|
|
| (One intermediate revision by one other user not shown) |
| Line 1: |
Line 1: |
| {{Calculus |Integral}}
| | Although not especially African American skin care products, the formulas that contain the ingredients I have listed here will work extremely well in alleviating the common problems you suffer. To help confirm the diagnosis, the doctor may take a small sample (biopsy) of the affected skin to examine under a microscope. As mentioned above, this plant has anti-bacterial properties. These autoimmune diseases include hyperthyroidism (an overactive thyroid gland), adrenocortical insufficiency (the adrenal gland does not produce enough of the hormone called corticosteroid), alopecia areata (patches of baldness), and pernicious anemia (a low level of red blood cells caused by failure of the body to absorb vitamin B-12). When these toxins pollute and overpower the body, Vitiligo result depending on the genetic predisposition. <br><br>In people who have vitiligo, the immune cells (bodily cells that fight against infection) attack the melanocytes and destroy them in addition to eliminating infections. They claim the treatment plan might also help people vitiligo. He “did not have a history of heart problems. These white patches do not look and you eventually lose your self confidence. Herbal medicines useful for vitiligo are: Manjishtha (Rubia cordifolia), Saariva (Hemidesmus indicus), Triphala (Three fruits), Haridra (Curcuma longa), Daruharidra (Berberis aristata), Khadir (Acacia catechu), Vidanga (Embilia ribes) and Bavachi (Psoralia corylifolia). <br><br>For complete vitiligo cure, this is an excellent herbal product. About a gm of psoralea powder can be taken along with fresh milk, regularly once for a month. In vitiligo, the skin sample will usually show a complete absence of pigment-producing melanocytes. Before making a choice of buying an anti-vitiligo herbal supplement make sure that you know the ingredients that are curative in their real terms are present in the product. Otherwise, the condition is completely asmptomatis with no structural change or loss of sensation in the affected areas. <br><br>You'll be able to lessen the issue of wanting to remember to normal water them very quickly, by developing your herbs in a very window box or in trays on your patio or deck, so that you just observe them each time you carry a break on your patio. It is a well known fact, through the World Health Organization's Cancer Research Division, that ultraviolet (UV) rays can cause premature skin aging and skin cancer. There may be required a skin biopsy only in very rare cases though that is not common. He invented the Young Volumizer which is a safe procedure that is less invasive in reversing aging signs to reveal a natural and younger looking face. Alternative types include linear, segmental, trachoma and inflammatory vitiligo. <br><br>People with dark skin experience more prominent patches. In most of the cases, a blend of transmissible, immunologic and ecological factors can trigger Vitiligo. When you then combine these foods with garlic and a few berries you may have a incredibly powerful Vitiligo diet regime that can provide you with relief by enhancing your immune method, lessen free radicals and boost melanin production in your skin. In short, zinc is an antioxidant that wrestles free radicals and antagonizes heavy metals. Clinically, it is distinguished by white macules on the skin that can be few or numerous. <br><br>It is thus necessary to find a cure that can mitigate the condition and provide relief. They found that the build up of hydrogen peroxide was caused by a reduction of an enzyme that breaks up hydrogen peroxide into water and oxygen (catalase). The weather, the environment, stress, radiation from technology, your mood are also other factors that can lead to spotty skin. For those who are looking for just an additional supplement to help them recover there are specified studies being carried out for them, vitamin C, PABA, folic acid with vitamin B12 and L phenylalanine are few to name. Afterwards, skin color microscopic cells could possibly be grown inside science lab and employed to take care of stricken components.<br><br>When you have any kind of questions about where by and also how to work with [http://discover-prague.info/sitemap/ pc kus treatment for vitiligo], you'll be able to call us with the web site. |
| | |
| In [[calculus]], and more generally in [[mathematical analysis]], '''integration by parts''' is a theorem that relates the [[integral (mathematics)|integral]] of a [[product (mathematics)|product]] of functions to the integral of their derivative and antiderivative. 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 derived in one line simply by integrating the [[product rule]] of [[derivative|differentiation]].
| |
| | |
| If ''u'' = ''u''(''x''), ''v'' = ''v''(''x''), and the [[differential of a function|differential]]s ''du'' = ''u ''<nowiki>'</nowiki>(''x'') ''dx'' and ''dv'' = ''v''<nowiki>'</nowiki>(''x'') ''dx'', then integration by parts states that
| |
| | |
| :<math>\int u(x) v'(x) \, dx = u(x) v(x) - \int u'(x) v(x) \, dx </math>
| |
| | |
| or more compactly:
| |
| | |
| :<math>\int u \, dv=uv-\int v \, du.\!</math>
| |
| | |
| More general formulations of integration by parts exist for the [[Riemann–Stieltjes integral#Properties and relation to the Riemann integral|Riemann–Stieltjes integral]] and [[Lebesgue–Stieltjes integral#Integration by parts|Lebesgue–Stieltjes integral]]. The discrete analogue for sequences is called [[summation by parts]].
