Partial integration

The partial integration (partial integration, integration by parts, lat. Integratio per partes), including product integration called, is in the calculus a way to calculate definite integrals and determination of primitive functions . It can be understood as an analogue to the product rule of differential calculus . The Gaussian integral theorem from vector analysis with some of its special cases is a generalization of partial integration for functions of several variables .

Partial integration rule

If there is an interval and there are two continuously differentiable functions on , then applies ${\displaystyle [a,b]}$ ${\displaystyle f,\,g\colon [a,b]\to \mathbb {R} }$ ${\displaystyle [a,b]}$

${\displaystyle {\begin{aligned}\int _{a}^{b}f'(x)\cdot g(x)\,\mathrm {d} x&={\Big [}f(x)\cdot g(x){\Big ]}_{a}^{b}-\int _{a}^{b}f(x)\cdot g'(x)\,\mathrm {d} x\\&=f(b)\cdot g(b)-f(a)\cdot g(a)-\int _{a}^{b}f(x)\cdot g'(x)\,\mathrm {d} x\,.\end{aligned}}}$

This rule is called partial integration. It got its name because when it is used, only part of the integral is determined on the left-hand side of the equal sign, namely , and the second expression, namely , still contains an integral. This rule is therefore useful when the antiderivative is too known or easy to calculate, and when the integral expression on the right is easier to calculate. ${\displaystyle [f(x)\cdot g(x)]_{a}^{b}}$ ${\displaystyle -\int _{a}^{b}f(x)\cdot g'(x)\,\mathrm {d} x}$ ${\displaystyle f'}$

example

As an example, the integral

${\displaystyle \int _{2}^{10}x\cdot \ln(x)\,\mathrm {d} x}$

considered, where is the natural logarithm . If you place and , you get ${\displaystyle \ln(x)}$ ${\displaystyle f'(x)=x}$ ${\displaystyle g(x)=\ln(x)}$

${\displaystyle f(x)={\frac {1}{2}}x^{2}}$and . ${\displaystyle g'(x)={\frac {1}{x}}}$

This then results

${\displaystyle {\begin{aligned}\int _{2}^{10}x\cdot \ln(x)\,\mathrm {d} x&=\left[{\frac {1}{2}}{x^{2}}\cdot \ln(x)\right]_{2}^{10}-\int _{2}^{10}{\frac {1}{2}}{x^{2}}\cdot {\frac {1}{x}}\,\mathrm {d} x\\&=\left[{\frac {1}{2}}{x^{2}}\cdot \ln(x)\right]_{2}^{10}-\left[{\frac {1}{4}}{x^{2}}\right]_{2}^{10}\\&=50\ln(10)-2\ln(2)-24\,.\end{aligned}}}$

For more examples, see the Indefinite Integrals and Partial Integration section of this article. In contrast to this example, only indefinite integrals are calculated there. This means that there are no limits on the integrals, which are then inserted into the function in the last step, as is done here in the example.

history

A geometric shape of the control of the partial integration is already found in Blaise Pascal's work Traité des Trilignes Rectangles et de leurs Onglets ( treatise on curve triangles and their, adjoint body ' ), which in 1658 as part of the Lettre de A. Benedetto Ville à M. Carcavy appeared . Since the term integral was not yet coined at that time, this rule was not described by means of integrals, but by summing infinitesimals.

Gottfried Wilhelm Leibniz , who, together with Isaac Newton, is considered to be the inventor of differential and integral calculus , proved the statement made in modern notation

${\displaystyle \int _{a}^{b}y(x)\mathrm {d} x={\frac {1}{2}}\left({\Big [}x\cdot y(x){\Big ]}_{a}^{b}+\int _{a}^{b}\left(y(x)-x\cdot y'(x)\right)\mathrm {d} x\right)\,.}$

It is a special case of the partial integration rule. Leibniz called this rule the transmutation theorem and communicated it to Newton in his letter, which he sent to England in response to the epistola prior , Newton's first letter . With the help of this theorem, Leibniz examined the area of a circle and was able to use the formula

${\displaystyle {\frac {\pi }{4}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\dotsb =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}}$

to prove. Today it is called the Leibniz series .

