coming to this article will also be hoping to use it to "solve" integrals, but we should at least use proper language. Something like "integration by parts is often used to find antiderivatives in terms of elementary functions" would be ok.

Some other, mostly related, issues:

• The lead paragraph says that integration by parts "is a rule that transforms...". Despite the fact that the word "rule" is not uncommon in mathematics (e.g. "product rule"!) it really doesn't mean much, so "theorem" should be used here. "Transforms" is simply incorrect: it's like saying that "the equality $\sin ^{2}x+\cos ^{2}x=1$ transforms trigonometric functions to a constant". These things aren't being transformed, they are simply equal.
• The lead paragraph focuses on the use as a way to find antiderivatives. This is important to say here, but should not take up the whole lead and we shouldn't start with it (we should state what integration by parts is before we say how it often used). Something like "integration by parts is a theorem that relates the integral of a product of functions to the integral of their derivative and antiderivative". Or even just "is a theorem about integration stating" and then the formula would do.
• The sections that that describe how to use integration by parts to find antiderivatives (strategy, LIATE rule and examples) should be made subsections and grouped under a single section heading of "use to find antiderivatives", perhaps with a short introductory blurb.
• I have no idea what the subsection "Interchange of the order of integration" is doing here, especially under "examples". I think it should just be removed.
• The "recursive integration by parts formula" needs to make it clear that the stated formula is only true when the next derivative of u is zero.
• The "tabular integration by parts" seems to imply that it is a related but different technique to recursive integration by parts. In fact it is just a different layout for writing down the same thing. I question whether it should be included at all. If it must, it should at least be a subsection of the recursion section.
• I quite like the visualisation section proposed by Igny above on this talk page. I think we should include that.
• An example of the use of integration by parts to show something other than finding an antiderivative should be included. For example, using it to show the following inequality for the absolute integral of a function's Fourier transform might be a good one (this is the 1-dimensional version):

$\int _{-\infty }^{\infty }|{\hat {f}}(\xi )|\,\mathrm {d} \xi \leq {\frac {1}{2}}\int _{-\infty }^{\infty }|f(x)+f''(x)|\,\mathrm {d} x.$

Quietbritishjim (talk) 16:12, 4 April 2012 (UTC)

I quickly implemented some of these changes in this edit, as I agree with them: nice guidelines =) (and apologies for my stuck-up behaviour below). The Interchange of the order of integration section was removed since its off-topic. F = q(E+v×B) ⇄ ∑_ic_i 20:15, 14 May 2012 (UTC)

clean up

Main changes:

1. Rewrote the lead for clarity and compactness, it seems almost circular to integrate the product rule then derive it again using the fundamental theorem of calculus (that segemnt didn't lead anywhere anyway)
2. yet another article with a mix of italic and upright d's for differentials, changed all to italic for one-style consistency (WP:MOS),
3. why introduce u = f(x) and v = g(x)?? It just dpulicates the number of letters readers have to worry about, the fewer symbols there are the easier to keep track of what's happening. I changed all f to u and g to v, since they are easier + quicker to write, and fairly common in this context.
4. added a short note on products of n factors, why not mention?

F = q(E+v×B) ⇄ ∑_ic_i 08:44, 29 April 2012 (UTC)

Some of your changes are good (I agree there's no point using both u,v and f,g, and the d's should be consistent), but some are mathematically incorrect. Also, I think the lead should contain a statement of the theorem, and preferably a hint at its proof as it did before your changes, because "The lead should be able to stand alone as a concise overview" (WP:LEAD). Since mathematical correctness is more important than style I have reverted them completely for now, but I will try to merge the correct changes in when I have time in the next couple of days (if you don't do it yourself, of course).