| |
| | |
| == Theorem ==
| |
| | |
| ===Product of two functions===
| |
| The theorem can be derived as follows. Suppose ''u''(''x'') and ''v''(''x'') are two [[continuously differentiable]] [[function (mathematics)|functions]]. The [[product rule]] states (in [[Notation for differentiation#Leibniz's notation|Leibniz’ notation]]):
| |
| | |
| :<math>\frac{d}{dx}\left(u(x)v(x)\right) = v(x) \frac{d}{dx}\left(u(x)\right) + u(x) \frac{d}{dx}\left(v(x)\right).\!</math>
| |
| | |
| Integrating both sides with respect to ''x'', over an [[interval (mathematics)|interval]] ''a'' ≤ ''x'' ≤ ''b'':
| |
| | |
| :<math>\int_a^b \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx </math>
| |
| | |
| then applying the [[fundamental theorem of calculus]],
| |
| | |
| :<math>\int_a^b \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \left[u(x)v(x)\right]_a^b</math>
| |
| | |
| gives the formula for '''integration by parts''':
| |
| | |
| :<math>\left[u(x)v(x)\right]_a^b = \int_a^b u'(x)v(x)\,dx + \int_a^b u(x)v'(x)\,dx. </math>
| |
| | |
| Since ''du'' and ''dv'' are [[differentials of a function]] of one variable ''x'',
| |
| | |
| :<math>du=u'(x)dx, \quad dv=v'(x)dx. </math>
| |
| | |
| The original integral ∫''uv''′ ''dx'' contains ''v''′ ([[derivative]] of ''v''); in order to apply the theorem, ''v'' ([[antiderivative]] of ''v''′) must be found, and then the resulting integral ∫''vu''′ ''dx'' must be evaluated.
| |
| | |
| ===Product of many functions===
| |
| | |
| Integrating the product rule for three multiplied functions, ''u''(''x''), ''v''(''x''), ''w''(''x''), gives a similar result:
| |
| | |
| :<math>\int_a^b u v \, dw = u v w - \int_a^b u w \, dv - \int_a^b v w \, du.</math>
| |
| | |
| In general for ''n'' factors
| |
| | |
| :<math>\frac{d}{dx} \left(\prod_{i=1}^n u_i(x) \right)= \sum_{j=1}^n \prod_{i\neq j}^n u_i(x) \frac{du_j(x)}{dx}, </math>
| |
| | |
| which leads to
| |
| | |
| :<math> \Bigl[ \prod_{i=1}^n u_i(x) \Bigr]_a^b = \sum_{j=1}^n \int_a^b \prod_{i\neq j}^n u_i(x) \, du_j(x), </math>
| |
| | |
| where the [[product (mathematics)|product]] is of all functions except for the one differentiated in the same term.
| |
| | |
| == Visualization ==
| |
| [[Image:Integration by parts v1.jpg|thumb|280px |Graphical interpretation of the theorem. The pictured curve is parametrized by the variable t.]]
| |
| Define a parametric curve by (''x'', ''y'') = (''f''(''t''), ''g''(''t'')). Assuming that the curve is locally [[Injective function|one-to-one]], we can define
| |
| :<math>x(y) = f(g^{-1}(y))</math>
| |
| :<math>y(x) = g(f^{-1}(x))</math>
| |
| | |
| The area of the blue region is
| |
| | |
| :<math>A_1=\int_{y_1}^{y_2}x(y)dy</math>
| |
| | |
| Similarly, the area of the red region is
| |
| :<math>A_2=\int_{x_1}^{x_2}y(x)dx</math>
| |
| | |
| The total area ''A''<sub>1</sub> + ''A''<sub>2</sub> is equal to the area of the bigger rectangle, ''x''<sub>2</sub>''y''<sub>2</sub>, minus the area of the smaller one, ''x''<sub>1</sub>''y''<sub>1</sub>:
| |
| | |
| :<math>\overbrace{\int_{y_1}^{y_2}x(y)dy}^{A_1}+\overbrace{\int_{x_1}^{x_2}y(x)dx}^{A_2}=\biggl.x_iy_i\biggl|_{i=1}^{i=2}</math>
| |
| | |
| Assuming the curve is smooth within a neighborhood, this generalizes to indefinite integrals:
| |
| :<math>\int xdy + \int y dx = xy</math>
| |
| Rearranging:
| |
| :<math>\int xdy = xy - \int y dx</math>
| |
| Thus integration by parts may be thought of as deriving the area of the blue region from the total area and that of the red region.
