Incalculus,and more generally inmathematical analysis,integration by partsorpartial integrationis a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.It is frequently used to transform the antiderivative of a product of functions into an antiderivative for which a solution can be more easily found. The rule can be thought of as an integral version of theproduct ruleofdifferentiation;it is indeed derived using the product rule.
The integration by parts formula states:
Or, lettingandwhileandthe formula can be written more compactly:
It is important to note that the former expression is written as a definite integral and the latter is written as an indefinite integral. Applying the appropriate limits to the latter expression should yield the former, but the latter is not necessarily equivalent to the former.
This is to be understood as an equality of functions with an unspecified constant added to each side. Taking the difference of each side between two valuesandand applying thefundamental theorem of calculusgives the definite integral version:
The original integralcontains thederivativev';to apply the theorem, one must findv,theantiderivativeofv',then evaluate the resulting integral
It is not necessary forandto be continuously differentiable. Integration by parts works ifisabsolutely continuousand the function designatedisLebesgue integrable(but not necessarily continuous).[3](Ifhas a point of discontinuity then its antiderivativemay not have a derivative at that point.)
If the interval of integration is notcompact,then it is not necessary forto be absolutely continuous in the whole interval or forto be Lebesgue integrable in the interval, as a couple of examples (in whichandare continuous and continuously differentiable) will show. For instance, if
is not absolutely continuous on the interval[1, ∞),but nevertheless
so long asis taken to mean the limit ofasand so long as the two terms on the right-hand side are finite. This is only true if we chooseSimilarly, if
is not Lebesgue integrable on the interval[1, ∞),but nevertheless
with the same interpretation.
One can also easily come up with similar examples in whichandarenotcontinuously differentiable.
Further, ifis a function of bounded variation on the segmentandis differentiable onthen
wheredenotes the signed measure corresponding to the function of bounded variation,and functionsare extensions oftowhich are respectively of bounded variation and differentiable.[citation needed]
Graphical interpretation of the theorem. The pictured curve is parametrized by the variable t.
Consider a parametric curve by (x,y) = (f(t),g(t)). Assuming that the curve is locallyone-to-oneandintegrable,we can define
The area of the blue region is
Similarly, the area of the red region is
The total areaA1+A2is equal to the area of the bigger rectangle,x2y2,minus the area of the smaller one,x1y1:
Or, in terms oft,
Or, in terms of indefinite integrals, this can be written as
Rearranging:
Thus integration by parts may be thought of as deriving the area of the blue region from the area of rectangles and that of the red region.
This visualization also explains why integration by parts may help find the integral of an inverse functionf−1(x) when the integral of the functionf(x) is known. Indeed, the functionsx(y) andy(x) are inverses, and the integral ∫xdymay be calculated as above from knowing the integral ∫ydx.In particular, this explains use of integration by parts to integratelogarithmandinverse trigonometric functions.In fact, ifis a differentiable one-to-one function on an interval, then integration by parts can be used to derive a formula for the integral ofin terms of the integral of.This is demonstrated in the article,Integral of inverse functions.
Integration by parts is aheuristicrather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate this single function into a product of two functionsu(x)v(x) such that the residual integral from the integration by parts formula is easier to evaluate than the single function. The following form is useful in illustrating the best strategy to take:
On the right-hand side,uis differentiated andvis integrated; consequently it is useful to chooseuas a function that simplifies when differentiated, or to choosevas a function that simplifies when integrated. As a simple example, consider:
Since the derivative of ln(x) is1/x,one makes (ln(x)) partu;since the antiderivative of1/x2is −1/x,one makes1/x2partv.The formula now yields:
The antiderivative of −1/x2can be found with thepower ruleand is1/x.
Alternatively, one may chooseuandvsuch that the productu′ (∫vdx) simplifies due to cancellation. For example, suppose one wishes to integrate:
If we chooseu(x) = ln(|sin(x)|) andv(x) = sec2x, thenudifferentiates to 1/ tanxusing thechain ruleandvintegrates to tanx;so the formula gives:
The integrand simplifies to 1, so the antiderivative isx.Finding a simplifying combination frequently involves experimentation.
In some applications, it may not be necessary to ensure that the integral produced by integration by parts has a simple form; for example, innumerical analysis,it may suffice that it has small magnitude and so contributes only a small error term. Some other special techniques are demonstrated in the examples below.
Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative timesis also known.
The function which is to bedvis whichever comes last in the list. The reason is that functions lower on the list generally have simplerantiderivativesthan the functions above them. The rule is sometimes written as "DETAIL", whereDstands fordvand the top of the list is the function chosen to bedv.An alternative to this rule is the ILATE rule, where inverse trigonometric functions come before logarithmic functions.
