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For a locally Lipschitz function $T$ between Euclidean spaces of the same dimension, the area formula integrates its absolute Jacobian determinant against the multiplicity of its fibers. In particular, for an injective $C^1$ map $T:X\to {\mathbb{R}}^d$ and nonnegative measurable $a$,This injective version extends the change of variables formula to maps with a possibly singular derivative.

For a positive random variable $X$, write its base-ten logarithm as an integer part plus a fractional part:The floor function handles values below one as well: for $X=0.0042$, $K=-3$ and the decimal significand is $10^Z=4.2$. Since $10^Z\in [1,10)$, the first nonzero decimal digit is $D=\lfloor 10^Z\rfloor$. The monotonicity of the logarithm gives the exact event equivalenceIf $Z$ has the uniform distribution on $[0,1)$, the probability of this interval is its length. Consequently $\mathbb{P}(D=d)={\log }_{10}(d+1)-{\log }_{10}d$, proving the Benford law. This argument assumes uniformity of the fractional part of the logarithm, rather than uniformity of $X$ itself. See the logarithmic-coordinate explanation in Peter Sheridan Dodds, Benford's Law, essential story.

An exact constructed example is the log-uniform distribution on $[1,10)$, with probability density function $f_X(x)=1/(x\ln 10)$ there and zero elsewhere. Its total mass is ${\int }_1^{10}dx/(x\ln 10)=1$. The change of variables formula with $x=10^z$ gives $f_Z(z)=f_X(10^z)(\ln 10)10^z=1$ for $0\le z<1$. Equivalently, integrating directly givesFor example, the probability of first digit four is ${\log }_{10}(5/4)\approx 0.096910013$. By contrast, an ordinary uniform distribution for $X$ on $[1,10)$ has probability density function $1/9$ there, so $\mathbb{P}(D=d)={\int }_d^{d+1}dx/9=1/9$ for every digit. This is a concrete counterexample to any claim that the Benford law holds for every positive random variable.

More generally, a log-uniform distribution on $[A,B)$ with $0<A<B$ has uniform $Y={\log }_{10}X$ on $[{\log }_{10}A,{\log }_{10}B)$. If its logarithmic span ${\log }_{10}(B/A)=m$ is a positive integer, reduction to the fractional part wraps an interval of length $m$ around the unit interval exactly $m$ times, making $Z$ uniform even when $A$ is not a power of ten. Arbitrary partial-decade truncations generally do not have this property: for a log-uniform distribution on $[1,2)$, the first digit is always one.

Under the uniform fractional-logarithm assumption, multiplication by any constant $c>0$ replaces $Z$ by $Z^{\prime }=\{Z+{\log }_{10}c\}$. This is a rotation modulo one, which preserves the uniform distribution and therefore preserves the Benford law; see scale invariance of decimal significands. For the constructed $[1,10)$ example, multiplying by $0.001$ gives a log-uniform distribution on $[0.001,0.01)$ with the same digit probabilities. The claim concerns the decimal significand law: there is no proper probability density function proportional to $1/x$ over all of $(0,\infty )$, since its integral diverges. The existing scale-invariant prior describes that unnormalizable measure, while a finite log-uniform distribution is normalizable.

For $a,b>0$, this probability density function is obtained from a beta distribution variable $U$ by $X=U/(1-U)$. Indeed $u=x/(1+x)$ and $du/dx=(1+x{)}^{-2}$, and the change of variables formula gives the displayed density. In particular, the ratio of two independent equal-rate exponential distributions has $a=b=1$, density $(1+x{)}^{-2}$ and cumulative distribution function $x/(1+x)$.

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]

If $Y=g(X)$ for a differentiable bijection, thenThis is the change of variables formula applied to probability density.

For a random variable $X$ with probability density function $f_X$ on an interval, let $Y=h(X)$, where $h$ is a diffeomorphism onto another interval. The change of variables formula givesIndeed, integrating the right-hand side over an outcome interval gives the probability that $X$ lies in its inverse image. The absolute derivative preserves positive probability when $h$ reverses orientation. A gravity deflection of an atomic beam is a physical example: its speed-to-height transformation changes the probability density and its mode.