| |
| | |
| This visualisation also explains why integration by parts may help find the integral of an inverse function ''f''<sup>−1</sup>(''x'') when the integral of the function ''f''(''xv'') is known. Indeed, the functions ''x''(''y'') and ''y''(''x'') are inverses, and the integral ∫''x dy'' may be calculated as above from knowing the integral ∫''y dx''.
| |
| | |
| ==Application to find antiderivatives==
| |
| | |
| ===Strategy===
| |
| Integration by parts is a [[heuristic]] rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate it into a product of two functions ''u''(''x'')''v''(''x'') such that the integral produced by the integration by parts formula is easier to evaluate than the original one. The following form is useful in illustrating the best strategy to take:
| |
| | |
| :<math>\int uv \,dx = u \int v \,dx - \int \left ( u' \int v \,dx \right )\,dx.\!</math>
| |
| | |
| Note that on the right-hand side, ''u'' is differentiated and ''v'' is integrated; consequently it is useful to choose ''u'' as a function that simplifies when differentiated, and/or to choose ''v'' as a function that simplifies when integrated. As a simple example, consider:
| |
| | |
| :<math>\int \frac{\ln x}{x^2}\,dx.\!</math>
| |
| | |
| Since the derivative of ln ''x'' is 1/''x'', we make (ln ''x'') part of ''u''; since the antiderivative of 1/''x''<sup>2</sup> is −1/''x'', we make (1/''x''<sup>2</sup>) part of ''v''. The formula now yields:
| |
| | |
| :<math>\int \frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} - \int \biggl( \frac{1}{x} \biggr) \biggl( -\frac{1}{x} \biggr) \, dx.\!</math>
| |
| | |
| The antiderivative of −1/''x''<sup>2</sup> can be found with the [[power rule]] and is 1/''x''.
| |
| | |
| Alternatively, we may choose ''u'' and ''v'' such that the product ''u''' (∫''v dx'') simplifies due to cancellation. For example, suppose we wish to integrate:
| |
| | |
| :<math>\int\sec^2x\ln|\sin x|dx.</math>
| |
| | |
| If we choose ''u''(''x'') = ln |sin ''x''| and ''v''(''x'') = sec<sup>2</sup>x, then ''u'' differentiates to 1/ tan ''x'' using the [[chain rule]] and ''v'' integrates to tan ''x''; so the formula gives:
| |
| | |
| :<math>\int\sec^2x\ln|\sin x|dx=\tan x\ln|\sin x|-\int\tan x\frac{1}{\tan x}dx.</math>
| |
| | |
| The integrand simplifies to 1, so the antiderivative is ''x''. 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, in [[numerical 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.
| |
| | |
| ===Examples===
| |
| ;Polynomials and trigonometric functions
| |
| | |
| In order to calculate
| |
| | |
| :<math>I=\int x\cos (x) \,dx\,,</math> | |
| | |
| let:
| |
| :<math>u = x \Rightarrow d u = dx</math>
| |
| :<math>dv = \cos(x)\,dx \Rightarrow v = \int\cos(x)\,dx = \sin x</math>
| |
| | |
| then:
| |
| | |
| :<math>
| |
| \begin{align}
| |
| \int x\cos (x) \,dx & = \int u \, dv \\
| |
| & = uv - \int v \, du \\
| |
| & = x\sin (x) - \int \sin (x) \,dx \\
| |
| & = x\sin (x) + \cos (x) + C,
| |
| \end{align}
| |
| \!</math>
| |
| | |
| where ''C'' is an [[arbitrary constant of integration]].
| |
| | |
| For higher powers of ''x'' in the form
| |
| | |
| :<math>\int x^n e^x \,dx,\,\int x^n\sin (x) \,dx,\,\int x^n\cos (x) \,dx\,,</math>
| |
| | |
| repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of ''x'' by one.