Here are the mathematical issues: It does not make sense to "cancel the dx" as you've said, since these are just artifacts of the notation, not actual factors. I'm not sure that you've realised that $u'(x)$ is just another notation $du(x)/dx$, so your argument using $du/dx\,dx=du$ was circular, and in any case "$du$" is just a shorthand notation for $du/dx\,dx$ (see the lead section of integration by substitution), so all three things are trivially equal. You've also removed the mention of the fundamental theorem of calculus, and your comment above says something about having already derived it from the product rule. You need to use BOTH the product rule and the fundamental theorem of calculus to get integration by parts. The product rule gives you

$\int _{a}^{b}{\frac {d}{dx}}{\Bigl (}u(x)v(x){\Bigr )}\,dx=\int _{a}^{b}u'(x)v(x)\,dx+\int _{a}^{b}u(x)v'(x)\,dx$

and the fundamental theorem of calculus gives you

$\int _{a}^{b}{\frac {d}{dx}}{\Bigl (}u(x)v(x){\Bigr )}\,dx={\Bigl [}u(x)v(x){\Bigr ]}_{a}^{b}.$

Combining these and rearranging gives the theorem. I think you didn't see the need for the second one because you thought it followed by the "cancelling the dx's" in the first integral. The reason the lead only mentioned the product rule and not the fundamental theorem of calculus was because it was only giving a hint at the proof, not a complete argument, and the product rule is really the main idea. Quietbritishjim (talk) 12:47, 29 April 2012 (UTC)

Actually I wrote "du = u '(x) dx where du and dx are differentials" and u' was already defined as a derivative. Why is that circular? Isn't that particular point mathematically correct, the definition of a differential of a function? Above you wrote the circular du = (du/dx)dx which can confuse the derivative of u as a ratio of differential quantities. Yes I do know the notations are equivalant. It doesn't matter. Forget it... F = q(E+v×B) ⇄ ∑_ic_i 21:20, 30 April 2012 (UTC)

I tried again, incorperating your suggestions. F = q(E+v×B) ⇄ ∑_ic_i 22:37, 30 April 2012 (UTC)

I'm sorry I never replied to this! I was going to wait until I had time to really look at your changes properly before replying, and that time never materialised. You significantly improved this article; it was confusing and disorganised before, and now it's clear and well-organised. Thanks for your efforts! Quietbritishjim (talk) 14:42, 30 March 2013 (UTC)

Weasel Wording and the Infinite Congruence Theorem Section

According to user 88.88.245.32 the Infinte Congruence Theorem section contains weasel wording. I'm not sure if this is the correct use of the tag "weasel wording". Possibly one could remove the tag and instead look into citing some of the information in the section.

Of course, one could question the relevance of the entire section. I don't see how the cited link has anything to do with integartion by parts, nor does it have any mention of an "infinite congruence theorem". If someone really wants to talk about Infinite Congruence Theorems they could start an article on it or maybe contribute to the Linear congruence theorem article which is severly lacking in terms of verifiablity. — Preceding unsigned comment added by 128.175.16.246 (talk • contribs) 23:33, 17 February 2013‎ (UTC)

Relevance of entire section is indeed questionable - have removed it. If the infinite congruence theorem is a notable example of the implementation of integration by parts (which isn't short on applications) then it needs to be rewritten in a way that has some meaning and brings out the connection. It should also be referenced by a page that supports what is being said, not a blog post with no connection to anything in the section, let alone integration by parts. — Preceding unsigned comment added by 86.135.154.236 (talk) 19:22, 29 March 2013 (UTC)

If there's some dispute about that section, and it has no good reference, then of course the right thing to do is take it out. I also agree that if it's primary importance here is as an example of the use of integration by parts then that connection should be made explicit. I don't know anything about that theorem so I certainly can't fix it.

However, I disagree that the article "isn't short on applications". There is only one application shown: finding antiderivatives of specific integrals. This is a very important application, but it's also important to show that integration by parts has uses in pure mathematics in proving theorems too. The section removed seems like it would be a good example of that. I suggested another one on this talk page (bound on the absolute integral of the Fourier transform), although I haven't added it due lack of time and lack of a good reference (it seems to be a bit of a folk theorem). Perhaps "uses in pure mathematics" or something would be a good section to add, if we could collect a few applications. Or maybe "uses in Fourier analysis" would be better, since that's where many pure applications seem to be (another example: the Riemann–Lebesgue lemma). Quietbritishjim (talk) 14:37, 30 March 2013 (UTC)

I agree that some applications would be helpful. I was referring to the method of integration by parts as not being short on applications, not the article. Descriptions of applications would have to make at least some sense, though. The example removed was very definitely not a good example. (Apologies for unsigned edits earlier - I wasn't logged in) Bobathon71 (talk) 22:00, 30 March 2013 (UTC)

Derivation

Can the given derivation of by parts formula be replaced by this?

