Part 0 · The Toolkit — Chapter 0.2

Integration and Accumulation

The second half of calculus, and the one physics actually spends most of its time doing.

Tools you'll need  — Chapter 0.1, specifically the idea that a differentiable function is locally linear with an error that dies faster than the displacement. Everything below is that one fact, run in reverse or run in bulk.

1 · Accumulation: the integral as a limit of sums

Start with something you can picture. A car has velocity $v(t)$, which is not constant. How far does it travel between $t=a$ and $t=b$?

If $v$ were constant the answer would be $v\cdot(b-a)$ and there would be no chapter. Since it isn't, do the only honest thing available. Chop the trip into $N$ short intervals, each of length $\Delta t = (b-a)/N$. On a short enough interval the velocity barely changes, which is exactly the content of continuity. So on the $i$-th interval the distance covered is close to $v(t_i)\,\Delta t$ for any $t_i$ inside it. Now add them up:

Geometrically each term is the area of a rectangle of height $v(t_i)$ and width $\Delta t$, so the sum is the area of a staircase drawn under the graph of $v$. The approximation is bad for small $N$ and improves as the steps get finer, because the amount by which $v$ can wander inside one interval shrinks with the interval.

The definite integral is what (0.2.1) converges to. Written out properly, for a function $f$ on $[a,b]$: choose a partition $a = x_0 \lt x_1 \lt \cdots \lt x_N = b$, write $\Delta x_i = x_i - x_{i-1}$, define the mesh $\|P\| = \max_i \Delta x_i$, and pick a tag $x_i^{*}\in[x_{i-1},x_i]$ in each cell. Then

whenever that limit exists.

1.1 · What "the limit exists" is doing here

As in Chapter 0.1, the escape clause is the interesting part. The demand in (0.2.2) is severe: there must be a single number $I$ such that every partition with small enough mesh, tagged any way you like, gives a sum within $\varepsilon$ of $I$. You do not get to choose convenient tags. A function that passes this test is called Riemann integrable.

Continuous functions on a closed bounded interval pass, and the reason is worth seeing because it is the only place in this chapter where compactness does real work. On $[a,b]$ a continuous function is uniformly continuous (⚑ Heine–Cantor, quoted): for any $\varepsilon$ there is a single $\delta$ that works everywhere at once. So take any partition of mesh less than $\delta$. Inside cell $i$ the values of $f$ all lie within $\varepsilon$ of each other, so the largest possible sum and the smallest possible sum on that cell differ by at most $\varepsilon\,\Delta x_i$. Adding up,

Every tagged sum is trapped between the upper and lower sums, and (0.2.3) squeezes that gap to zero. So all of them converge to the same number. Continuity was not decoration. It is the hypothesis that makes the definition work.

Now a function that fails. Let $\chi(x)$ be $1$ when $x$ is rational and $0$ when it is irrational. Every interval, however short, contains both kinds of number. So on any partition you can tag entirely at rationals and get $\sum 1\cdot\Delta x_i = b-a$, or tag entirely at irrationals and get $0$. The sums do not converge to anything. So $\chi$ is not Riemann integrable, at any mesh, ever.

rectangles N

N = 8

left midpoint right

sum = 7.251701 exact = 7.239992

error = 1.171e-2

N × error = 0.0937 N² × error = 0.7493

1.2 · Why midpoint wins, and why you already knew

The figure showed midpoint tagging beating the other two by a whole power of $N$. That is worth explaining, because the reason is one you already have from Chapter 0.1.

The left rule approximates $f$ on each cell by a constant, $f(x_i)$. Chapter 0.1 tells you the error in a constant approximation is first order in the cell width: $f$ drifts by about $f'(x_i)\Delta x$ across the cell. Each cell therefore contributes an error of order $\Delta x^2$, and there are $N \sim 1/\Delta x$ of them, so the total error is order $\Delta x \sim 1/N$.