| |
| | |
| ;Exponentials and trigonometric functions
| |
| | |
| An example commonly used to examine the workings of integration by parts is
| |
| | |
| :<math>I=\int e^{x} \cos (x) \,dx. </math>
| |
| | |
| Here, integration by parts is performed twice. First let
| |
| | |
| :<math>u = \cos(x) \Rightarrow du = -\sin(x)\,dx</math>
| |
| :<math>dv = e^x \, dx \Rightarrow v = \int e^x \,dx = e^x</math>
| |
| | |
| then:
| |
| | |
| :<math>\int e^{x} \cos (x) \,dx = e^{x} \cos (x) + \int e^{x} \sin (x) \,dx.\!</math>
| |
| | |
| Now, to evaluate the remaining integral, we use integration by parts again, with:
| |
| | |
| :<math>u = \sin(x) \Rightarrow du = \cos(x)\, dx</math>
| |
| :<math>dv = e^x \,dx \Rightarrow v = \int e^x \,dx = e^x.</math>
| |
| | |
| Then:
| |
| | |
| : <math>\int e^x \sin (x) \,dx = e^x \sin (x) - \int e^x \cos (x) \,dx. </math>
| |
| | |
| Putting these together,
| |
| | |
| :<math>\int e^x \cos (x) \,dx = e^x \cos (x) + e^x \sin (x) - \int e^x \cos (x) \,dx. </math>
| |
| | |
| The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get
| |
| | |
| :<math>2 \int e^{x} \cos (x) \,dx = e^{x} ( \sin (x) + \cos (x) ) + C\!</math>
| |
| | |
| which rearranges to:
| |
| | |
| :<math>\int e^x \cos (x) \,dx = {e^x ( \sin (x) + \cos (x) ) \over 2} + C'\!</math>
| |
| | |
| where again ''C'' (and ''C''<nowiki>'</nowiki> = ''C''/2) is an [[arbitrary constant of integration]].
| |
| | |
| A similar method is used to find the [[integral of secant cubed]].
| |
| | |
| ;Functions multiplied by unity
| |
| | |
| 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 times ''x'' is also known.
| |
| | |
| The first example is ∫ ln(''x'') d''x''. We write this as:
| |
| | |
| :<math>I=\int \ln (x) \cdot 1 \,dx.\!</math>
| |
| | |
| Let:
| |
| | |
| :<math>u = \ln(x) \Rightarrow du = \frac{dx}{x}</math>
| |
| :<math>dv = dx \Rightarrow v = x\, </math>
| |
| | |
| then:
| |
| | |
| : <math>
| |
| \begin{align}
| |
| \int \ln (x) \,dx & = x \ln (x) - \int \frac{x}{x} \,dx \\
| |
| & = x \ln (x) - \int 1 \,dx \\
| |
| & = x \ln (x) - x + C
| |
| \end{align}
| |
| </math>
| |
| | |
| where ''C'' is the [[constant of integration]].
| |
| | |
| The second example is the [[inverse tangent]] function arctan(''x''):
| |
| | |
| :<math>I=\int \arctan (x) \cdot 1 \,dx. </math>
| |
| | |
| Rewrite this as
| |
| | |
| :<math>\int \arctan (x) \cdot 1 \,dx. </math>
| |
| | |
| Now let:
| |
| | |
| : <math>u = \arctan(x) \Rightarrow du = \frac{dx}{1 + x^2}</math>
| |
| | |
| : <math>dv =dx \Rightarrow v = x</math>
| |
| | |
| then
| |
| | |
| : <math>
| |
| \begin{align}
| |
| \int \arctan (x) \,dx
| |
| & = x \arctan (x) - \int \frac{x}{1 + x^2} \,dx \\[8pt]
| |
| & = x \arctan (x) - {1 \over 2} \ln \left( 1 + x^2 \right) + C
| |
| \end{align}
| |
| </math>
| |
| | |
| using a combination of the [[inverse chain rule method]] and the [[natural logarithm integral condition]].
| |
| | |
| ===LIATE rule===
| |
| A [[rule of thumb]] proposed by Herbert Kasube of Bradley University advises that whichever function comes first in the following list should be ''u'':<ref>{{Cite journal|jstor=2975556|first=Herbert E. |last=Kasube|title=A Technique for Integration by Parts |journal=[[The American Mathematical Monthly]] |volume=90 |issue=3|year=1983 |pages=210–211|doi=10.2307/2975556}}</ref>
| |
| | |
| :'''L''' - [[Logarithmic function]]s: ln ''x'', log<sub>''b''</sub> ''x'', etc.
| |
| :'''I''' - [[Inverse trigonometric function]]s: arctan ''x'', arcsec ''x'', etc.
| |
| :'''A''' - [[Polynomial|Algebraic functions]]: ''x''<sup>2</sup>, 3''x''<sup>50</sup>, etc.
| |
| :'''T''' - [[Trigonometric functions]]: sin ''x'', tan ''x'', etc.