Indefinite integrals and partial integration

The partial integration can also be used to prevent indefinite integrals to calculate - so to primitives to determine. For this purpose, the integral limits are usually deleted for partial integration, so the integration constant must now be observed.

rule

Let and be two continuously differentiable functions and if the antiderivative of is known, then we can use the rule for partial integration ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle f\cdot g'}$

${\displaystyle \int f'(x)\cdot g(x)\,\mathrm {d} x=f(x)\cdot g(x)-\int f(x)\cdot g'(x)\,\mathrm {d} x}$

an antiderivative to be found. ${\displaystyle f'\cdot g}$

Examples

This section uses two examples to show how an antiderivative is determined with the help of partial integration. In the first example no antiderivative is determined. This example shows that when determining an antiderivative with partial integration, one must also pay attention to the integration constant. In the second example the antiderivative of the logarithm and in the third example an antiderivative to a fractional-rational function is determined.

Reciprocal function

In this example the indefinite integral of is considered and partially integrated. Although not helpful for the concrete determination of the antiderivative of , it makes it clear that the constant of integration must be observed. It applies ${\displaystyle {\tfrac {1}{x}}}$ ${\displaystyle {\tfrac {1}{x}}}$

${\displaystyle {\begin{aligned}\int 1\cdot {\frac {1}{x}}\,\mathrm {d} x&=x\cdot {\frac {1}{x}}-\int x\cdot {\frac {-1}{x^{2}}}\,\mathrm {d} x\\&=1+\int x\cdot {\frac {1}{x^{2}}}\,\mathrm {d} x\\&=1+\int {\frac {1}{x}}\,\mathrm {d} x\,.\end{aligned}}}$

In the sense of indefinite integrals this equation is correct, because the functions and are both antiderivatives of the function . If one were to regard this expression as a definite integral with the limits , the middle (the integral-free) term would be omitted, because it holds ${\displaystyle \textstyle \int 1\cdot {\frac {1}{x}}\,\mathrm {d} x}$ ${\displaystyle \textstyle 1+\int {\frac {1}{x}}\,\mathrm {d} x}$ ${\displaystyle {\tfrac {1}{x}}}$ ${\displaystyle 0<a<b}$

${\displaystyle \left[x\cdot {\frac {1}{x}}\right]_{a}^{b}={\frac {b}{b}}-{\frac {a}{a}}=0}$.

Logarithmic function

If there is only one term in the integrand whose antiderivative cannot be easily deduced without a table value, one can occasionally integrate partially by inserting the factor . This works, for example, with the logarithm function . In order to determine the antiderivative of , the logarithm is differentiated during partial integration and the antiderivative is formed from the one function. So it applies ${\displaystyle 1}$ ${\displaystyle \ln }$ ${\displaystyle \ln }$

${\displaystyle {\begin{aligned}\int \ln(x)\,\mathrm {d} x&=\int 1\cdot \ln(x)\,\mathrm {d} x\\&=x\cdot \ln(x)-\int x\cdot {1 \over x}\,\mathrm {d} x\\&=x\cdot \ln(x)-\int 1\,\mathrm {d} x\\&=x\cdot \ln(x)-x+C\,.\end{aligned}}}$

Product of sine and cosine functions

Sometimes you can take advantage of the fact that after several steps of partial integration, the original integral returns on the right-hand side of the equals sign, which can then be combined with the original integral on the left-hand side by equivalence conversion .

As an example, the indefinite integral

${\displaystyle \int \sin(x)\cdot \cos(x)\,\mathrm {d} x}$

calculated. To do this, you put and , so it turns out ${\displaystyle f(x)=\cos(x)}$ ${\displaystyle g'(x)=\sin(x)}$

${\displaystyle f'(x)=-\sin(x)}$ and ${\displaystyle g(x)=-\cos(x)}$

and one receives

${\displaystyle \int \sin(x)\cdot \cos(x)\,\mathrm {d} x=-\cos ^{2}(x)-\int \sin(x)\cdot \cos(x)\,\mathrm {d} x}$.