The midpoint rule also uses a constant, but it uses the value at the centre. That one change is the whole story. The linear part of $f$, measured from the centre, is odd: it overshoots on the right of the midpoint by exactly as much as it undershoots on the left. Those two errors cancel exactly, not approximately. So the midpoint rule is, silently, as accurate as the best linear approximation on each cell. Its leading error is therefore the first thing linearisation misses, which is the quadratic term $\tfrac12 f''h^2$ that Chapter 0.1 identified as the leftover. Cell error of order $\Delta x^3$, times $N$ cells, gives $\Delta x^2\sim 1/N^2$.

This is worth internalising because the move recurs constantly: put your sample point where the linear error is antisymmetric and it cancels. It is the same reason odd moments of a Gaussian vanish in §4, the same reason a symmetric difference quotient is second-order accurate, and the same reason the leading correction in a saddle-point expansion (Chapter 5.4) is quadratic rather than linear.

Grind box — the error constants, exactly, so you can check them against the figure

Work on one cell of width $d$ and expand $f$ using Chapter 0.1's quadratic form.

Left rule. With the cell $[x, x+d]$, the exact contribution is

$$\int_0^d f(x+s)\,\dd s = \int_0^d\!\Big[f(x) + f'(x)s + O(s^2)\Big]\dd s = f(x)\,d + \tfrac12 f'(x)\,d^{2} + O(d^{3}),$$

while the estimate is $f(x)d$. So the per-cell error (estimate minus truth) is $-\tfrac12 f'(x)d^{2}$. Sum over cells and recognise the leftover sum as another Riemann sum:

$$E_{\text{left}} \approx -\tfrac12 d^{2}\sum_i f'(x_i) = -\tfrac{d}{2}\sum_i f'(x_i)\,d \;\longrightarrow\; -\frac{d}{2}\int_a^b f' = -\frac{(b-a)\big[f(b)-f(a)\big]}{2N}.$$

For the figure's $f$ on $[0,3]$: $f(3)-f(0) = 1.64112$, so $N\times E_{\text{left}} \to -3(1.64112)/2 = -2.4617$. That is the number in the readout. The right rule is the same computation with the opposite sign, $+2.4617$.

Midpoint. Expand about the centre $m$ and integrate symmetrically. The odd term integrates to zero:

$$\int_{-d/2}^{d/2}\!\Big[f(m)+f'(m)s+\tfrac12 f''(m)s^{2}\Big]\dd s = f(m)\,d + 0 + \frac{f''(m)\,d^{3}}{24},$$

using $\int_{-d/2}^{d/2}s\,\dd s = 0$ and $\int_{-d/2}^{d/2}s^{2}\dd s = d^{3}/12$. The estimate is $f(m)d$, so the per-cell error is $-f''(m)d^{3}/24$ and

$$E_{\text{mid}} \to -\frac{d^{2}}{24}\int_a^b f'' = -\frac{(b-a)^{2}\big[f'(b)-f'(a)\big]}{24\,N^{2}}.$$

Here $f'(t) = \tfrac12+\cos t$, so $f'(3)-f'(0) = -1.98999$ and $N^{2}\times E_{\text{mid}} \to -9(-1.98999)/24 = +0.7462$. Also in the readout.

Trapezoid, and a bonus. Averaging the two endpoint values gives, after expanding both about $m$, an estimate $f(m)d + f''(m)d^{3}/8$, hence a per-cell error of $+f''(m)d^{3}/12$, which is exactly $-2$ times the midpoint error. So the combination

$$S = \tfrac13\big(2M + T\big)$$

annihilates the $d^{3}$ term identically. That is Simpson's rule, and it is not a lucky guess: it is the unique weighting that kills the leading error of two rules whose errors are in a known fixed ratio. Combining two approximations to cancel a known leading error is a manoeuvre with a name of its own, Richardson extrapolation, and you meet its physics incarnation in Chapter 5.11, where the "known leading error" is a divergence and cancelling it is renormalisation.

2 · The Fundamental Theorem, both halves, derived

Nothing so far connects (0.2.2) to derivatives. The definition is about areas and sums, while Chapter 0.1 was about local slopes. Watching those two collapse into one statement is the point of this section.