| |
| :'''E''' - [[Exponential function]]s: ''e''<sup>''x''</sup>, 19<sup>''x''</sup>, etc.
| |
| | |
| The function which is to be ''dv'' is whichever comes last in the list: functions lower on the list have easier [[antiderivative]]s than the functions above them. The rule is sometimes written as "DETAIL" where ''D'' stands for ''dv''.
| |
| | |
| To demonstrate the LIATE rule, consider the integral
| |
| | |
| :<math>\int x\cos x \, dx.\!</math>
| |
| | |
| Following the LIATE rule, ''u'' = ''x'' and ''dv'' = cos ''x'' dx, hence ''du'' = ''dx'' and ''v'' = sin ''x'', which makes the integral become
| |
| :<math>x\sin x - \int 1\sin x \, dx\!</math>
| |
| which equals
| |
| :<math> x\sin x + \cos x+C.\!</math>
| |
| | |
| In general, one tries to choose ''u'' and ''dv'' such that ''du'' is simpler than ''u'' and ''dv'' is easy to integrate. If instead cos ''x'' was chosen as ''u'' and ''x'' as ''dv'', we would have the integral
| |
| | |
| :<math> \frac{x^2}2\cos x + \int \frac{x^2}2\sin x\, dx,</math>
| |
| | |
| 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
| |
| | |
| :<math>\int x^3e^{x^2}\, dx,</math>
| |
| | |
| one would set
| |
| | |
| :<math>u=x^2, \quad dv=xe^{x^2}\, dx,</math>
| |
| | |
| so that | |
| | |
| : <math> du = 2x\,dx, \quad v = \frac12 e^{x^2}. </math>
| |
| | |
| Then
| |
| | |
| : <math> \int x^3e^{x^2}\, dx = \int \left(x^2\right) \left( xe^{x^2} \right) \, dx = \int u \, dv
| |
| = uv - \int v\,du = \frac12 x^2 e^{x^2} - \int xe^{x^2}\,dx.</math>
| |
| | |
| Finally, this results in
| |
| :<math>\int x^3e^{x^2}\, dx=\frac{1}{2}e^{x^2}(x^2-1)+C.</math>
| |
| | |
| ==Applications in pure mathematics==
| |
| | |
| Integration by parts is often used as a tool to prove theorems in [[mathematical analysis]]. This section gives a few of examples.
| |
| | |
| ===Use in special functions===
| |
| | |
| The [[gamma function]] is an example of a [[special function]], defined as an [[improper integral]]. Integration by parts illustrates it to be an extension of the [[factorial]]:
| |
| | |
| :<math>\begin{align}
| |
| \Gamma(z) & = \int_0^\infty d\lambda e^{-\lambda} \lambda^{z-1} \\
| |
| & = - \int_0^\infty d\left(e^{-\lambda}\right) \lambda^{z-1} \\
| |
| & = - \left[e^{-\lambda}\lambda^{z-1}\right]_0^\infty + \int_0^\infty d\left(\lambda^{z-1}\right) e^{-\lambda} \\
| |
| & = 0 + \int_0^\infty d\lambda\left(z-1\right) \lambda^{z-2} e^{-\lambda} \\
| |
| & = (z-1)\Gamma(z-1) \\
| |
| \end{align} </math>
| |
| | |
| yielding the famous identity
| |
| | |
| :<math>\Gamma(z) = (z-1)\Gamma(z-1)\,.</math>
| |
| | |
| For integer ''z'', applying this formula repeatedly gives the factorial (denoted by the !):
| |
| | |
| :<math>\Gamma(z+1) = z!</math>
| |
| | |
| ===Use in harmonic analysis===
| |
| | |
| Integration by parts is often used in [[harmonic analysis]], particularly [[Fourier analysis]], to show that quickly oscillating integrals with sufficiently smooth integrands decay quickly. The most common example of this is its use in showing that the decay of function's Fourier transform depends on the smoothness of that function, as described below.
