Let $a>0$ and $t>0$. Near zero, the complex integral ${\int }_0{x}^{-1}{e}^{-tx}/(ia+\log x)dx$ fails both absolute and ordinary convergence through its real part, which behaves as $dx/(x\log x)$. Taking the imaginary part first givesThe regularized integrand $e^{-tx}/[x(a^2+{\log }^{2}x)]$ is integrable at zero: setting $u=-\log x$ changes its tail to a function comparable to $du/u^2$. Exponential decay controls infinity, and $a>0$ prevents an interior singularity at $x=1$.

A family with positive Mellin transform exponent can consequently have a finite imaginary-part boundary limit even when its full complex limit does not exist. Taking the limit of the full divergent expression and only then selecting its imaginary part is not justified. Endpoint-expansion coefficients can also change nonuniformly: a sine zero may cancel a gamma function pole and leave a nonzero constant term. This explains the regularization needed in the logarithmic expansion of Ramanujan's function.

The change of variables $x=e^u$, using the exponential function, turns the Mellin transform intoThe factor $dx=e^{u}du$ changes $x^{s-1}$ to $e^{su}$; missing this factor would shift every exponent by one. With $h_{\sigma }(u)=e^{\sigma u}f(e^u)$ and the Fourier transform convention $\hat{h}(\omega )={\int }_{\mathbb{R}}h(u)e^{-i\omega u}du$, the identity isChoosing $\sigma$ inside the Mellin fundamental strip makes $h_{\sigma }$ a Lebesgue integrable function. The minus sign in the Fourier transform argument follows from the stated convention. For the Mellin transform of an interval indicator with $[b,c]=[1,4]$ and $\sigma =0$, $h_0=1_{[0,\log 4]}$: the Fourier transform of an interval indicator is literally the same computation in a different coordinate.

Splitting the real line at $u=0$ also connects the Mellin transform to two unilateral Laplace transforms. If $k_{+}(v)=f(e^v)$ and $k_{-}(v)=f(e^{-v})$ for $v\ge 0$, thenwhere both integrals converge absolutely. Their Laplace convergence regions impose opposite endpoint conditions, explaining the two sides of the Mellin fundamental strip. Finally, $f(a{e}^u)=f(e^{u+\log a})$ shows that a change of scale is a translation in $u$. Weighted Fourier transforms of this translation give exactly the factor $a^{-s}$ in the scaling property of the Mellin transform. For a smooth function with compact support inside $(0,\infty )$, the Mellin inversion theorem follows from Fourier inversion; the existing formula states the corresponding hypotheses.

For $a>0$, let $g(x)=f(ax)$. Whenever the Mellin transform of $f$ is absolutely convergent,The Mellin fundamental strip is unchanged: the corresponding integrals of absolute values differ by the finite positive factor $a^{-\sigma }$. The change of variables $y=ax$ gives $x^{s-1}dx=a^{-s}{y}^{s-1}dy$, proving the rule. For $s=\sigma +i\tau$, the transform's absolute value changes by $a^{-\sigma }$, while the factor $e^{-i\tau \log a}$ changes its phase. On the line $\sigma =0$, wherever absolute convergence holds, the transform's absolute value is preserved.

For the Mellin transform of an interval indicator, replacing $1_{[1,4]}(x)$ by $1_{[1,4]}(2x)=1_{[1/2,2]}(x)$ givesAt the removable singularity $s=0$, both sides have value $\log 4$. Scaling leaves the width measured by the natural logarithm unchanged. Likewise, the exponential function $e^{-ax}$ has Mellin transform $a^{-s}\Gamma (s)$ on $\mathrm{Re}s>0$.

The fundamental strip of a Mellin transform is the interior of the set of real parts $\sigma$ for whichViewed in the complex plane, it is a vertical strip $\alpha <\mathrm{Re}s<\beta$; either bound may be infinite, and the interior may be empty. Both endpoints of the improper integral must be checked. Convergence on a boundary line, whether absolute convergence or conditional convergence, requires a separate test. The Mellin transform is holomorphic throughout a nonempty fundamental strip.^[ref]

A practical sufficient test uses Big O notation: if $f(x)=O(x^p)$ near zero and $f(x)=O(x^{-q})$ near infinity, with real $p,q$, then comparison with elementary improper integrals givesThus $-p<\sigma <q$ guarantees absolute convergence. These upper bounds on $|f|$ need not locate the maximal Mellin fundamental strip; faster actual decay can widen it. The strip property itself follows by bounding the weight at any intermediate $\sigma$ by the sum of two weights already known to be integrable. On a closed narrower strip, neighboring integrable weights also dominate the natural logarithm factors introduced by differentiation under the integral sign, making the transform a holomorphic function.