We take it in two halves, in the order they are usually needed. First we ask what happens when you differentiate an accumulation. Then we use the answer to evaluate an integral without summing anything at all.

2.1 · First half: differentiating an accumulation

Let $f$ be continuous on $[a,b]$ and define the accumulation function

$F(x)$ is the total accumulated between the fixed left end $a$ and the moving right end $x$. Ask Chapter 0.1's question about it: how fast does $F$ change as you slide $x$?

Push $x$ along by $h$. The accumulation from $a$ to $x+h$ is the accumulation from $a$ to $x$ plus whatever the sliver from $x$ to $x+h$ contributes. That is nothing more than additivity of the sum in (0.2.2) over adjacent blocks of cells:

Now estimate the sliver. Because $f$ is continuous at $x$, write $f(t) = f(x) + \eta(t)$ where $\eta(t) = f(t)-f(x)$, and let

That last statement is the definition of continuity at $x$, nothing more. Our goal now is to separate the sliver into a clean rectangle plus a leftover small enough to ignore, so split it in two:

The bound on the leftover follows straight from (0.2.2): every tagged sum for $\int\eta$ is at most $\varepsilon(h)$ times the total width $|h|$, so the limit is too. Combining (0.2.5) and (0.2.7):

Let's look at what that last line is actually saying. It is literally Chapter 0.1's boxed equation: a value, plus something linear in $h$, plus an error that dies faster than $h$. The coefficient of $h$ here is $f(x)$. And Chapter 0.1 established that the coefficient with that property is unique, and that it is called the derivative. So we can read the derivative straight off:

No trickery, no area formulas, no pictures required. The sliver is a thin rectangle of height $f(x)$ and width $h$. The error you make by pretending it is exactly a rectangle is second order, so it dies. Same move as the corner square in the product rule, same move as everything in Chapter 0.1.

2.2 · Second half: evaluating an integral

(0.2.9) is a statement about a function you build by integrating. The version you actually compute with runs the other way. Suppose you can find any function $G$ with $G' = f$, and call such a function an antiderivative. Then $G$ and the $F$ of (0.2.4) have the same derivative, so their difference has a vanishing derivative:

That a function with vanishing derivative on an interval must be constant is not free. It comes from the mean value theorem, which is derived in the grind box below. We do not know the constant $C$, and we do not need to, because our next move makes it cancel. Evaluate at both ends and subtract:

using $F(a)=\int_a^a f = 0$, since an integral whose two limits coincide accumulates nothing. What is left is the formula you actually compute with, and it deserves a box of its own:

This is the reason you can compute integrals at all. The limit of sums in (0.2.2) is an infinite process with no obvious shortcut. (0.2.12) replaces it with two evaluations of a function you found by running the differentiation rules backwards. Every integral you have ever computed by hand was this theorem being spent.

Grind box — the fine print: mean value theorem, and where $F' = f$ fails

Zero derivative implies constant. First Rolle's theorem: if $g$ is continuous on $[a,b]$, differentiable inside, and $g(a)=g(b)$, then $g'(c)=0$ somewhere inside. Why: a continuous function on a closed bounded interval attains a maximum and a minimum. If both occur at the endpoints then $g$ is constant and any interior $c$ works. Otherwise there is an interior extremum at $c$. There the difference quotient is $\ge 0$ approaching from one side and $\le 0$ from the other, and since the derivative exists both one-sided limits equal it, forcing $g'(c)=0$.

Now the mean value theorem. Given $G$ on $[a,b]$, apply Rolle to

$$g(x) = G(x) - G(a) - \frac{G(b)-G(a)}{b-a}\,(x-a),$$

which satisfies $g(a)=g(b)=0$. The conclusion $g'(c)=0$ reads $G(b)-G(a) = G'(c)\,(b-a)$. If $G'\equiv 0$ then every such difference vanishes, so $G$ takes the same value everywhere: constant. That is the step used in (0.2.10).

Where the first half breaks. (0.2.9) used continuity of $f$ at the point $x$. Drop it and the theorem drops with it. Take $f(t) = \operatorname{sign}(t)$ and $a=-1$. Then $F(x) = |x| - 1$, which is perfectly continuous everywhere but has no derivative at $0$, which is precisely Chapter 0.1's corner. The accumulation function exists. It just isn't differentiable where the integrand jumps.