| |
| | |
| ;Fourier transform of derivative
| |
| | |
| If ''f'' is a ''k''-times continuously differentiable function and all derivatives up to the ''k''th one decay to zero at infinity, then its [[Fourier transform]] satisfies
| |
| | |
| :<math>(\mathcal{F}f^{(k)})(\xi) = (2\pi i\xi)^k \mathcal{F}f(\xi),</math>
| |
| | |
| where {{nowrap|''f''<sup>(''k'')</sup>}} is the ''k''th derivative of ''f''. (The exact constant on the right depends on the [[Fourier transform#Other conventions|convention of the Fourier transform used]].) This is proved by noting that
| |
| | |
| :<math>\frac{d}{dy} e^{-2\pi iy\xi} = -2\pi i\xi e^{-2\pi iy\xi},</math>
| |
| | |
| so using integration by parts on the Fourier transform of the derivative we get
| |
| | |
| :<math>\begin{align}
| |
| (\mathcal{F}f')(\xi) &= \int_{-\infty}^\infty e^{-2\pi iy\xi} f'(y)\,dy \\
| |
| &=\left[e^{-2\pi iy\xi} f(y)\right]_{-\infty}^\infty - \int_{-\infty}^\infty (-2\pi i\xi e^{-2\pi iy\xi}) f(y)\,dy \\
| |
| &=2\pi i\xi \int_{-\infty}^\infty e^{-2\pi iy\xi} f(y)\,dy \\
| |
| &=2\pi i\xi \mathcal{F}f(\xi).
| |
| \end{align}</math>
| |
| | |
| Applying this [[Mathematical induction|inductively]] gives the result for general ''k''. A similar method can be used to find the [[Laplace transform]] of a derivative of a function.
| |
| | |
| ;Decay of Fourier transform
| |
| | |
| The above result tells us about the decay of the Fourier transform, since it follows that if ''f'' and {{nowrap|''f''<sup>(''k'')</sup>}} are integrable then
| |
| | |
| :<math>\vert\mathcal{F}f(\xi)\vert \leq \frac{I(f)}{1+\vert 2\pi\xi\vert^k}</math>, where <math>I(f)=\int_{-\infty}^\infty\Bigl(\vert f(y)\vert + \vert f^{(k)}(y)\vert\Bigr) dy</math>.
| |
| | |
| In other words, if ''f'' satisfies these conditions then its Fourier transform decays at infinity at least as quickly as {{nowrap|1/{{!}}''ξ''{{!}}<sup>''k''</sup>}}. In particular, if {{nowrap|''k'' ≥ 2}} then the Fourier transform is integrable.
| |
| | |
| The proof uses the fact, which is immediate from the [[Fourier transform#Definition|definition of the Fourier transform]], that
| |
| :<math>\vert\mathcal{F}f(\xi)\vert \leq \int_{-\infty}^\infty \vert f(y) \vert \,dy.</math>
| |
| Using the same idea on the equality stated at the start of this subsection gives
| |
| :<math>\vert(2\pi i\xi)^k \mathcal{F}f(\xi)\vert \leq \int_{-\infty}^\infty \vert f^{(k)}(y) \vert \,dy.</math>
| |
| Summing these two inequalities and then dividing by {{nowrap|1 + {{!}}2''πξ''<sup>''k''</sup>{{!}}}} gives the stated inequality.
| |
| | |
| ===Use in operator theory===
| |
| | |
| One use of integration by parts in [[operator theory]] is that it shows that the {{nowrap|-∆}} (where ∆ is the [[Laplace operator]]) is a [[positive operator]] on {{nowrap|''L''<sup>2</sup>}} (see [[Lp space|''L''<sup>''p''</sup> space]]). If ''f'' is smooth and compactly supported then, using integration by parts, we have
| |
| | |
| :<math>\begin{align}
| |
| \langle -\Delta f, f \rangle_{L^2} &= -\int_{-\infty}^\infty f''(x)\overline{f(x)}\,dx \\
| |
| &=-\left[f'(x)\overline{f(x)}\right]_{-\infty}^\infty + \int_{-\infty}^\infty f'(x)\overline{f'(x)}\,dx \\
| |
| &=\int_{-\infty}^\infty \vert f'(x)\vert^2\,dx \geq 0.
| |
| \end{align}</math>
| |
| | |
| ===Other applications===
| |
| <!---INCLUDING DERIVATIONS HERE WOULD BE TOO LENGTHLY, IDEALLY KEEP THIS AS A LIST--->
| |
| * For determining [[boundary condition]]s in [[Sturm–Liouville theory]]
| |
| * Deriving the [[Euler–Lagrange equation]] in the [[calculus of variations]]
| |
| | |
| ==Recursive integration by parts==
| |
| Integration by parts can often be applied [[recursion|recursive]]ly on the ∫ ''v du'' term to provide the following formula
| |
| | |
| :<math>\int uv = u v_1 - u' v_2 + u'' v_3 - \cdots + (-1)^{n-1}\ u^{(n-1)} \ v_{n} + (-1)^n \int{u^{(n)}v_{n}}.\!</math>
| |
| | |
| Here, ''u''′ is the first derivative of ''u'' and ''u''′′ is the second derivative. Further, ''u''<sup>(''n'')</sup> is a notation to describe its ''n''th derivative with respect to the independent variable. Another notation approved in the calculus theory has been adopted:
| |
| | |
| :<math>v_{n+1}(x)=\int\! \int\ \cdots \int v \ (dx)^{n+1}.\!</math>
| |
| | |
| There are ''n'' + 1 integrals.