An exact two-endpoint example isHere both bounds are sharp: at $\sigma =-2$ the zero-end absolute convergence test contains ${\int }_0^{1}dx/x$, and at $\sigma =3$ the infinity-end test contains ${\int }_1^{\infty }dx/x$. The displayed rational function continues meromorphically outside the Mellin fundamental strip and has simple poles at $-2$ and $3$. Its other finite continued values outside the strip are values of a meromorphic continuation, not of a convergent defining integral. More general endpoint information leads to meromorphic continuation of a Mellin transform from an asymptotic expansion.

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.

A posted price is a fixed price announced by a seller, who sells if an eligible buyer accepts. Suppose two buyers have independent values with common cumulative distribution function $F$, and a buyer accepts exactly when its value is at least the price $p\ge 0$. The no-sale probability is $F(p-{)}^2$, where $F(p-)=\mathrm{Pr}(V<p)$, so expected seller revenue isFor a continuous value distribution this becomes $p(1-F(p{)}^2)$. With an atom at $p$, using $F(p)$ instead would exclude buyers who are indifferent and agree to buy. A convention that indifferent buyers decline has that strict-acceptance formula instead. Selecting among several accepting buyers does not affect revenue when each pays the same posted price.

The expectation of the seller's total receipts over the buyers' valuation types and any randomization in the sales mechanism. In an auction it is the sum of the bidders' expected payments; with a posted price it is price times sale probability. It does not subtract selling costs or the seller's opportunity cost of retaining the item, so it is revenue rather than profit.

For constant masses, a time-independent potential energy $V$ and forces ${\mathbf{F}}_i=-{\nabla }_{i}V+{\mathbf{R}}_i$ with zero total extra power ${\sum }_i{\mathbf{R}}_i\cdot {\mathbf{v}}_i=0$, Newton's second law gives $dT/dt=-dV/dt$. Thus mechanical energy is conserved. Smooth stationary constraint reactions and a purely magnetic Lorentz force supply zero power. This is more restrictive than conservation of energy in general, which may include thermal or other internal forms.

An observed dissipative channel combined with direct feedback can create a modified Lindblad operator and a compensating Quantum Hamiltonian. Choosing those parameters can make a prescribed pure state stationary and globally attractive. At exceptional parameter values, the dissipator can instead leave an entire family of states stationary, so algebraic target stationarity does not establish stabilization. Checking the Affine Bloch equation and its transverse eigenvalues distinguishes these possibilities.

A path graph state on three qubits simulates two successive gates $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$. Measure vertex one in the equatorial qubit measurement basis $|v_r(\alpha )⟩=(|0⟩+(-1{)}^r{e}^{i\alpha }|1⟩)/\sqrt{2}$, then vertex two at angle $(-1{)}^r\beta$, obtaining bits $r,s$. Two one-bit teleportations and $U(\theta )X=e^{-i\theta }ZU(-\theta )$ leave vertex three in $X^s{Z}^{r}U(\beta )U(\alpha )|+⟩$, up to global phase. A computational output bit $z$ is corrected to $z\oplus s$; the $Z$ factor affects only phase. Adaptive sign choice is required for general angles, although the final bit correction is purely classical.

For a measure space $(X,\mathcal{F},\mu )$ and $1\le p<\infty$, $L^p(\mu )$ consists of equivalence classes of measurable functions satisfyingTwo representatives are identified when they agree almost everywhere. For $p=\infty$, use the essential supremum norm, not an infinite power inside the integral. The quotient is essential: a function supported on a null set has zero integral norm even if it is not identically zero.^[ref]

These are Banach spaces; L2 space also has the integral inner product making it a Hilbert space. On $(0,1)$ with Lebesgue measure, $f(x)=x^{-\alpha }$, $\alpha >0$, belongs to $L^p$ exactly when $\alpha p<1$. This shows why membership depends on the exponent: a function can be integrable while its square is not. For $0<p<1$, the same power expression defines a quasi-norm rather than a norm, as explained under Lq quasi-norm.