Alternative derivation

The theorem can be derived as follows. Suppose u(x) and v(x) are two continuously differentiable functions. The product rule states (in Leibniz’ notation): ${\frac {d}{dx}}\left[u(x)v(x)\right]=u(x){\frac {d}{dx}}\left[v(x)\right]+{\frac {d}{dx}}\left[u(x)\right]v(x).\!$ or,  $u(x){\frac {d}{dx}}\left[v(x)\right]={\frac {d}{dx}}\left[u(x)v(x)\right]-{\frac {d}{dx}}\left[u(x)\right]v(x)\!$ Integrating both sides of the equation with respect to x, gives: $\int u(x){\frac {d}{dx}}\left[v(x)\right]\,dx=\int \left[{\frac {d}{dx}}\left[u(x)v(x)\right]-v(x){\frac {d}{dx}}\left[u(x)\right]\right]dx$ or,  $\int u(x){\frac {d}{dx}}\left[v(x)\right]\,dx=\int \left[{\frac {d}{dx}}\left[u(x)v(x)\right]\right]\,dx-\int v(x)\left[{\frac {d}{dx}}\left[u(x)\right]\right]dx$ then by the definition of antiderivative,  $\int {\frac {d}{dx}}\left[u(x)v(x)\right]\,dx=u(x)v(x)$ this gives: $\int u(x){\frac {d}{dx}}\left[v(x)\right]\,dx=u(x)v(x)-\int v(x){\frac {d}{dx}}\left[u(x)\right]\,dx$ Since du and dv are differentials of a function of one variable x, and $du={\frac {d}{dx}}u(x)\,dx,dv={\frac {d}{dx}}v(x)dx,$ Therefore, $\int u(x)\,dv=u(x)v(x)-\int v(x)\,du$, which is the formula for integration by parts. The original integral ∫u v ′ dx contains v ′ (derivative of v); in order to apply the theorem, v (antiderivative of v ′) must be found, and then the resulting integral ∫v u ′ dx must be evaluated.

— Syεd Шαмiq Aнмεd Hαsнмi (тαlк) 19:43, 11 April 2013 (UTC)

I don't like this version. Can you explain why you think it's an improvement? Using antiderivatives instead of definite integrals gives a weaker theorem, and the way you've written the proof hides the fact that the fundamental theorem of calculus is still being used (it's needed to show that the antiderivatives of these functions exist). I suppose that both the definite integral and antiderivative versions should be stated. Maybe best would be (a) statement of both forms of the theorem (b) proof of the definite integral version (c) explanation of why the antiderivative version follows immediately. Quietbritishjim (talk) 22:14, 15 April 2013 (UTC)

So you did not like my version... That’s not a problem. I was just asking if this version with the indefinite integrals was better. If you do not like it, you may not use it or may modify it to your liking and include it there in the article or in any other way you wish! You are most welcome for that. I just asked for an opinion for this version.

However, I actually get your point and understand the problems with this. I would very much like it if you yourself change it as you think suitable, as I don’t think I’m capable enough myself to do anything as such. Hope you understand this. Well, thanks a lot for pointing my error. Regards...

— Syɛd Шαмiq Aнмɛd Hαsнмi (тαlк) 05:16, 16 April 2013 (UTC)

My reply wasn't meant to be an authoritative smack down! Sorry if it sounded that way, and please don't be put off of editing by me. In particular when I said "Can you explain why you think it's an improvement?", that was a serious honest question. If your main concern is that indefinite integrals as well as definite ones are discussed in the "theorem" section, I agree. I am happy for addition (perhaps with reorganisation), it's just didn't want your changes to replace the current version. The definite integral version is more important for theoretical purposes so it's important that it's there too. Quietbritishjim (talk) 16:43, 17 April 2013 (UTC)

No! You need not apologise... I did not decide to give up editing due to that reply. I would never have changed this article myself anyway (at least for the present). You see, I’m still just a Matriculate (having a certificate equivalent to a GCSE in Britain now), when many people here are PhD’s. That what was I meant by my incompetence. And that is why I said you are most welcome to come forward and edit the article yourself. However, as I have already said, I see the problem of addition and reorganisation here. Better go ahead and edit it yourself accordingly (for best results)... Regards.