| |
| | |
| Note that the integrand above (''uv'') differs from the previous equation. The ''dv'' factor has been written as ''v'' purely for convenience.
| |
| | |
| The above mentioned form is convenient because it can be evaluated by differentiating the first term and integrating the second (with a sign reversal each time), starting out with ''uv''<sub>1</sub>. It is very useful especially in cases when ''u''<sup>(''k'' + 1)</sup> becomes zero for some ''k'' + 1. Hence, the integral evaluation can stop once the ''u''<sup>(''k'')</sup> term has been reached.
| |
| | |
| ===Tabular integration by parts===
| |
| While the aforementioned [[recursion|recursive]] definition is correct, it is often tedious to remember and implement. A much easier visual representation of this process is often taught to students and is dubbed either "the tabular method",<ref>{{Cite journal|url=http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1125.pdu|first=Sanjay K. |last=Khattri|title=FOURIER SERIES AND LAPLACE TRANSFORM THROUGH TABULAR INTEGRATION|journal=[[The Teaching of Mathematics]] |volume=XI |issue=2|year=2008 |pages=97–103}}</ref> "the ''[[Stand and Deliver]]'' method",<ref>{{Cite journal|url=http://www.maa.org/pubs/Calc_articles/ma035.pdu|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> "rapid repeated integration" or "the tic-tac-toe method". This method works best when one of the two functions in the product is a polynomial, that is, after differentiating it several times one obtains zero. It may also be extended to work for functions that will repeat themselves.
| |
| | |
| For example, consider the integral
| |
| | |
| :<math>\int x^3 \cos x \, dx.\!</math>
| |
| | |
| Let ''u'' = ''x''<sup>3</sup>. Begin with this function and list in a column all the subsequent derivatives until zero is reached. Secondly, begin with the function ''v'' (in this case cos(''x'')) and list each integral of ''v'' until the size of the column is the same as that of ''u''. The result should appear as follows.
| |
| | |
| :{| class="wikitable" style="text-align:center"
| |
| ! Derivatives of ''u'' (Column A) !! Integrals of ''v'' (Column B)
| |
| |-
| |
| | <math> x^3 \, </math> || <math>\cos x \, </math>
| |
| |-
| |
| | <math> 3x^2 \, </math> || <math>\sin x \, </math>
| |
| |-
| |
| | <math> 6x \, </math> || <math>-\cos x \, </math>
| |
| |-
| |
| | <math> 6 \, </math> || <math>-\sin x \, </math>
| |
| |-
| |
| | <math> 0 \, </math> || <math>\cos x \, </math>
| |
| |}
| |
| | |
| Now simply pair 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 '''alternating''' signs (beginning with the positive sign). Do so until further pairing leads to sums of zeros. The result is the following (notice the alternating signs in each term):
| |
| | |
| :<math>(+)(x^3)(\sin x) - (3x^2)(-\cos x) + (6x)(-\sin x) - (6)(\cos x) + C \,. </math>
| |
| | |
| Which, with simplification, leads to the result
| |
| | |
| :<math>x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + C. \, </math>
| |
| | |
| With proper understanding of the tabular method, it can be extended. Consider
| |
| | |
| :<math>\int e^x \cos x \,dx. </math>
| |
| | |
| :{| class="wikitable" style="text-align:center"
| |
| ! Derivatives of ''u'' (Column A) !! Integrals of ''v'' (Column B)
| |
| |-
| |
| | <math> e^x \, </math> || <math>\cos x \, </math>
| |
| |-
| |
| | <math> e^x \, </math> || <math>\sin x \, </math>
| |
| |-
| |
| | <math> e^x \, </math> || <math>-\cos x \, </math>
| |
| |}
| |
| | |
| In this case in the last step it is necessary to integrate the product of the two bottom cells obtaining:
| |
| | |
| :<math> \int e^x \cos x \,dx = e^x\sin x + e^x\cos x - \int e^x \cos x \,dx, </math>
| |
| | |
| which leads to
| |
| | |
| :<math> 2 \, \int e^x \cos x \,dx = e^x\sin x + e^x\cos x, </math>
| |
| | |
| and yields the result: | |
| | |
| :<math>\int e^x \cos x \,dx = {e^x ( \sin x + \cos x ) \over 2} + C.\!</math>
| |
| | |
| ==Higher dimensions==
| |
| The formula for integration by parts can be extended to functions of several variables. Instead of an interval one needs to integrate over an ''n''-dimensional set. Also, one replaces the derivative with a [[partial derivative]].
