Editing Integration by parts (section)

==Repeated integration by parts==
{{Further|Cauchy formula for repeated integration}}
Considering a second derivative of <math>v</math> in the integral on the LHS of the formula for partial integration suggests a repeated application to the integral on the RHS:
<math display="block">\int u v''\,dx = uv' - \int u'v'\,dx = uv' - \left( u'v - \int u''v\,dx \right).</math>

Extending this concept of repeated partial integration to derivatives of degree {{mvar|n}} leads to
<math display="block">\begin{align}
\int u^{(0)} v^{(n)}\,dx &= u^{(0)} v^{(n-1)} - u^{(1)}v^{(n-2)} + u^{(2)}v^{(n-3)} - \cdots + (-1)^{n-1}u^{(n-1)} v^{(0)} + (-1)^n \int u^{(n)} v^{(0)} \,dx.\\[5pt]
&= \sum_{k=0}^{n-1}(-1)^k u^{(k)}v^{(n-1-k)} + (-1)^n \int u^{(n)} v^{(0)} \,dx.
\end{align}</math>

This concept may be useful when the successive integrals of <math>v^{(n)}</math> are readily available (e.g., plain exponentials or sine and cosine, as in  [[Laplace transform|Laplace]] or [[Fourier transform]]s), and when the {{mvar|n}}th derivative of <math>u</math> vanishes (e.g., as a polynomial function with degree <math>(n-1)</math>). 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 
<math display="block">\int u^{(0)} v^{(n)}\,dx \quad</math> and <math>\quad \int u^{(\ell)} v^{(n-\ell)}\,dx \quad</math> and <math>\quad \int u^{(m)} v^{(n-m)}\,dx \quad\text{ for } 1 \le m,\ell \le n</math>
get related. This may be interpreted as arbitrarily "shifting" derivatives between <math>v</math> and <math>u</math> within the integrand, and proves useful, too (see [[Rodrigues' formula]]).

===Tabular integration by parts===
The essential process of the above formula can be summarized in a table; the resulting method is called "tabular integration"<ref>{{Cite book |first1=G. B. |last1=Thomas |author-link=George B. Thomas |first2=R. L. |last2=Finney |title=Calculus and Analytic Geometry |publisher=Addison-Wesley |location=Reading, MA |edition=7th |year=1988 |isbn=0-201-17069-8 }}</ref> and was featured in the film ''[[Stand and Deliver]]'' (1988).<ref>{{Cite journal |url=https://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Horowitz307-311.pdf |first=David |last=Horowitz |title=Tabular Integration by Parts |journal=[[The College Mathematics Journal]] |volume=21 |issue=4 |year=1990 |pages=307–311 |doi=10.2307/2686368 |jstor=2686368}}</ref>

For example, consider the integral

<math display="block">\int x^3 \cos x \,dx \quad</math> and take <math>\quad u^{(0)} = x^3, \quad v^{(n)} = \cos x.</math>

Begin to list in column '''A''' the function <math>u^{(0)} = x^3</math> and its subsequent derivatives <math>u^{(i)}</math> until zero is reached. Then list in column '''B''' the function <math>v^{(n)} = \cos x</math> and its subsequent integrals <math>v^{(n-i)}</math> until the size of column '''B''' is the same as that of column '''A'''. The result is as follows:

:{| class="wikitable" style="text-align:center"
!# ''i'' !! Sign !! A: derivatives <math>u^{(i)}</math>  !! B: integrals <math>v^{(n-i)}</math>
|-
| 0 || + || <math>x^3</math> || <math>\cos x</math>
|-
| 1 || − || <math>3x^2</math> || <math>\sin x</math>
|-
| 2 || + || <math>6x</math> || <math>-\cos x</math>
|-
| 3 || − || <math>6</math> || <math>-\sin x</math>
|-
| 4 || + || <math>0</math> || <math>\cos x</math>
|}

The product of the entries in {{nowrap|row {{mvar|i}}}} of columns '''A''' and '''B''' together with the respective sign give the relevant integrals in {{nowrap|step {{mvar|i}}}} in the course of repeated integration by parts. {{nowrap|Step {{math|''i'' {{=}} 0}}}} yields the original integral. For the complete result in {{nowrap|step {{math|''i'' > 0}}}} the {{nowrap|{{mvar|i}}th integral}} must be added to all the previous products ({{math|0 ≤ ''j'' < ''i''}}) of the {{nowrap|{{mvar|j}}th entry}} of column A and the {{nowrap|{{math|(''j'' + 1)}}st entry}} of 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 given {{nowrap|{{mvar|j}}th sign.}} This process comes to a natural halt, when the product, which yields the integral, is zero ({{math|''i'' {{=}} 4}} in the example).  The complete result is the following (with the alternating signs in each term):

<math display="block">\underbrace{(+1)(x^3)(\sin x)}_{j=0} + \underbrace{(-1)(3x^2)(-\cos x)}_{j=1} + \underbrace{(+1)(6x)(-\sin x)}_{j=2} +\underbrace{(-1)(6)(\cos x)}_{j=3}+ \underbrace{\int(+1)(0)(\cos x) \,dx}_{i=4: \;\to \;C}.</math>

This yields

<math display="block">\underbrace{\int x^3 \cos x \,dx}_{\text{step 0}} = x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + C. </math>

The repeated partial integration also turns out useful, when in the course of respectively differentiating and integrating the functions <math>u^{(i)}</math> and  <math>v^{(n-i)}</math>  their product results in a multiple of the original integrand. In this case the repetition may also be terminated with this index {{mvar|i.}}This can happen, expectably, with exponentials and trigonometric functions. As an example consider

<math display="block">\int e^x \cos x \,dx. </math>

:{| class="wikitable" style="text-align:center"
!# ''i'' !! Sign !! A: derivatives <math>u^{(i)}</math> !! B: integrals <math>v^{(n-i)}</math>
|-
| 0 || + || <math>e^x</math> || <math>\cos x</math>
|-
| 1 || − || <math>e^x</math> || <math>\sin x</math>
|-
| 2 || + || <math>e^x</math> || <math>-\cos x</math>
|}

In this case the product of the terms in columns '''A''' and '''B''' with the appropriate sign for index {{math|''i'' {{=}} 2}} yields the negative of the original integrand (compare {{nowrap|rows {{math|''i'' {{=}} 0}}}} {{nowrap|and {{math|''i'' {{=}} 2}}).}}

<math display="block"> \underbrace{\int e^x \cos x \,dx}_{\text{step 0}} = \underbrace{(+1)(e^x)(\sin x)}_{j=0} + \underbrace{(-1)(e^x)(-\cos x)}_{j=1} + \underbrace{\int(+1)(e^x)(-\cos x) \,dx}_{i= 2}. </math>

Observing that the integral on the RHS can have its own constant of integration <math>C'</math>, and bringing the abstract integral to the other side, gives:

<math display="block"> 2 \int e^x \cos x \,dx = e^x\sin x + e^x\cos x + C', </math>

and finally:

<math display="block">\int e^x \cos x \,dx = \frac 12 \left(e^x ( \sin x + \cos x ) \right) + C,</math>

where <math>C = \frac{C'}{2}</math>.