To demonstrate the LIATE rule, consider the integral
Following the LIATE rule,u=x,anddv= cos(x) dx,hencedu=dx,andv= sin(x), which makes the integral become
which equals
In general, one tries to chooseuanddvsuch thatduis simpler thanuanddvis easy to integrate. If instead cos(x) was chosen asu,andx dxasdv,we would have the integral
which, after recursive application of the integration by parts formula, would clearly result in an infinite recursion and lead nowhere.
Although a useful rule of thumb, there are exceptions to the LIATE rule. A common alternative is to consider the rules in the "ILATE" order instead. Also, in some cases, polynomial terms need to be split in non-trivial ways. For example, to integrate
one would set
so that
Then
Finally, this results in
Integration by parts is often used as a tool to prove theorems inmathematical analysis.
The above result tells us about the decay of the Fourier transform, since it follows that ifandare integrable then
In other words, ifsatisfies these conditions then its Fourier transform decays at infinity at least as quickly as1/|ξ|k.In particular, ifthen the Fourier transform is integrable.
One use of integration by parts inoperator theoryis that it shows that the−∆(where ∆ is theLaplace operator) is apositive operatoron(seeLpspace). Ifis smooth and compactly supported then, using integration by parts, we have
Considering a second derivative ofin the integral on the LHS of the formula for partial integration suggests a repeated application to the integral on the RHS:
Extending this concept of repeated partial integration to derivatives of degreenleads to
This concept may be useful when the successive integrals ofare readily available (e.g., plain exponentials or sine and cosine, as inLaplaceorFourier transforms), and when thenth derivative ofvanishes (e.g., as a polynomial function with degree). The latter condition stops the repeating of partial integration, because the RHS-integral vanishes.
In the course of the above repetition of partial integrations the integrals
andand
get related. This may be interpreted as arbitrarily "shifting" derivatives betweenandwithin the integrand, and proves useful, too (seeRodrigues' formula).
The essential process of the above formula can be summarized in a table; the resulting method is called "tabular integration"[5]and was featured in the filmStand and Deliver(1988).[6]
For example, consider the integral
and take
Begin to list in columnAthe functionand its subsequent derivativesuntil zero is reached. Then list in columnBthe functionand its subsequent integralsuntil the size of columnBis the same as that of columnA.The result is as follows:
#i
Sign
A: derivatives
B: integrals
0
+
1
−
2
+
3
−
4
+
The product of the entries inrowiof columnsAandBtogether with the respective sign give the relevant integrals instepiin the course of repeated integration by parts.Stepi= 0yields the original integral. For the complete result instepi> 0theith integralmust be added to all the previous products (0 ≤j<i) of thejth entryof column A and the(j+ 1)st entryof column B (i.e., multiply the 1st entry of column A with the 2nd entry of column B, the 2nd entry of column A with the 3rd entry of column B, etc....) with the givenjth sign.This process comes to a natural halt, when the product, which yields the integral, is zero (i= 4in the example). The complete result is the following (with the alternating signs in each term):
This yields
The repeated partial integration also turns out useful, when in the course of respectively differentiating and integrating the functionsandtheir product results in a multiple of the original integrand. In this case the repetition may also be terminated with this indexi.This can happen, expectably, with exponentials and trigonometric functions. As an example consider
#i
Sign
A: derivatives
B: integrals
0
+
1
−
2
+
In this case the product of the terms in columnsAandBwith the appropriate sign for indexi= 2yields the negative of the original integrand (comparerowsi= 0andi= 2).
Observing that the integral on the RHS can have its own constant of integration,and bringing the abstract integral to the other side, gives
Integration by parts can be extended to functions of several variables by applying a version of the fundamental theorem of calculus to an appropriate product rule. There are several such pairings possible in multivariate calculus, involving a scalar-valued functionuand vector-valued function (vector field)V.[7]
whereis the outward unit normal vector to the boundary, integrated with respect to its standard Riemannian volume form.Rearranging gives:
or in other words
Theregularityrequirements of the theorem can be relaxed. For instance, the boundaryneed only beLipschitz continuous,and the functionsu,vneed only lie in theSobolev space.
Consider the continuously differentiable vector fieldsand,whereis thei-th standard basis vector for.Now apply the above integration by parts to eachtimes the vector field:
Summing overigives a new integration by parts formula:
Hoffmann, Laurence D.; Bradley, Gerald L. (2004).Calculus for Business, Economics, and the Social and Life Sciences(8th ed.). pp. 450–464.ISBN0-07-242432-X.
Willard, Stephen (1976).Calculus and its Applications.Boston: Prindle, Weber & Schmidt. pp. 193–214.ISBN0-87150-203-8.
Washington, Allyn J. (1966).Technical Calculus with Analytic Geometry.Reading: Addison-Wesley. pp. 218–245.ISBN0-8465-8603-7.