Notice the direction of that asymmetry, because it matters later. Differentiating makes a function rougher: a corner in $F$ becomes a jump in $f$. Integrating makes it smoother: a jump in $f$ becomes a mere corner in $F$. This is why integral formulations of physical laws are better behaved than differential ones, why "weak solutions" of PDEs are defined by integrating against test functions, and why the Dirac delta of Chapter 0.9, an object with no pointwise meaning at all, is perfectly well defined the moment it appears inside an integral.

Fine print on the second half. (0.2.12) as derived assumed $f$ continuous, so that $F$ exists and $F'=f$. A slightly stronger version holds: if $f$ is merely Riemann integrable and happens to have some antiderivative $G$, then $\int_a^b f = G(b)-G(a)$ still, by applying the mean value theorem to $G$ on each cell of a partition and recognising the result as a tagged Riemann sum. You will not need the stronger version, but you should know the weaker one has hypotheses.

3 · Techniques, derived as reversed rules

Because of (0.2.12), integrating means recognising something as a derivative. Every differentiation rule from Chapter 0.1 therefore becomes an integration technique when read backwards. There are only two that matter, and one of them matters enormously.

3.1 · Substitution is the chain rule, reversed

Let $F' = f$ and let $g$ be differentiable. The chain rule says

Now read (0.2.13) from right to left. It says that the awkward-looking product on its right-hand side has a known antiderivative, namely $F\circ g$. An antiderivative is exactly what (0.2.12) asks for, so feed it in:

That is substitution, derived rather than asserted. Two things are worth noticing. First, the informal ritual "let $u=g(x)$, so $\dd u = g'(x)\dd x$" is a compressed way of writing exactly (0.2.14), and the licence to manipulate $\dd u$ that way is the same licence Chapter 0.1 discussed under "one deliberate abuse". Second, the limits transform too. Forgetting that is the most common error in the entire technique. It is also why $\int_{-1}^{1}x\,\ee^{-x^2}\dd x$ is zero: under $u=x^2$ both limits become $1$, and any integral whose two limits coincide is zero. The symmetry argument and the substitution argument agree, as they must.

3.2 · Integration by parts is the product rule, reversed

Take the product rule $(uv)' = u'v + uv'$, integrate both sides from $a$ to $b$, and apply (0.2.12) to the left (whose integrand is manifestly a derivative):

We want the integral we cannot do sitting alone on one side, so move the other one across. That rearrangement is integration by parts:

Read (0.2.16) as an instruction rather than a formula: you may move a derivative from one factor onto the other, at the cost of a minus sign and a boundary term. That sentence is, without much exaggeration, the most-used manipulation in theoretical physics. It is not primarily a trick for computing $\int x\ee^{x}$. It is the mechanism by which:

• An action becomes a field equation. In Chapter 1.2 you vary $S=\int L(q,\dot q)\,\dd t$ and get a term proportional to $\delta\dot q$, which is a variation of a derivative and cannot be set to zero independently. By parts moves the time derivative off $\delta \dot q$ and onto the coefficient, leaving everything multiplying $\delta q$ alone. That single step is what turns "the action is stationary" into the Euler–Lagrange equation, and in Chapter 5.2 the identical step in four dimensions turns a field Lagrangian into a field equation. Every field equation in this book is produced by (0.2.16).
• An operator gets an adjoint. In Chapter 4.4, $\int \psi^{*}\,\partial_x\phi\,\dd x = -\int(\partial_x\psi)^{*}\phi\,\dd x$ when boundary terms vanish. The minus sign is why $\hat p = -\ii\hbar\,\partial_x$ is Hermitian and $\partial_x$ alone is not. The factor of $\ii$ exists to absorb the minus sign that (0.2.16) generates. That is the entire origin of the $\ii$ in the momentum operator.
• A derivative gets defined for objects that have none. The derivative of a distribution in Chapter 0.9 is defined by declaring (0.2.16) to hold.