If one adds the output integral on both sides of the equation, it follows

${\displaystyle 2\int \sin(x)\cdot \cos(x)\,\mathrm {d} x=-\cos ^{2}(x)}$ .

If we now divide by 2 on both sides, we get

${\displaystyle \int \sin(x)\cdot \cos(x)\,\mathrm {d} x=-{\tfrac {1}{2}}\cos ^{2}(x)}$

and one has found an antiderivative. All primitive functions therefore look like this:

${\displaystyle \int \sin(x)\cdot \cos(x)\,\mathrm {d} x=-{\tfrac {1}{2}}\cos ^{2}(x)+C}$.

If you swap the roles of and in partial integration, the result is analogous ${\displaystyle f}$ ${\displaystyle g}$

${\displaystyle \int \sin(x)\cdot \cos(x)\,\mathrm {d} x={\tfrac {1}{2}}\sin ^{2}(x)+C}$.

which is also obtained by inserting into the first formula found. One can therefore specify both and as an antiderivative with the same authorization , both differ only in one constant. ${\displaystyle \cos ^{2}(x)=1-\sin ^{2}(x)}$ ${\displaystyle -{\tfrac {1}{2}}\cos ^{2}(x)}$ ${\displaystyle {\tfrac {1}{2}}\sin ^{2}(x)}$

Product of polynomial and exponential functions

In the case of some indefinite integrals, it makes sense to choose a term that does not change or changes only insignificantly during integration, for example the natural exponential function or the trigonometric functions . ${\displaystyle f'}$

As an example, the indefinite integral

${\displaystyle \int e^{x}\cdot \left(2-x^{2}\right)\,\mathrm {d} x}$

considered. If one sets below the integral for each partial integration step and for the rest of the term, the result is ${\displaystyle f'(x)=e^{x}}$ ${\displaystyle g}$

${\displaystyle {\begin{aligned}\int e^{x}\cdot \left(2-x^{2}\right)\,\mathrm {d} x&=e^{x}\cdot \left(2-x^{2}\right)-\int e^{x}\cdot (-2x)\,\mathrm {d} x\\&=e^{x}\cdot \left(2-x^{2}\right)+e^{x}\cdot 2x-\int 2\cdot e^{x}\,\mathrm {d} x\\&=e^{x}\cdot \left(2-x^{2}\right)+e^{x}\cdot 2x-2\cdot e^{x}+C\\&=e^{x}\cdot \left(2-x^{2}+2x-2\right)+C\\&=e^{x}\cdot \left(2x-x^{2}\right)+C\,.\end{aligned}}}$

Derivation

The product rule from differential calculus says that for two continuous differentiable functions and equality ${\displaystyle f}$ ${\displaystyle g}$

${\displaystyle (f\cdot g)'=f'\cdot g+f\cdot g',}$

applies and follows by means of term rewriting

${\displaystyle f'\cdot g=(f\cdot g)'-f\cdot g'\,.}$

Using the main theorem of integral and differential calculus follows

${\displaystyle {\begin{aligned}\int _{a}^{b}f'\cdot g&=\int _{a}^{b}(f\cdot g)'-\int _{a}^{b}f\cdot g'\\&={\Big [}f\cdot g{\Big ]}_{a}^{b}\,-\int _{a}^{b}f\cdot g'\,,\end{aligned}}}$

which is the rule

${\displaystyle \int _{a}^{b}f'(x)\cdot g(x)\,\mathrm {d} x=f(b)\cdot g(b)-f(a)\cdot g(a)-\int _{a}^{b}f(x)\cdot g'(x)\,\mathrm {d} x}$

for partial integration results.

Partial integration using a table (DI method)

If you want to determine indefinite integrals with the help of partial integration, you can work with a table. The derivatives of are written in the left column and antiderivatives of in the right column until one of the following three conditions is met: ${\displaystyle f}$ ${\displaystyle g}$

1. One derivative is zero,
2. the indefinite integral of a row (the product of the associated cells) is known or
3. a line is repeated

Case 1: One derivative is zero

Example: ${\displaystyle \int \cos(x)\cdot x^{2}\,\mathrm {d} x}$

Since is easier to integrate than , we choose ${\displaystyle \cos(x)}$ ${\displaystyle x^{2}}$

${\displaystyle f=x^{2},\quad g=\cos(x)}$.

