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A constant added to a function changes its reference level without changing its derivative:Geometrically, it is a vertical translation of the graph. For $F(x)=x^3/3$, changing $C$ moves every plotted value by the same amount while preserving every tangent slope. This is the geometric reason that antiderivatives form a family.

When $F$ represents a physical quantity, the physical dimension of $C$ must match that of $F$. When position is obtained from velocity, $C$ is a position offset. An unknown starting position remains unknown even when the rate of change is known exactly.

When describing the family of antiderivatives on an interval, the arbitrary additive constant is the constant of integration. On a disconnected open domain, the corresponding freedom is a locally constant function, as in antiderivatives on disconnected domains, rather than necessarily one constant on the whole domain.

Let $U$ be an open subset of the real line, and let $F,G$ be antiderivatives of the same function on $U$. Each connected component $I_j$ is an open interval. On that component, $(G-F{)}^{\prime }=0$, so the mean value theorem gives $G-F=C_j$. The constants on different components are independent: there is no subinterval inside $U$ joining them. Conversely, adding any chosen constant on each component preserves the derivative everywhere on $U$.

For example, the domain of $g(x)=1/x$ is $U=\mathbb{R}\setminus \{0\}$. Every antiderivative on this domain has the formwith arbitrary real $C_{-},C_{+}$. The logarithm of the absolute value differentiates to $1/x$ on each side. The function with $C_{-}=2$ and $C_{+}=-1$ is therefore a valid antiderivative, although it cannot be written as $\log |x|+C$ with one global constant.

An initial condition at $x=1$ fixes $C_{+}$ and leaves $C_{-}$ free. Fixing both constants requires one reference value in each connected component, or additional conditions that couple the components. The locally constant function $F(x)-\log |x|$ captures this remaining freedom.

For constant mass density $\rho$ on the lower half of the spherical shell $a\le |{\mathbf{r}}^{\prime }|\le b$, with $0<a<b$, the displayed Newtonian gravitational potential applies at $(0,0,h)$ for every $h>0$, with zero at infinity. Use source spherical coordinates with polar angle $\alpha \in [\pi /2,\pi ]$. For a fixed source radius $s$,Multiplication by $-2\pi G\rho s^2$ and integration from $a$ to $b$ leaves the antiderivative $s^3/3+h{s}^2/2-(s^2+h^2{)}^{3/2}/3$, proving the formula. The entire positive axis is outside the matter, even for $a<h<b$, so there are no interior/exterior radial cases. The limit at the origin is $-\pi G\rho (b^2-a^2)$; far away the leading term is $-GM/h$, with total mass $M=2\pi \rho (b^3-a^3)/3$.

The change of variables formula expresses an integral in coordinates that make its integrand or its region simpler. Three ingredients change together: the region, the function being integrated, and the element of area or volume. The last change is measured by the absolute Jacobian determinant; forgetting it amounts to counting transformed cells as though their sizes had stayed the same.

In one dimension, substitution is the integrated chain rule. If $g$ is continuously differentiable on $[a,b]$ and $f$ is continuous on its image, thenThese are directed definite integrals: if $g$ decreases, the transformed endpoints reverse. This identity does not require $g$ to be injective, because an antiderivative composed with $g$ already accounts for any backtracking.^[ref]

For a concrete computation, set $u=1+x^2$, so $du=2xdx$. The endpoints $x=0,1$ become $u=1,2$:The numerator is precisely the derivative of the expression in the denominator. Substitution reveals the familiar natural logarithm hidden inside this rational function. Keeping the original endpoints after changing the integrand would describe a different integral.

For the elementary two-dimensional theorem, let $T:U\to V$ be a $C^1$ diffeomorphism between open sets in the plane, let $S\subset U$ be a bounded region with piecewise smooth boundary, and let $f$ be continuous on $T(S)$, with the displayed integrals finite. ThenHere $DT$ is the Jacobian matrix. The diffeomorphism assumption means a bijection with a continuously differentiable inverse, and ensures a nonzero Jacobian determinant. This excludes overlapping sheets; the more general area formula handles multiplicity.^[ref]

Take $S=[0,1{]}^2$ and the linear mapIts inverse is $(u,v)=((x-y)/2,y)$. The square becomes a parallelogram with vertices $(0,0),(2,0),(3,1),(1,1)$: base two, height one, area two. Consequently ${\iint }_{T(S)}1dxdy={\iint }_{S}2dudv=2$. A nonconstant integrand changes too: ${\iint }_{T(S)}xdxdy={\int }_0^1{\int }_0^1(2u+v)2dudv=3$.