Grind box — the bookkeeping, and why "the boundary term vanishes" is physics, not maths

Worked mechanics. $\int_0^{\infty} x\,\ee^{-x}\dd x$. Choose $u=x$ and $v'=\ee^{-x}$, so $u'=1$ and $v=-\ee^{-x}$. Then

$$\int_0^{\infty} x\,\ee^{-x}\dd x = \Big[-x\ee^{-x}\Big]_0^{\infty} + \int_0^{\infty}\ee^{-x}\dd x = 0 + 1 = 1.$$

The boundary term vanished at $\infty$ because $\ee^{-x}$ beats $x$, and at $0$ because $x$ beats $\ee^{-x}$. That is a check, performed, not an assumption. Choosing $u$ and $v'$ the other way round gives $\int x\ee^{-x} = \big[\tfrac12x^2\ee^{-x}\big] + \tfrac12\int x^2\ee^{-x}$, which is true and useless: the remaining integral is worse than the one you started with. The craft in by parts is entirely in choosing which factor gets differentiated. Pick the one that gets simpler.

The move that builds physics. Here is the Chapter 1.2 step, done now so it is not new later. Let $S[q]=\int_{t_1}^{t_2} L(q,\dot q,t)\,\dd t$ and perturb the path $q \to q + \delta q$. To first order,

$$\delta S = \int_{t_1}^{t_2}\!\left[\pdv{L}{q}\,\delta q + \pdv{L}{\dot q}\,\delta\dot q\right]\dd t.$$

The second term is stuck: $\delta\dot q$ is not independent of $\delta q$, it is its derivative. Use (0.2.16) with $u = \partial L/\partial\dot q$ and $v' = \delta \dot q$, noting $\delta\dot q = \dv{}{t}\delta q$:

$$\int_{t_1}^{t_2}\pdv{L}{\dot q}\,\dv{}{t}(\delta q)\,\dd t = \left[\pdv{L}{\dot q}\,\delta q\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}\dv{}{t}\!\left(\pdv{L}{\dot q}\right)\delta q\,\dd t.$$

The boundary term dies because we only compare paths with the same endpoints, so $\delta q(t_1)=\delta q(t_2)=0$. What is left is

$$\delta S = \int_{t_1}^{t_2}\left[\pdv{L}{q} - \dv{}{t}\pdv{L}{\dot q}\right]\delta q\,\dd t,$$

and now $\delta q$ is arbitrary and unconstrained, so demanding $\delta S=0$ for all of them forces the bracket to vanish pointwise. That bracket is the Euler–Lagrange equation. Notice that by parts did all the work: it converted an unusable variation of a derivative into a usable variation of the coordinate.

The honest warning. In field theory you will constantly see $\int \partial_\mu\phi\,\partial^\mu\phi \to -\int\phi\,\Box\phi$ with the boundary term "dropped". That is not a mathematical triviality. It is the physical assumption that the fields fall off fast enough at infinity. Sometimes they don't, and then the boundary term is the interesting object rather than the discarded one:

• In general relativity the Einstein–Hilbert action needs an explicit boundary term (the Gibbons–Hawking term) added by hand, precisely because the naive one does not vanish (Chapter 3.6).
• Boundary terms at a black hole horizon are where the entropy lives (Chapter 7.9).
• Topological terms in gauge theory are total derivatives, which is to say pure boundary, and they still change the physics. That is how the strong CP problem exists at all (Chapter 6.5).

So: drop boundary terms freely, but know each time that you have made a physical claim about what happens far away, and be ready to be asked to defend it.

4 · The Gaussian integral

We now derive the single most-used integral in this book. It appears in the normalisation of every quantum wave packet (Chapter 4.6) and in the free-particle propagator (Chapter 5.4). And this is not an overstatement: the entire perturbative expansion of quantum field theory consists of Gaussian integrals plus corrections (Chapter 5.7). Learn it once, properly.