The cofactor matrix consists of the signed minors of a matrix. It is the transpose of the adjugate matrix. The inverse expression applies to invertible matrices; the definition by minors also applies to singular matrices. Cofactors transform oriented hypersurface area vectors in a change of variables formula.

For a conformal bijection $f:D\to \tilde{D}$ and smooth $u,v$ of compact support on $\tilde{D}$, $\boxed{(u\circ f,v\circ f{)}_{\nabla ,D}=(u,v{)}_{\nabla ,\tilde{D}}}$. The Jacobian matrix of $f$ is a rotation times $|f^{\prime }|$. Consequently the chain rule introduces $|f^{\prime }{|}^2$ into the gradient pairing, while the change of variables formula introduces precisely the same factor into area. They cancel. The identity extends by completion to the Dirichlet energy spaces; it does not preserve the $L^2$ term of an inhomogeneous Sobolev norm.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

For real $a\ne 0$ and $b$, the change of variables formula gives $f_{aX+b}(ax+b)=f_X(x)/|a|$. ThereforeThe absolute value also covers a decreasing transformation. If a numerical length changes from metres to centimetres, $a=100$ and the differential entropy changes by $\ln 100\approx 4.60517$ nats, although the physical uncertainty is the same. The reference coordinate scale must therefore accompany an absolute differential entropy value.

Let $X$ have a probability density function on an open interval, and let $g$ be continuously differentiable, strictly increasing there, with $g^{\prime }(x)>0$. The change of variables formula givesIf the differential entropy and the expected absolute logarithm of $g^{\prime }(X)$ are finite, taking the expectation of the negative logarithm proves the displayed formula, for a fixed base $b>1$. Local stretching reduces probability density; its contribution to differential entropy is the expected logarithmic stretch. The map is invertible, so this change does not mean that the transformation destroyed information about $X$.

A double integral integrates a function over a two-dimensional region. A change of variables formula can replace the region and area element using a Jacobian determinant.

It is a multiple integral with two integration variables.

In vacuum, $\nu =c/\lambda$, where $c$ is the speed of light. Increasing wavelength decreases frequency, so the ordinary signed differentials satisfy $d\nu =-c{\lambda }^{-2}d\lambda$. A spectral slice's positive contribution to radiance must be unchanged. The change of variables formula therefore gives the absolute Jacobian determinantThese formulas use $L_{\lambda }$ per metre and $L_{\nu }$ per hertz. The factor $c/{\nu }^2$ has units $\mathrm{m}\mathrm{s}=\mathrm{m}\mathrm{H}{\mathrm{z}}^{-1}$, which converts the former density into the latter. If ${\ell }_{\mathrm{nm}}$ denotes the numerical density per nanometre in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$, its per-metre value is $10^9{\ell }_{\mathrm{nm}}$. Conversely, the per-nanometre value is $10^{-9}$ times the per-metre value. Likewise, a density per terahertz is $10^{12}$ times its per-hertz value because $1\mathrm{THz}=10^{12}\mathrm{Hz}$.

For a concrete stipulated spectrum, let $L_{\lambda }=C=10^9\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$ between ${\lambda }_1=500\mathrm{nm}$ and ${\lambda }_2=600\mathrm{nm}$, and zero outside. This is exactly $1\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$ throughout the band. Using the exact International System of Units value $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the corresponding frequency bounds areThe transformed spectral radiance is $L_{\nu }=Cc/{\nu }^2$ on this interval. At ${\nu }_1$ it is approximately $1.200830743\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, and at ${\nu }_2$ it is $8.339102380\times 10^{-13}$ in the same units. These are respectively $1.200830743$ and $0.833910238$ per terahertz. Thus a flat wavelength density becomes a decreasing frequency density; relabelling the horizontal axis alone would miss both the shape and the units. ^[ref].

