What is integration by parts, really?

Question

Integration by parts comes up a lot - for instance, it appears in the definition of a weak derivative / distributional derivative, or as a tool that one can use to turn information about higher derivatives of a function into information about an integral of that function. Concrete examples of this latter category include: proving that $f \in C^2(S^1)$ implies that the Fourier series of $f$ converges absolutely and uniformly, and the Taylor series expansion with the integral formula for remainder.

However, I don't feel like I really understand what integration by parts is really doing. To me, it is just an algebraic trick that follows from the fundamental theorem of calculus and the product rule. Is there some more conceptual way to think about it?

How do you think about this useful idea?

Answer 1

I've always found it helpful to think about it like this: (picture source)

The area of the gray areas combined is $u_2v_2 - u_1 v_1$, which is where the $uv$ term comes from.

Answer 2

Integration by parts is a corollary of the product rule:

$(uv)' = uv' + u'v$

Take the integral of both sides to get $uv = \int u \ dv + \int v \ du$.

If you were supposed to remember it separately from the product rule then it's not as easy to work with as you have to make guesses as to what to assign $u$ and what to assign $dv$ (usually $dv = f(t) dt$). But if you have knowledge of product rule then you take the integrand (in terms of $t$) call it $F(t)$ and use the product rule on it first. Then your choices of $u, dv$ are readily obvious.

Answer 3

One idea is that integration by parts expresses the fact that the adjoint of $\frac{d}{dx}$ is $-\frac{d}{dx}$ (in a setting where boundary terms vanish). In the multivariable case, integration by parts expresses the fact that the adjoint of $\nabla$ is $-\text{div}$.

Answer 4

Might not be rigorous but this takes the cake for me:

$$\sum_{k=m}^n f_k(g_{k+1}-g_k) = [f_{n+1}g_{n+1} - f_m g_m] - \sum_{k=m}^n g_{k+1}(f_{k+1}- f_k).$$

$$\int u\,dv = uv - \int v\,du$$

The first formula takes a sum that includes a $f\Delta g$ term and transforms it to a sum containing a $g\Delta f$. Integration by parts takes an integral with a $u\,dv$ term and transforms it to an integral with a $v\,du$ term.

(Wikipedia- Summation by Parts)

Answer 5

Consider the integral $I=\int f(x)g(x)dx$, ie. the integral of the product of two functions.

Now imagine sliding $f(x)$ a small distance $\epsilon$ along the $x$-axis relative to $g$. With reasonable assumptions about differentiability, the integral becomes $$\int f(x+\epsilon)g(x)dx=\int (f(x)+\epsilon f'(x))g(x)dx+O(\epsilon^2)$$ which becomes $$I+\epsilon\int f'(x)g(x)dx+O(\epsilon^2).$$

So $\int f'(x)g(x)dx$ tells you how $I$ changes as you slide $f$. But this must be the opposite of sliding $g$ the other way. So it should also equal $\int f(x)g'(x)dx$, modulo some bits that fall off the end as you slide if your integration region has endpoints.

If you look at many applications of integration by parts, you may find this explanation fits well. It can help make the derivation of the Euler-Lagrange equations clearer and gives insight into its frequent use in domains like electromagnetism and quantum mechanics. For example it becomes completely obvious that the $p$ operator from quantum mechanics is Hermitian and it's clear how this is directly related to its role as the generator of translations.

Answer 6

I like to think of integration by parts as Fubini's Theorem. So if $$ F(x) = \int_a^x f(y) \, dy, \quad G(x) = \int_a^x g(y) \, dy ,$$ then $$ \int_a^b F(x) g(x) \, dx + \int_a^b f(x) G(x) \, dx $$ $$ = \int_{x=a}^b \int_{y=a}^x f(y) g(x) \, dy \, dx + \int_{x=a}^b \int_{y=a}^x f(x) g(y) \, dy \, dx $$ $$ = \int_{x=a}^b \int_{y=x}^b f(x) g(y) \, dy \, dx + \int_{x=a}^b \int_{y=a}^x f(x) g(y) \, dy \, dx $$ where in the first half I switched the roles of $x$ and $y$, and then interchanged the order of integration $$ = \int_{x=a}^b \int_{y=a}^b f(x) g(y) \, dy \, dx = F(b) G(b) = [ F(x)G(x) ]_{x=a}^b $$ remembering that $F(a) = G(a) = 0$. (I know in essence this is the same as Henry Swanson's answer, but this is a different perspective.)

Answer 7

A mathematical idea is useful for what you can use it for. A "conceptual" explanation may be attractive, but is relatively worthless for actually using integration by parts to do anything.

Instead, you should think of integration by parts in terms of how it lets you manipulate integrals; e.g. when you have $x$ or $\ln x$ or $\arctan x$ in an integrand, you can arrange to apply integration by parts differentiate it away into $1$ or $\frac{1}{x}$ or $\frac{1}{1+x^2}$ respectively, which potentially gives an integrand simpler than what you started with, depending on what the antiderivative of the cofactor is.

Or as another example, in the case of the distributional derivative, it let's you remove a derivative from one factor in an integrand by differentiating the other factor.

This isn't really something one understands a priori: instead, by working through problems, one gains experience and eventually an intuitive understanding of how it can be used to simplify an integrand.

Answer 8

I learned this point of view recently:

"Integration by parts is a consequence of the translation invariance of the Lebesgue measure"

What this means is the following:

$\int \frac{f(x + h) - f(x)}{h} g(x) dx = \frac{1}{h} [\int f(x + h)g(x) dx - \int_{\mathbb{R}} f(x) g(x) dx] = \frac{1}{h} [ \int_{\mathbb{R}} f(y)g(y - h) dy - \int_{\mathbb{R}} f(y)g(y) dy ] = \int_{\mathbb{R}} f(y) \frac{ g(y - h) - g(y)}{h} dy$

Translation invariance appears in the second equality, where we substitute $x + h = y$ into the first integral on the RHS.

Sending $h \to 0$ gives:

$\int_{\mathbb{R}} f'(x) g(x) dx = - \int_{\mathbb{R}} f(y) g'(y) dy$.

The boundary conditions can be obtained in the following way:

$$\int_a^b \frac{f(x + h) - f(x)}{h} g(x) dx = \frac{1}{h} [\int_a^b f(x + h)g(x) dx - \int_a^b f(x) g(x) dx] = \frac{1}{h} [ \int_{a + h}^{b + h} f(y)g(y - h) dy - \int_a^b f(y)g(y) dy ]$$ $$= \frac{1}{h} [ \int_{a}^{b} f(y)g(y - h) dy - \int_a^{a+h} f(y)g(y -h) dy + \int_b^{b+h} f(y) g(y - h) dy - \int_a^b f(y)g(y) dy ]$$ $$ = \frac{1}{h} [ \int_{a}^{b} f(y)[g(y - h) - g(y)] dy - \int_a^{a+h} f(y)g(y -h) dy + \int_b^{b+h} f(y) g(y - h) dy $$

Whereupon sending $h \to 0$ gives $\int_a^b f(x) g'(x) dx = - \int_a^b f(y) g'(y) dy- f(a)g(a) + f(b) g(b) $.

The broader picture that this story fits into is that this shows what must change in order to obtain the integration by parts formula in the Malliavin calculus. This also shows how to obtain an integration for parts formula against a measure whose translation in a particular direction is absolutely continuous.