First, does it even converge? For $|x|\ge 1$ we have $x^{2}\ge|x|$, so $\ee^{-x^{2}}\le \ee^{-|x|}$, and $\int_1^{\infty}\ee^{-x}\dd x = \ee^{-1}$ is finite. On $[-1,1]$ the integrand is bounded by $1$. So $I$ is finite and positive, and we are entitled to go looking for its value.

We cannot find it the usual way, because $\ee^{-x^{2}}$ has no elementary antiderivative. The callout after §5 makes that precise, and it is a theorem rather than a failure of effort. So we do something that looks like cheating and isn't: square it.

Renaming the second dummy variable $y$ is free. Combining the two one-dimensional integrals into one two-dimensional integral is the real step, and it is legal because the integrand is positive (grind box). Now the point. The quantity $x^{2}+y^{2}$ is the squared distance from the origin, so the integrand depends only on distance. That is exactly the situation polar coordinates are for, and our next job is to find out what $\dd x\,\dd y$ turns into when we change to them.

4.1 · The area element, derived

Set $x = r\cos\theta$, $y=r\sin\theta$, so $x^{2}+y^{2}=r^{2}$. The one thing that needs care is what $\dd x\,\dd y$ becomes. A cell of the polar grid runs from $r$ to $r+\dd r$ and from $\theta$ to $\theta+\dd\theta$. Its area is the difference of two circular sectors, and a sector of radius $\rho$ and angle $\dd\theta$ has area $\tfrac12\rho^{2}\dd\theta$ (that is what radian measure means):

That is exact, and the leftover in (0.2.19) is second order in the cell size, so it dies in the limit. This is the same corner-square argument as the product rule in Chapter 0.1, wearing a different hat. Hence $\dd x\,\dd y = r\,\dd r\,\dd\theta$. Chapter 0.6 will produce the same factor $r$ mechanically, as the Jacobian determinant. It is worth having seen it geometrically first, because the Jacobian is nothing but "how much a coordinate change stretches area", and here you can watch it stretch.

4.2 · The value

With the area element settled, we can make the substitution. The whole plane is covered by $r\in[0,\infty)$ and $\theta\in[0,2\pi)$, so

And that integral is elementary, because the stray factor of $r$ is exactly the $g'$ that substitution wants: with $u=r^{2}$, $\dd u = 2r\,\dd r$,

So $I^{2} = 2\pi\cdot\tfrac12 = \pi$. We established at the start that $I \gt 0$, so we may take the positive square root, and the Gaussian integral is ours:

Let's pause on why this worked, because the mechanism is more useful than the answer. In one dimension $\ee^{-x^{2}}$ has no elementary antiderivative. In two dimensions, after going polar, the integrand acquires a factor of $r$ from the area element, and $r\,\ee^{-r^{2}}$ does have one. The trick is not a trick about exponentials. It is the observation that the geometry of the plane supplies the missing factor. Nothing in the one-dimensional problem could have told you that $\pi$ was going to appear. It came from the angular integral, which is to say from the fact that a circle exists.

Grind box — is squaring the integral legal, and the polar limit

Two things were waved at in (0.2.18) and both are fine, but you should know why.

Turning a product of integrals into a double integral. ⚑ Quoted: this is Fubini's theorem, and for a non-negative integrand it holds with no extra hypotheses at all (that version is Tonelli's theorem). Our integrand $\ee^{-(x^2+y^2)}$ is positive everywhere, so we are safe. For oscillatory integrands the same step can fail spectacularly, and in Chapter 5.4 the Gaussian we need is $\ee^{\ii a x^{2}}$, which is purely oscillatory and does not decay at all. The standard repair is to give $a$ a small imaginary part, do the honest convergent integral, and take the limit at the end. That deformation has a name, the $\ii\epsilon$ prescription, and it is not bookkeeping: it is what selects the causal propagator over the anticausal one.