The direction of the map matters when selecting the factor. Here $T$ goes from the new coordinates to the original coordinates, so the factor is two. Its inverse has Jacobian determinant $1/2$, appropriate for converting area in the opposite direction. Writing down the map before computing its Jacobian matrix prevents a reciprocal-factor error.

Now reflect the image vertically: $\tilde{T}(u,v)=(2u+v,-v)$ has Jacobian determinant $-2$, but still produces area two. The sign records reversal of the orientation of a vector space; ordinary area uses its absolute value. For oriented differential forms, the pullback of a differential form instead retains the signed determinant.^[ref]

The geometric beauty is that a nonlinear differentiable map has a local linear approximation: tiny rectangles behave approximately like these parallelograms. Summing their scaled contributions gives the change of variables formula. In polar coordinates this factor is $r$, explaining $dxdy=rdrd\theta$; equal angular steps cover more area farther from the origin. Choosing suitable coordinates can simplify both the boundary and the integrand.^[ref]

Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

On a connected open interval, if $F^{\prime }=g$, every antiderivative of $g$ has the form $F+C$, where $C$ is a constant function. Indeed, two solutions satisfy $(G-F{)}^{\prime }=0$: the zero derivative on a connected open set principle makes their difference constant. In one dimension, the mean value theorem proves this on any closed subinterval. ^[ref]

An initial condition or boundary condition can determine $C$. If a particular antiderivative is $F_0$ and $F(x_0)=y_0$, thenFor $g(x)=x^2$, $F_0(x)=x^3/3$ and $F(1)=2$, this gives $C=5/3$. Without that reference value, choosing $C=0$ simply selects one member of the family.

The freedom has a physical meaning in a simple kinematic model. A velocity $v(t)=at$, with constant acceleration $a$, determines position only as $x(t)=a{t}^2/2+C$. The initial condition $x(0)=x_0$ selects $C=x_0$; changing the starting position does not change the prescribed velocity.

For the differential equation $T^{\prime }=-B/r^2$ on $r>0$, the antiderivative is $T=B/r+C$. The boundary condition $T(R_s)=T_s$ gives $C=T_s-B/R_s$. Discarding the constant of integration therefore needs a specified condition or an approximation that quantifies its effect. On a domain with several connected components, the freedom is larger: antiderivatives on disconnected domains require a separate constant on each component.

Let $H$ be any antiderivative of the initial velocity, $H^{\prime }=g$. One choice of profiles isOpposite additive constants in $F$ and $G$ cancel, so the displacement is independent of the chosen antiderivative. Even for compact support of $g$, these separate profiles can approach nonzero constants; cancellation in their sum is essential to support widening for the wave equation.

For a short derivation, the characteristic coordinates $\xi =x-ct$, $\eta =x+ct$ give $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$ by the chain rule. Thus $u=F(\xi )+G(\eta )$, and the Cauchy data imply $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$, giving the profiles above.

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

The complex exponential functions $e^{int}$, for integers $n$, are orthogonal under the normalized integral inner product over a complete period. If $n=m$, the integrand $e^{i(n-m)t}$ is one; otherwise its antiderivative has equal endpoint values, giving zero. For the real modes, ${\int }_{-\pi }^{\pi }{\cos }^2(nt)dt={\int }_{-\pi }^{\pi }{\sin }^2(nt)dt=\pi$ for $n\ge 1$, while every mixed sine-cosine integral and every unequal-frequency pairing vanishes.

This makes the coefficient formulas useful: multiply a trigonometric polynomial by one mode and integrate, and all other modes cancel. For $f(x)=1/2+\cos x+\frac{1}{2}\sin (2x)$, integration against $\cos x$ gives $\pi$, and against $\sin (2x)$ gives $\pi /2$, recovering $a_1=1$ and $b_2=1/2$. Division by the squared norm of the mode supplies the normalization. A full period matters; truncating the integration interval generally destroys these cancellations.

Let $L>0$ and real $\kappa \ne 0$. Extend $e^{\kappa x}$ with period $2L$. In the Fourier series convention $a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L)]$, put $q_n=n\pi /L$. Direct integration givesFor example, the required antiderivatives are $e^{\kappa x}(\kappa \cos q_{n}x+q_n\sin q_{n}x)/({\kappa }^2+q_n^2)$ and $e^{\kappa x}(\kappa \sin q_{n}x-q_n\cos q_{n}x)/({\kappa }^2+q_n^2)$. The Dirichlet-Jordan convergence theorem gives the exponential value away from the jumps and their midpoint $\cosh (\kappa L)$ at odd multiples of $L$. For $\kappa =0$, the function is identically one, with $a_0=2$ and all other Fourier coefficients zero.