| |
| | |
| More specifically, suppose Ω is an [[Open set|open]] [[bounded set|bounded subset]] of ℝ<sup>''n''</sup> with a [[piecewise smooth]] [[boundary (topology)|boundary]] Γ. If ''u'' and ''v'' are two [[smooth function|continuously differentiable]] functions on the [[Closure (topology)|closure]] of Ω, then the formula for integration by parts is
| |
| :<math>\int_{\Omega} \frac{\partial u}{\partial x_i} v \,d\Omega = \int_{\Gamma} u v \, \nu_i \,d\Gamma - \int_{\Omega} u \frac{\partial v}{\partial x_i} \, d\Omega,</math>
| |
| where <math>\hat{\mathbf{\nu}}</math> is the outward unit [[surface normal]] to Γ, ''ν<sub>i</sub>'' is its ''i''-th component, and ''i'' ranges from 1 to ''n''.
| |
| | |
| By replacing ''v'' in the above formula with ''v''<sub>''i''</sub> and summing over ''i'' gives the vector formula
| |
| :<math> \int_{\Omega} \nabla u \cdot \mathbf{v}\, d\Omega = \int_{\Gamma} u (\mathbf{v}\cdot \hat{\nu})\, d\Gamma - \int_\Omega u\, \nabla\cdot\mathbf{v}\, d\Omega,</math>
| |
| where '''v''' is a vector-valued function with components ''v''<sub>1</sub>, ..., ''v''<sub>''n''</sub>.
| |
| | |
| Setting ''u'' equal to the constant function 1 in the above formula gives the [[divergence theorem]]
| |
| :<math> \int_{\Gamma} \mathbf{v} \cdot \hat{\nu}\, d\Gamma = \int_\Omega \nabla\cdot\mathbf{v}\, d\Omega.</math> | |
| For <math>\mathbf{v}=\nabla v</math> where <math>v\in C^2(\bar{\Omega})</math>, one gets
| |
| :<math> \int_{\Omega} \nabla u \cdot \nabla v\, d\Omega = \int_{\Gamma} u\, \nabla v\cdot\hat{\nu}\, d\Gamma - \int_\Omega u\, \nabla^2 v\, d\Omega,</math>
| |
| which is the [[Green's identities|first Green's identity]].
| |
| | |
| The [[Differentiability class|regularity]] requirements of the theorem can be relaxed. For instance, the boundary Γ need only be [[Lipschitz continuous]]. In the first formula above, only ''u'', ''v'' ∈ ''H''<sup>1</sup>(Ω) is necessary (where ''H''<sup>1</sup> is a [[Sobolev space]]); the other formulas have similarly relaxed requirements.
| |
| | |
| ==See also==
| |
| * [[Lebesgue–Stieltjes_integral#Integration_by_parts|Integration by parts for the Lebesgue–Stieltjes integral]]
| |
| * [[Quadratic_variation#Semimartingales|Integration by parts]] for [[semimartingale]]s, involving their quadratic covariation.
| |
| * [[Integration by substitution]]
| |
| * [[Legendre transformation]]
| |
| | |
| ==Notes==
| |
| <references/>
| |
| | |
| ==References==
| |
| * {{Cite book| first=Lawrence C. | last=Evans | title=Partial Differential Equations | publisher=American Mathematical Society | location=Providence, Rhode Island | year=1998 | isbn=0-8218-0772-2}}
| |
| * {{Cite book| first=Todd | last=Arbogast | coauthors=Jerry Bona | title=Methods of Applied Mathematics | url=http://www.math.utexas.edu/users/arbogast/appMath08.pdu |format=Pdu| year=2005}}
| |
| *{{Cite journal| last = Horowitz | first = David | title = Tabular Integration by Parts | journal = The College Mathematics Journal | volume = 21 | issue = 4 | pages = 307–311 |date=September 1990 | doi = 10.2307/2686368 | jstor = 2686368}}
| |
| | |
| ==External links==
| |
| {{wikibooks|Calculus|Integration techniques/Integration by Parts|Integration by parts}}
| |
| * {{springer|title=Integration by parts|id=p/i051730}}
| |
| * [http://mathworld.wolfram.com/IntegrationbyParts.html Integration by Parts — From MathWorld]
| |
| | |
| [[Category:Integral calculus]]
| |
| | |
| [[es:Métodos de integración#Método de integración por partes]]
| |