The infinite region. Strictly, $\iint_{\R^2}$ means a limit of integrals over expanding bounded regions, and one should check the answer does not depend on how you expand. Squares of side $2L$ give $I_L^{2}$ where $I_L=\int_{-L}^{L}\ee^{-x^2}\dd x$. Discs of radius $R$ give $\pi(1-\ee^{-R^{2}})$ by the same polar computation. Since every disc of radius $L$ sits inside the square of side $2L$, which sits inside the disc of radius $L\sqrt2$,

$$\pi\big(1-\ee^{-L^{2}}\big) \;\le\; I_L^{2} \;\le\; \pi\big(1-\ee^{-2L^{2}}\big),$$

and both ends go to $\pi$. Squeezed. The answer is independent of the exhaustion, which for a positive integrand it always is.

4.3 · The general Gaussian

The Gaussian we meet in physics has a width, and usually a linear term as well, so let's generalise in two steps. First the width. For $a \gt 0$ put $u = \sqrt a\,x$, so $\dd u = \sqrt a\,\dd x$, and the limits are unchanged because rescaling the whole line leaves it the whole line:

Now add a linear term, which is where it starts being useful. The move is to complete the square:

The last term is a constant as far as $x$ is concerned, so it comes out of the integral. The first is a shifted Gaussian, and shifting the integration variable over the whole real line changes nothing, since the whole line is what we integrate over either way. Hence

Commit (0.2.25) to memory. In Chapter 5.7 the variable $x$ becomes an entire field configuration, $a$ becomes a differential operator, and $b$ becomes an external source $J$. With almost no change of shape, the formula becomes $Z[J] = Z[0]\exp\big(\tfrac12\!\int\!\!\int J\,\Delta\,J\big)$, the generating functional of a free quantum field theory. $\Delta$ is the propagator, and it sits exactly where $1/4a$ sits here.

4.4 · Differentiating under the integral sign

What about $\int x^{2}\ee^{-ax^{2}}\dd x$? You could hunt for a substitution. Don't. Notice instead that $a$ appears in (0.2.23) as a parameter, and that differentiating the integrand with respect to it pulls down exactly the factor you want:

Suppose for the moment that we may swap the order of $\partial_a$ and $\int\dd x$. The grind box below shows that we may here, and shows what has to go wrong for the swap to fail. Granting it, differentiating both sides of (0.2.23) with respect to $a$ hands us an integral we never had to solve:

Nothing stops us doing that again. Each further derivative in $a$ brings down another factor of $-x^{2}$, so every even moment comes out for free:

while every odd moment vanishes because $x^{2n+1}\ee^{-ax^{2}}$ is an odd function integrated over a symmetric interval. That is the same cancellation that made the midpoint rule good in §1.2.

This technique deserves a name and a place in your hands: parameter differentiation (sometimes "Feynman's trick", though it long predates him). You introduce a parameter the original problem did not have, differentiate or integrate with respect to it, and trade a hard integral for an easy one plus some calculus. Problem 2 below is the cleanest possible example.

4.5 · The same number, two ways — and why the second way is quantum field theory

There is a second route to (0.2.27) that looks like a detour and is actually the whole of Chapter 5.7. Instead of differentiating with respect to $a$, differentiate (0.2.25) with respect to the source $b$. Since $\partial_b\,\ee^{bx} = x\,\ee^{bx}$, each derivative in $b$ brings down one power of $x$:

Do the left side from the closed form: $\partial_b^{2}\big[\sqrt{\pi/a}\;\ee^{b^{2}/4a}\big] = \sqrt{\pi/a}\,\ee^{b^{2}/4a}\big[\tfrac{1}{2a} + \tfrac{b^{2}}{4a^{2}}\big]$, which at $b=0$ is $\tfrac{1}{2a}\sqrt{\pi/a}$. Same answer as (0.2.27). ✓

Now let's read what just happened structurally. We built a function of a source, $Z(b)$. We then differentiated it with respect to the source and switched the source off, and out came an average of a power of the integration variable. Rename $Z(b)$ the generating functional, rename $x$ the field, rename "average of a power" a correlation function, and you have the working method of quantum field theory, in which every Feynman diagram is a term produced by differentiating $\ee^{b^{2}/4a}$ enough times. The whole apparatus is (0.2.25) plus bookkeeping.