Now we can set up the table

sign	D (for differentiation)	I (for integration)
+	${\displaystyle x^{2}}$	${\displaystyle \cos(x)}$
-	${\displaystyle 2x}$	${\displaystyle \sin(x)}$
+	${\displaystyle 2}$	${\displaystyle -\cos(x)}$
-	${\displaystyle 0}$	${\displaystyle -\sin(x)}$

The fourth line has a zero as a derivative, i.e. H. we can end the table after four rows. In order to calculate the indefinite integral, we have to multiply the individual cells diagonally, observing the signs

${\displaystyle \int \cos(x)\cdot x^{2}\,\mathrm {d} x=x^{2}\sin(x)+2x\cos(x)-2\sin(x)+C}$

${\displaystyle =\sin(x)\,(x^{2}-2)+2x\cos(x)+C}$

Case 2: A line can be integrated

Example: ${\displaystyle \int 3x(\ln(x)-1)\cdot \left(x^{5}-4x^{4}\right)\,\mathrm {d} x}$

In this case it is easier to integrate the polynomial, so we choose

${\displaystyle f=3x(\ln(x)-1),\quad g=x^{5}-4x^{4}}$

sign	D.	I.
+	${\displaystyle 3x(\ln(x)-1)}$	${\displaystyle x^{5}-4x^{4}}$
-	${\displaystyle 3\ln(x)}$	${\displaystyle {\frac {1}{6}}x^{6}-{\frac {4}{5}}x^{5}}$
+	${\displaystyle {\frac {3}{x}}}$	${\displaystyle {\frac {1}{42}}x^{7}-{\frac {2}{15}}x^{6}}$

We have to multiply diagonally again

${\displaystyle \int 3x(\ln(x)-1)\cdot \left(x^{5}-4x^{4}\right)\,\mathrm {d} x=\left({\frac {1}{2}}x^{7}-{\frac {12}{5}}x^{6}\right)\cdot (\ln(x)-1)-3\ln(x)\cdot \left({\frac {1}{42}}x^{7}-{\frac {2}{15}}x^{6}\right)+\int \left({\frac {1}{14}}x^{6}-{\frac {2}{5}}x^{5}\right)\,\mathrm {d} x}$

We can compute an antiderivative for the part to be integrated

${\displaystyle \int \left({\frac {1}{14}}x^{6}-{\frac {2}{5}}x^{5}\right)\,\mathrm {d} x={\frac {1}{98}}x^{7}-{\frac {1}{15}}x^{6}}$

and summarize the result

${\displaystyle \int 3x(\ln(x)-1)\cdot \left(x^{5}-4x^{4}\right)\,\mathrm {d} x=-{\frac {24}{49}}x^{7}+{\frac {7}{3}}x^{6}+\ln(x)\cdot \left({\frac {3}{7}}x^{7}-2x^{6}\right)+C}$

Case 3: A line is repeated

Example: ${\displaystyle \int e^{-x}\cdot \cos(x)\,\mathrm {d} x}$

We vote

${\displaystyle f=\cos(x),\quad g=e^{-x}}$

sign	D.	I.
+	${\displaystyle \cos(x)}$	${\displaystyle e^{-x}}$
-	${\displaystyle -\sin(x)}$	${\displaystyle -e^{-x}}$
+	${\displaystyle -\cos(x)}$	${\displaystyle e^{-x}}$

The third line essentially corresponds to the first line, only that column D has a different sign.

We have to make an equation

${\displaystyle \int e^{-x}\cdot \cos(x)\,\mathrm {d} x=-e^{-x}\cdot \cos(x)+e^{-x}\sin(x)-\int e^{-x}\cdot \cos(x)\,\mathrm {d} x}$

and after switching ${\displaystyle \int e^{-x}\cdot \cos(x)\,\,dx}$

${\displaystyle \int e^{-x}\cdot \cos(x)\,\mathrm {d} x={\frac {e^{-x}}{2}}\cdot (\sin(x)-\cos(x))}$.