The equality of band radiances has a short exact proof:The reversed bounds and the Jacobian determinant preserve the same spectral measure. Each half-band, $500$–$550\mathrm{nm}$ and $550$–$600\mathrm{nm}$, carries $50\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, although the corresponding frequency widths are about $54.508$ and $45.423\mathrm{THz}$.

A spectrum's peak is consequently a statement about a particular density and coordinate. The frequency and wavelength peaks of Planck's law differ for precisely this reason. Integrated radiance over a specified physical band supplies a comparison that survives the change of variables.

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.

The Jacobian determinant is the determinant of the square Jacobian matrix. Its absolute value is the local volume-scaling factor in the change of variables formula.

The Jacobian matrix and determinant describe respectively the linear derivative and its signed volume factor.

Let $\varphi :I\to J$ be a $C^1$ diffeomorphism between open intervals, which may be unbounded. The change of variables formula holds for every nonnegative Lebesgue measurable function $f:J\to [0,\infty ]$, allowing infinite integrals. Applying it to $|f|$ also proves equivalence of Lebesgue integrability of $f$ and of the transformed integrand; for an integrable signed or complex $f$ the same identity then holds by splitting into positive and negative, or real and imaginary, parts. The absolute value of the derivative is essential for a decreasing substitution such as $\varphi (x)=1/x$. Endpoints can be omitted because they are null sets.

A multiple integral accumulates a function over a region with several coordinates. The Fubini theorem and Tonelli theorem give conditions under which it equals successive one-dimensional integrals. A change of variables formula expresses it in new coordinates using the absolute Jacobian determinant.

If a region is invariant under one coordinate reflection and an integrable function changes sign under that reflection, its multiple integral is zero. Apply the change of variables formula to the reflection, whose absolute Jacobian determinant is one: the integral equals its negative. This applies to odd coordinate powers in symmetric volumes and to the off-diagonal entries of an inertia tensor when the mass density shares the coordinate-reflection symmetries.

Assign any value, for example zero, at $(0,0)$, where the displayed quotient is undefined. A single point is a Lebesgue null set, so this choice does not affect Lebesgue integrability. Elsewhere the function is continuous, hence it is Lebesgue measurable on the plane after this extension.

In polar coordinates, $x^2+y^2=r^2$ and the area element is $rdrd\theta$. Applying the nonnegative change of variables formula to the absolute value givesFor $0<r\le 1$, $|\sin (r^2)|\le r^2$, so the radial integrand is bounded by $r^{3-2\alpha }$. This is integrable at zero when $\alpha <2$. For $r\ge 1$, $|\sin (r^2)|\le 1$, giving the integrable tail bound $r^{1-2\alpha }$ when $\alpha >1$. In particular,ThereforeEquivalently, the substitution rule $u=r^2$ reduces its exact L1 norm toThis is radial sine-power kernel integrability, reducing a two-dimensional question to the sine power kernel integrability thresholds at zero and infinity.

Fix $y\ne 0$. The function $x\mapsto f(x,y)$ is continuous on $[-1,1]$, because its denominator is bounded away from zero, and it is an odd function of $x$. Its integral over this symmetric interval is therefore zero. At $y=0$ the entire section is zero, including the specified value at the origin. Thus the inner $x$ integral exists for every $y$ and is identically zero. The same reasoning with $x$ and $y$ exchanged givesTo test Lebesgue integrability in two dimensions, examine the absolute value near the origin. For $0<\epsilon <1$, the annulus $\epsilon <\sqrt{x^2+y^2}<1$ lies inside the square. In polar coordinates,The change of variables formula on this annulus, where the integrand has no singularity, givesThe lower bound tends to infinity as $\epsilon \downarrow 0$. HenceMoreover reflection in either coordinate axis exchanges the positive part and negative part while preserving Lebesgue measure. Both parts therefore have infinite integral, so a signed double Lebesgue integral would involve the undefined expression $\infty -\infty$.

There is no contradiction with Fubini's theorem: its absolute integrability hypothesis is missing. This is an example of finite iterated integrals without joint integrability, with every individual one-dimensional section integrable despite the two-dimensional singularity.