Grind box — when you may differentiate under the integral sign, and when you may not

The swap in (0.2.27) is a genuine hypothesis, not a formality. It is a statement that a limit (the derivative in $a$) commutes with another limit (the integral), and limits do not generally commute.

A concrete failure. Let $h(x,a) = \dfrac{2x}{a^{2}}\ee^{-x^{2}/a^{2}}$ for $a\gt0$. Substituting $u = x^{2}/a^{2}$,

$$\int_0^{1} h(x,a)\,\dd x = \int_0^{1/a^{2}}\ee^{-u}\,\dd u = 1-\ee^{-1/a^{2}} \;\xrightarrow[a\to0]{}\; 1.$$

But for any fixed $x\gt0$, $h(x,a)\to 0$ as $a\to0$ (the exponential beats the prefactor). So $\lim\int = 1$ while $\int\lim = 0$. The mass has not vanished. It has escaped into an ever narrower, ever taller spike near the origin. Hold that image: it is the Dirac delta being born, and Chapter 0.9 makes it respectable.

The condition that saves us. The sufficient hypothesis (dominated convergence, proved properly in Chapter 4.3) is that the $a$-derivative of the integrand be bounded, uniformly in $a$ near the point of interest, by a single fixed integrable function. Check it for our case: fix $a_0\gt0$ and let $a$ range over $[a_0,\infty)$. Then

$$\left|\pdv{}{a}\ee^{-ax^{2}}\right| = x^{2}\ee^{-ax^{2}} \;\le\; x^{2}\ee^{-a_0x^{2}},$$

and the right-hand side is a fixed function of $x$ with finite integral (it is (0.2.27) itself, which we have just evaluated as finite). So the swap is justified for every $a\gt0$, which is every case we care about. The moral is the book's general one: the manipulation is legal, and you should be able to say why on demand.

width σ

σ = 0.398

peak = 1.0021 peak × σ = 0.39894

total area = 1.00000 area in ±σ = 0.68269

5 · Improper integrals, and when things diverge

Every integral in §4 ran to infinity, so we should be precise about what that means. An integral over an unbounded region is defined as a limit of integrals over bounded ones:

and it exists only if that limit does. The canonical test case is the power law, and it is worth doing once because it decides essentially every convergence question in physics.

For $p\neq1$, (0.2.12) gives

and everything hinges on the sign of $1-p$. If $p \gt 1$ then $1-p\lt0$, so $R^{1-p}\to0$ and the integral converges to $1/(p-1)$. If $p\lt1$ then $R^{1-p}\to\infty$ and it diverges. The borderline $p=1$ has to be done separately because the antiderivative changes character: $\int_1^R x^{-1}\dd x = \ln R \to \infty$. So

The second half of (0.2.32) is the identical computation with the limit taken at the other end ($\int_\epsilon^1 x^{-p}\dd x = (1-\epsilon^{1-p})/(1-p)$), and it says something you should hold onto: the same integrand can fail at either end, and for opposite reasons. $1/x$ diverges at both. Nothing converges at both ends by being a pure power law. You need different behaviour in the two regimes, and that is why real physical answers always have a scale in them somewhere.

Which brings us to the payoff. The quantum corrections of Chapter 5.10 look like $\int^{\Lambda}\dd^{4}k/k^{4}\sim\ln\Lambda$ and diverge as the cutoff is removed. That is not a new pathology of quantum field theory. It is exactly (0.2.32) in momentum space, failing at the large-$k$ ("ultraviolet") end, with the small-$k$ ("infrared") end able to fail independently for the other reason. Renormalisation is the disciplined handling of that: such an integral was never a physical quantity on its own, and what you actually measure is a combination in which the divergence cancels and a finite, scale-dependent answer survives.

6 · Worked examples

Two examples, each spending a different part of the chapter. The first turns an integral over an infinite region into a number you already know. The second gets an integral without ever doing it.

7 · Your turn

Four problems. The first two set integration by parts and parameter differentiation against each other on the same family of integrals, so you can see what each one costs. The third is a mass balance you already trust, redone as a boundary-term argument. The fourth builds most of the uncertainty principle out of this chapter alone.