Partial integration with only one function (case 2)

Example: ${\displaystyle \int \ln(x)\,\mathrm {d} x}$

We vote

${\displaystyle f=\ln(x)}$, ${\displaystyle g=1}$

sign	D.	I.
+	${\displaystyle \ln(x)}$	${\displaystyle 1}$
-	${\displaystyle {\frac {1}{x}}}$	${\displaystyle x}$

The second line can be integrated here according to case 2 and we can calculate

${\displaystyle \int \ln(x)\,\mathrm {d} x=x\cdot \ln(x)-\int 1\,\mathrm {d} x=x\cdot \ln(x)-x}$.

Sums display

If the -th derivative of a function vanishes , i. H. is a polynomial of degree , the repeated partial integration or the DI method can be written as follows: ${\displaystyle (n+1)}$ ${\displaystyle f}$ ${\displaystyle f}$ ${\displaystyle n}$

${\displaystyle \int _{a}^{b}f(x)g(x)\,\mathrm {d} x=\sum _{k=0}^{n}(-1)^{k}\left[f^{(k)}(x)\cdot \,^{(k+1)}\!g(x)\right]_{a}^{b}}$,

where denotes a -th antiderivative of . ${\displaystyle \,^{(j)}\!g}$ ${\displaystyle j}$ ${\displaystyle g}$

Example:

${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\,\mathrm {d} x=\sum _{k=0}^{n}(-1)^{k}\left[{\frac {n!}{(n-k)!}}x^{n-k}\cdot {\frac {1}{(-a)^{k+1}}}e^{-ax}\right]_{0}^{\infty }}$

The integral vanishes at infinity, and at 0 only in the case not: ${\displaystyle n-k=0}$

${\displaystyle =\sum _{k=0}^{n}(-1)^{k}\left[-{\frac {n!}{(n-k)!}}\cdot \delta _{nk}\cdot {\frac {1}{(-a)^{k+1}}}\right]=-(-1)^{n}\cdot {\frac {n!}{(n-n)!}}\cdot {\frac {1}{(-a)^{n+1}}}={\frac {n!}{a^{n+1}}}}$

Multi-dimensional partial integration

The partial integration in several dimensions is a special case of the Gaussian integral theorem : Be compact with a smooth edge in sections . Let the edge be oriented by an outer normal unit field . Furthermore, let a continuously differentiable vector field on an open neighborhood of and a continuously differentiable scalar field on . Then applies ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ ${\displaystyle \partial \Omega }$ ${\displaystyle {\vec {n}}}$ ${\displaystyle {\vec {v}}}$ ${\displaystyle \Omega }$ ${\displaystyle \varphi }$ ${\displaystyle \Omega }$

${\displaystyle \int _{\Omega }\operatorname {div} (\varphi {\vec {v}})\;\mathrm {d} V=\int _{\Omega }(\varphi \,\operatorname {div} \,{\vec {v}}+{\vec {v}}\cdot \operatorname {grad} \,\varphi )\;\mathrm {d} V=\int _{\partial \Omega }\varphi \,{\vec {v}}\cdot \mathrm {d} {\vec {S}}}$

with the abbreviation . Then follows the generalization of partial integration in several dimensions ${\displaystyle \mathrm {d} {\vec {S}}={\vec {n}}\;\mathrm {d} S}$

${\displaystyle \int _{\Omega }\varphi \,\operatorname {div} \,{\vec {v}}\;\mathrm {d} V=\int _{\partial \Omega }\varphi \,{\vec {v}}\cdot \mathrm {d} {\vec {S}}-\int _{\Omega }{\vec {v}}\cdot \operatorname {grad} \,\varphi \;\mathrm {d} V}$.