For $\beta >0$, periodically extend $\cosh (\beta x)$ and $\sinh (\beta x)$ from $-1<x\le 1$ with period $2$. Their Fourier series areand, away from odd integers,The first identity follows by integration of $\cosh (\beta x)\cos (n\pi x)$ on $[-1,1]$, using the antiderivativeFor the second identity, integration by parts and $d\cosh (\beta x)/dx=\beta \sinh (\beta x)$ show that its sine coefficient is $-n\pi /\beta$ times the first series' cosine coefficient. This avoids assuming that a Fourier series can always be differentiated term by term. The periodic hyperbolic cosine is continuous, so its series converges to it everywhere. The periodic hyperbolic sine has a jump discontinuity at each odd integer: the Dirichlet-Jordan convergence theorem gives the average of its two limiting values, namely zero, there. Both identities hold in the sense of mean-square convergence, which suffices for Parseval's identity.

The fundamental theorem of calculus says that accumulation and local change undo one another, under appropriate regularity conditions. If $f$ is a continuous function on $[a,b]$ andthen $A$ is differentiable for $a<x<b$ and $A^{\prime }(x)=f(x)$. Conversely, if $G$ is an antiderivative of $f$ on this interval, continuous at its endpoints, thenThe first statement turns an integral into a function with a known derivative; the second turns an entire accumulation into two endpoint values.

For $f(t)=1+2t$, accumulation from zero gives $A(x)=x+x^2$. At $x=2$, its derivative is $A^{\prime }(2)=5$, exactly the current value $f(2)$. The accumulated value there is $A(2)=6$, and by $x=3$ it has reached $A(3)=12$. The accumulated quantity and its rate are different functions, linked by differentiation.

This connection makes the theorem useful far beyond geometric area. It recovers displacement from velocity, gives a probability from a probability density function, and converts many differential equations into integral equations. The striking economy is that information spread across a whole interval can sometimes be recovered from an antiderivative at its two boundaries.

A short explanation of the first part is thatis the local average of $f$. Continuity makes this average approach $f(x)$. For the second part, $G-A$ has zero derivative and is constant by the mean value theorem. These are the ideas behind the proof; the endpoint and regularity conditions make them valid.

An early geometric form of this connection appears in Isaac Barrow's Lectiones geometricae (1670), Lecture X, Proposition 11. The original volume is catalogued by CESIMA, and Michael Nauenberg's historical research reproduces and examines the geometric argument.^[ref][ref]

The hyperbolic sine is the odd functionIts derivative is $\cosh x\ge 1$, so it increases strictly from $-\infty$ to $\infty$ and is a bijection of $\mathbb{R}$ onto itself. For example, $\sinh 0=0$, $\sinh 1\approx 1.175201$, and $\sinh (-1)\approx -1.175201$. It is the vertical coordinate on the unit hyperbola parametrized by $(\cosh x,\sinh x)$.

The Taylor seriesshows that $\sinh x\approx x$ near zero. Its tangent line there is $y=x$; unlike the ordinary sine, the hyperbolic sine grows without bound. Its antiderivative is $\cosh x+C$. In the catenary $y=a\cosh (x/a)$, the hyperbolic sine gives both the slope, through the derivative $y^{\prime }=\sinh (x/a)$, and the arc length $s(x)=a\sinh (x/a)$ from the lowest point for $x\ge 0$.

An integral adds up contributions of a function across a domain of integration. For a continuous function $f$ on $[a,b]$, the Riemann integral is the common limit of its Riemann sums as the largest subinterval width tends to zero:Each product combines a value with the width over which it acts. When $f$ is nonnegative, the result is the area beneath its graph; values below the horizontal axis contribute with the opposite sign. The signed area under a graph can therefore vanish even when the graph encloses nonzero geometric area.

For a velocity $v(t)=1+2t$ on $0\le t\le 3$, use $n$ equal time steps. The left and right Riemann sums areBecause the function is increasing, they bound the integral. With six steps they give $10.5$ and $13.5$; with sixty steps they give $11.85$ and $12.15$. Both approach the same total, $12$. This is accumulation made exact by refining finite measurements.