— Syɛd Шαмiq Aнмɛd Hαsнмi (тαlк) 18:43, 17 April 2013 (UTC)

@Raycheng200: Can you explain this edit [2] please? - SindHind (talk) 09:37, 4 September 2015 (UTC)

@SindHind: Hi, I am Raycheng200. In my opinion, indefinite integral is the "general" form of definite integral. If we prove the theorem "integration by parts" by indefinite integral first, we don't need to prove it again by definite integral. For example, I integrate 2x dx, and get x^2. Now, I am using the indefinite integral. And the same thing applies on the definite integral. which also contains upper limit and lower limit of the integral. Raycheng200 (talk) 14:44, 31 October 2015 (UTC)

@Raycheng200: Not really. The "indefinite integral" is so called because it is not a single function. It is some arbitrary member of the family of functions that have the same derivative. So, if you integrate two equal functions you don't necessarily get two equal functions. So the reasoning you have applied is not valid. - SindHind (talk) 23:12, 11 November 2015 (UTC)

Applications to pure mathematics

I said above that I'm disappointed that the only applications mentioned in this article are in finding expressions for antiderivatives, rather than some of the (many) theoretical applications. I added a section with two pure applications. It's not much, but a start. Any comments are appreciated. Quietbritishjim (talk) 02:36, 16 April 2013 (UTC)

Generalized to Sobolev functions?

I'm pretty sure the requirement that u and v be continuously differentiable is too strong and that the integration by parts formula is true when one or both functions are Sobolev functions of a certain type. I am not sure enough of a good sufficient/necessary condition to make the change. This seems to be done in the "Higher Dimensions" section but it should be done for the one-dimensional case separately. Gsspradlin (talk) 00:34, 2 July 2013 (UTC)

Differentials

What does $\int u\,dv$ mean? $u$ is not a function of $v$. - SindHind (talk) 09:48, 4 September 2015 (UTC)

It means $\int uv'\,dx,$ since $dv=v'\,dx.$ Which is clearly written in the article. Boris Tsirelson (talk) 10:22, 4 September 2015 (UTC)

Indeed, that has been said. But it assumes that dv is an expression that one substitute for. Is it an expression? Does it have a meaning of its own? More precisely, the differential page has a meaning for dv as an expression. But the antiderivative page doesn't. So, it seems as if you are mixing up two different notations of dv. - SindHind (talk) 12:40, 4 September 2015 (UTC)

From the article: "If u = u(x) and du = u′(x) dx, while v = v(x) and dv = v′(x) dx, then"... This is not about "expressions" (is there such mathematical notion? after all, only countably many functions have "personal names", other just exist in the set of all functions), this about two functions u and v.

From the article: $\int u(x)v'(x)\,dx=u(x)v(x)-\int v(x)\,u'(x)dx$ or more compactly: $\int u\,dv=uv-\int v\,du.$

The "more compact" notation is formally just a (convenient and widely used) shortcut for the "official" notation. But intuitively, one may think in terms of infinitesimal differentials (if one likes). Boris Tsirelson (talk) 13:50, 4 September 2015 (UTC)

See expression (mathematics), especially the section on "syntax". The $dv$ in $\int u\,dv$ is part of an expression, something like a closing parenthesis. You can't substitute it with something else. But I think I understand now what is being done, even though it is highly improper. - SindHind (talk) 15:38, 4 September 2015 (UTC)

Well, I understand your attitude, too. Still, this "highly improper" notation is convenient and widely used; one just has to understand that it does not fit (and is not intended to fit) the formal framework that you mean. By the way, the notation $\int _{A}f$ is also widely used (but not in elementary textbooks) for the integral of a function (not expression) f over a set A. In this notation, $\int _{[0,\pi ]}\sin =2,$ which may look strange for the beginner, but is correct. And, see also Riemann–Stieltjes_integral#Properties_and_relation_to_the_Riemann_integral. Boris Tsirelson (talk) 17:13, 4 September 2015 (UTC)