For nonnegative measurable functions,

A Boolean algebra of subsets contains the empty set and is closed under finite unions and complements. It is consequently closed under finite intersections, differences, and symmetric differences.

For $E=C\mu /(\beta +C\mu )<1$ and mean lifetime $\tau =(\beta +C\mu {)}^{-1}$, the normalized restoring matrix is $M={\tau }^{-1}\left(\begin{array}{cc}1& E\\ E& 1\end{array}\right)$ and the diffusion matrix is $D=(\tau \mu {)}^{-1}\left(\begin{array}{cc}2& E\\ E& 2\end{array}\right)$. Solving the normalized stationary fluctuation-dissipation relation for a reaction network gives the displayed covariance. The sum mode remains damped, but the difference mode's restoring rate is $\beta =(1-E)/\tau$. Its noise grows without bound relative to that restoring rate as $E\to 1$, giving strong negative correlations and invalidating small-fluctuation closure near exclusive joint removal.

// Target: mathematical-biology.bigb

If individual removal is absent, simultaneous removal of one molecule from each species leaves $X_1-X_2$ unchanged. Independent constant-rate births change it by plus or minus one, each at rate $\lambda$. The difference is therefore a symmetric continuous-time random walk on the integers with variance growing as displayed. It has no normalizable stationary law. This explains why a deterministic attracting line of equilibria does not provide a stationary stochastic distribution.

// Target: probability-and-statistics.bigb

For annually repeated forcing, a thermodynamic equilibrium is a periodic temperature field and thickness cycle. The yearly integral of net ice growth is zero, although substantial winter growth and summer ablation can each occur. Equilibrium thickness must specify an annual mean or a phase of the cycle. A change in forcing can remove a perennial-ice solution without making every transient thickness decrease at the same rate.

At fixed boundary temperatures, a snow cover decreases winter conductive heat loss through the resistance $h_s/k_s+h_i/k_i$ and inhibits sea ice growth. Its high albedo and persistence into summer reduce ablation. Therefore annual equilibrium thickness has competing snow sensitivities. The near cancellation at modest snow depth and thickening with persistent summer snow in the original Maykut-Untersteiner model are forcing-dependent responses, not a universal monotonically decreasing snow-thickness relation.

Attach two disks to a simply connected space $A$ along disjoint null-homotopic boundary circles, and additionally identify one interior point of each disk. Assume the attachments are CW complexes or admit compatible triangulations. The extra interior identification contributes a copy of $\mathbb{Z}$ to the first homology group, in addition to the two new second-homology classes. Indeed, after the first attachment $Y=A\cup D_1$, the intersection with $D_2$ has two components, $S^1\bigsqcup \{p\}$. In the Mayer–Vietoris sequence, the map $H_0(S^1\bigsqcup \{p\})\to H_0(Y)\oplus H_0(D_2)$ sends $(m,n)$ to $(m+n,-m-n)$, whose kernel is $\mathbb{Z}$. If $H_1(A)=0$, exactness gives $H_1(Y\cup D_2)=\mathbb{Z}$. The degree-two sequence is $0\to H_2(Y)\to H_2(Y\cup D_2)\to \mathbb{Z}\to 0$, which splits because $\mathbb{Z}$ is free. For $A=S^3$, the integral homology groups are $\mathbb{Z},\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero through three.

Identifying two distinct points of $S^2$ produces a space with homology groups $H_0=H_1=H_2=\mathbb{Z}$ and all higher groups zero. Cover the identified point by the image of two small disks, a contractible wedge of disks, and take as the other open set the complement of the point, a cylinder. Their intersection is two annuli. In the Mayer–Vietoris sequence, $H_1$ of this intersection maps onto $H_1$ of the cylinder with a rank-one kernel, producing $H_2=\mathbb{Z}$. The kernel at $H_0$ has rank one and produces $H_1=\mathbb{Z}$.

For a holomorphic scalar function on a neighborhood of a closed annulus, $M(r)$ is the maximum modulus on its radius-$r$ circle. The logarithm of its modulus is subharmonic; compare it with the harmonic function affine in $\log |z|$ which takes the logarithms of the two boundary maxima. The maximum principle for subharmonic functions proves the displayed bound. Zero boundary maxima are handled by adding a positive bound and taking its limit. The same bound holds for the norm of a Banach space-valued holomorphic function by applying scalar functionals and the Hahn-Banach theorem.