An integral retains units: velocity times time gives displacement, and a probability density function integrated over an interval gives a probability. Weighting a density by a function gives its expectation. The same operation serves different subjects because it expresses a common pattern, value multiplied by extent, then accumulated.

The fundamental theorem of calculus often evaluates a definite integral by finding an antiderivative, but a finite sum or a numerical approximation is still useful when no convenient antiderivative is available. An indefinite integral denotes a family of antiderivatives; a definite integral denotes an accumulated value between specified endpoints. More general constructions, including Lebesgue integration, extend the class of functions and domains that can be handled.

Integration by parts transfers a derivative from one factor of a product to the other, while keeping an endpoint contribution. Its usefulness comes from choosing which factor becomes simpler when differentiated, and which has an accessible antiderivative. The integral is then replaced by a boundary term minus another integral.

For real continuously differentiable functions $u,v$ on $[a,b]$, the precise definite integral formula isIn the shorter notation, $du=u^{\prime }(x)dx$ and $dv=v^{\prime }(x)dx$, giving $\int udv=uv-\int vdu$ for indefinite integrals, with a constant of integration understood. The product rule says $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$; the fundamental theorem of calculus turns its integral into the endpoint difference. Thus the formula packages a differentiation rule into a tool for integration.^[ref]

For example, choose $u=x$ and $dv=e^{x}dx$. Then $du=dx$ and $v=e^x$, since the exponential function is its own derivative:The boundary term is $e\approx 2.718282$; the remaining integral is $e-1\approx 1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its derivative includes $u^{\prime }v=e^x$ as well as the desired $u{v}^{\prime }=x{e}^x$.

In the figure, the area under $(x{e}^x{)}^{\prime }=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has area $e-1$, and its upper band has thickness $x{e}^x$ and area one. This decomposition makes the subtraction visible. The endpoint contribution is a difference of product values; it need not equal the area under either factor separately.

The same choice repeatedly lowers the degree of a polynomial when integrating a polynomial times an exponential function, while choosing $u=\ln x$ gives $\int \ln xdx=x\ln x-x+C$ for $x>0$. More broadly, moving a derivative onto a convenient test function is the mechanism behind a weak derivative. Always evaluate the boundary term: it disappears only when the relevant endpoint values justify doing so, often through a boundary condition.

If $F^{\prime }=f$ and $g$ is differentiable, the chain rule gives $(F\circ g{)}^{\prime }(x)=f(g(x))g^{\prime }(x)$. Taking antiderivatives gives the substitution ruleWriting $u=g(x)$ and $du=g^{\prime }(x)dx$ records this chain rule in reverse. For a continuous function $f$ and continuously differentiable $g$, the fundamental theorem of calculus gives the corresponding definite integral identityThe order of the transformed endpoints determines the sign. For example, ${\int }_0^{1}2x\cos (x^2)dx={\int }_0^1\cos udu=\sin 1$. If the substitution is used in the opposite direction by solving $x$ as a function of $u$, one must restrict to a suitable inverse branch; the tangent half-angle substitution illustrates why separate branches can be necessary even when the original integrand is smooth. The general change of variables formula extends substitution to higher dimensions.

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

The following scalar toy isolates one step in renormalization: removing a local divergence while retaining a finite dependence on a physical mass. For $M,\mu >0$ and a sharp cutoff regularization in Euclidean space with momentum bound $\Lambda$, define the dimensionless Feynman integralThis is a zero-external-momentum scalar bubble integral stripped of coupling and combinatorial factors. Using the three-sphere area $2{\pi }^2$ and $t=k_E^2$ reduces the integral toThe antiderivative follows by writing $t=(t+M^2)-M^2$. At large $\Lambda$ the divergent part is $\log {\Lambda }^2/(16{\pi }^2)$, with the logarithm's dimensionful argument understood only within the displayed ratios. It has no external-momentum dependence. In a four-scalar-field vertex this structure can therefore be removed by a local operator proportional to ${\varphi }^4$, rather than a momentum-dependent nonlocal interaction.