Rule of partial integration for Stieltjes integrals

Let and be two functions of finite variation , then we have ${\displaystyle g}$ ${\displaystyle f}$

${\displaystyle f(t)g(t)-f(0)g(0)=\int _{0}^{t}f(s)\;\mathrm {d} g(s)+\int _{0}^{t}g(s-)\;\mathrm {d} f(s)}$

or written differently

${\displaystyle f(t)g(t)-f(0)g(0)=\int _{0}^{t}f(s-)\;\mathrm {d} g(s)+\int _{0}^{t}g(s-)\;\mathrm {d} f(s)+\sum _{0<s\leq t}\Delta f(s)g(s)}$.

Weak derivation

→ Main article : Weak derivative

In the theory of partial differential equations , a generalization of the derivation of a differentiable function was found using the method of partial integration .

If one considers a function differentiable on an open interval (classic) and a function differentiable as often as desired with a compact support in , then applies ${\displaystyle I={]a,b[}}$ ${\displaystyle f\colon I\to \mathbb {R} }$ ${\displaystyle \varphi \in C_{c}^{\infty }(I)}$ ${\displaystyle I}$

${\displaystyle \int _{I}f^{\prime }(t)\varphi (t)\,\mathrm {d} t=-\int _{I}f(t)\varphi ^{\prime }(t)\,\mathrm {d} t}$.

Partial integration was used here. The boundary term, i.e. the term without an integral, is missing because the function has a compact support and therefore and is true. ${\displaystyle \varphi }$ ${\displaystyle \varphi (a)=0}$ ${\displaystyle \varphi (b)=0}$

If the function is now chosen as a -function , then, even if it is not differentiable (more precisely: has no differentiable representative in the equivalence class), a function can exist which the equation ${\displaystyle f}$ ${\displaystyle L^{2}}$ ${\displaystyle f}$ ${\displaystyle g\in L^{2}(I)}$

${\displaystyle \int _{I}g(t)\varphi (t)\,\mathrm {d} t=-\int _{I}f(t)\varphi ^{\prime }(t)\,\mathrm {d} t}$

fulfilled for every function . Such a function is called the weak derivative of . The resulting set of weakly differentiable -functions is a vector space and it belongs to the class of Sobolev spaces . The smooth functions with compact support, whose vector space is denoted by, are called test functions . ${\displaystyle \varphi \in C_{c}^{\infty }(I)}$ ${\displaystyle g}$ ${\displaystyle f}$ ${\displaystyle L^{2}}$ ${\displaystyle C_{c}^{\infty }(I)}$

However, if there is no function with the required condition, a distribution can always be found so that the above condition is met in the distribution sense. Then the distribution is called the derivative of . ${\displaystyle g\in L^{2}(I)}$ ${\displaystyle g}$ ${\displaystyle g}$ ${\displaystyle f}$

See also

• Integration by substitution , another important rule for calculating integrals

Web links

Wikibooks: Math for Non-Freaks: Partial Integration  - Learning and Teaching Materials

• Video: partial integration, substitution rule, integration by partial fraction decomposition . Jörn Loviscach 2012, made available by the Technical Information Library (TIB), doi : 10.5446 / 9987 .

Individual evidence

1. ^ Konrad Königsberger : Analysis 1 . Springer-Verlag, Berlin et al., 2004, ISBN 3-540-41282-4 , p. 202.
2. ^ Yvonne Stry: Mathematics compact: for engineers and computer scientists . 3rd, arr. Edition, Springer-Verlag, 2010, ISBN 3642111912 , p. 314.
3. ^ Thomas Sonar: 3000 Years of Analysis , Springer, Berlin 2011, ISBN 978-3-642-17203-8 , p. 273.
4. ^ Thomas Sonar: 3000 Years of Analysis , Springer, Berlin 2011, ISBN 978-3-642-17203-8 , pp. 418-421.
5. ↑ Otto Forster : Analysis Volume 1: Differential and integral calculus of a variable. Vieweg-Verlag, 8th edition 2006, ISBN 3-528-67224-2 , p. 210.
6. ↑ Mark Zegarelli: Calculus II For Dummies . Weinheim 2009, ISBN 978-3-527-70509-2 , pp. 152 .