Choose the fixed reference mass $\mu$ and subtract $I_{\Lambda }(\mu )$. If the loop contribution to a chosen four-point vertex coefficient is $c{\lambda }^2{I}_{\Lambda }(M)$, choose the additive counterterm in that same coefficient as $\delta {\lambda }_{\Lambda }=-c{\lambda }^2{I}_{\Lambda }(\mu )$, where $c$ includes its stated diagram factors. This defines a finite subtraction prescription:The renormalization condition for this toy loop contribution is $I_R(\mu ;\mu )=0$. The reference mass stays fixed when $M$ is varied; finite dependence on $M$ survives. A different finite subtraction changes the renormalization scheme, and the defined renormalized coupling must change consistently.

For $M=2\mu$, the limiting result is $16{\pi }^2{I}_R=-\log 4\approx -1.38629436$. At $\Lambda =10\mu$, the exact regulated difference is about $-1.32846343$; at $\Lambda =100\mu$, it is about $-1.38569459$. Expanding the exact expression gives the leading residual $2(M^2-{\mu }^2)/{\Lambda }^2$ in $16{\pi }^2[I_{\Lambda }(M)-I_{\Lambda }(\mu )]$. This explicitly separates finite-cutoff error from the chosen finite renormalization condition.

A sharp cutoff regularization is convenient for this scalar demonstration, but shifts of a divergent loop momentum can change its integration domain. A gauge theory calculation cannot assume that it preserves the Ward identities. The toy contains no interacting gauge bosons, ghost loops or overlapping ultraviolet subdivergences, so its successful subtraction does not prove Non-Abelian gauge theory renormalizability.

Take a wire segment on the $y$ axis, with source positions ${\mathbf{r}}^{\prime }=(0,s,0)$ for $s_1\le s\le s_2$, and positive electric current along $+y$. Observe at $\mathbf{r}=(a,0,0)$, where $a>0$ is the perpendicular distance to the wire's supporting line. Choose $x$ rightward and $y$ upward, so $+z$ points out of the page. ThenThus the magnetic field points into the page by the right-hand rule. The Biot-Savart law and an elementary antiderivative giveFor an original worked example, choose $s_1=-0.12\mathrm{m}$, $s_2=0.12\mathrm{m}$, $a=0.05\mathrm{m}$, and $I=3\mathrm{A}$. Use ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$ for these rounded calculations. Each endpoint contributes $0.12/0.13=12/13$, soThese are calculated inputs and output, not laboratory data. The two endpoint angles measured from the perpendicular each have sine $12/13$; this recovers the equivalent endpoint-angle formula without depending on a diagram's angle-sign convention.

For a segment centered on the perpendicular foot, $s_1=-L$, $s_2=L$. Its magnetic field magnitude isThe limit $L/a\to \infty$ gives $B_{\infty }={\mu }_{0}I/(2\pi a)$, consistent with Ampère's circuital law for an infinite straight wire. In the example, $B_{\infty }=12\mu \mathrm{T}$ and $B/B_{\infty }=12/13$: the infinite-wire value overestimates the segment value by $1/12$, or about $8.33\%$ of the segment value. For $a\gg L$, the isolated segment contribution instead behaves as $B\sim {\mu }_{0}I(2L)/(4\pi a^2)$. The complete localized electrical circuit has a different far-field behavior: return-path contributions cancel this leading segment term, leaving a magnetic dipole field when its magnetic dipole moment is nonzero. Segment and whole-circuit limits must be distinguished.

The SI units check directly: ${\mu }_{0}I/a$ has units $(\mathrm{T}m/A)\mathrm{A}/\mathrm{m}=\mathrm{T}$, the tesla. The apparent divergence as $a\to 0$ is a failure of the thin-wire model near the conductor, not a prediction of infinite magnetic field inside a real wire. This calculation gives only this segment's contribution; include the return wires before comparing with an apparatus.

For $0<b<c<\infty$, the indicator function $f(x)=1_{[b,c]}(x)$ has Mellin transformThe interval stays away from both zero and infinity, so the integral is absolutely convergent for every complex number $s$. Its Mellin fundamental strip is the whole complex plane. Evaluating the antiderivative $x^s/s$ proves the formula for $s\ne 0$; at $s=0$, integrating $1/x$ gives the real natural logarithm $\log (c/b)$. The apparent division by zero is a removable singularity, since $c^s-b^s=s(\log c-\log b)+O(s^2)$ by the Taylor series of the exponential function. The transform is an entire function.

For example, $1_{[1,4]}$ gives $(4^s-1)/s$ with value $\log 4$ at zero. On $s=i\tau$, the Mellin transform sums oscillations over the width $\log 4$ of the interval in natural logarithm coordinates; this is the same cancellation mechanism as in the Fourier transform of an interval indicator.