The Wallis integrals are $I_n={\int }_0^{\pi /2}{\sin }^{n}xdx$ for integers $n\ge 0$. Integration by parts gives $n{I}_n=(n-1)I_{n-2}$ for $n\ge 2$, with $I_0=\pi /2$, $I_1=1$. HenceBecause $0<\sin x<1$ in the integration interval, $0<I_n<I_{n-1}$ for $n\ge 1$. Combining this with the recurrence gives $2n/(2n+1)<I_{2n+1}/I_{2n}<1$, proving that the ratio tends to one and yielding the Wallis product.

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.

For $a<b$ and a real-valued function $f$ with a finite Riemann integral or Lebesgue integral on $[a,b]$, the signed area under a graph is its definite integral: geometric area above the horizontal axis contributes positively, and geometric area below it contributes negatively. If these two nonnegative areas are $A_{+}$ and $A_{-}$, thenThe second expression is the total geometric area under a graph. Taking the absolute value of the first expression does not usually give the second: cancellation has already removed information about how much area lies on each side.^[ref]

For an exact example, take the polynomial function $f(x)=x^2-1$ on $[-2,2]$. Its polynomial roots are $-1$ and $1$: it is negative between them and positive on the two outer intervals. An antiderivative is $F(x)=x^3/3-x$, so the fundamental theorem of calculus givesThus $A_{+}=8/3$ and $A_{-}=4/3$. The signed area under a graph is $4/3$, whereas the total geometric area under a graph is $4$. To compute the latter with antiderivatives, split at each sign-changing polynomial root and reverse the sign of the negative contribution.

Cancellation is useful information when the integrand represents a signed contribution. For the polynomial function $f(x)=x$ on $[-1,1]$, the two triangular areas give ${\int }_{-1}^{1}xdx=0$, although ${\int }_{-1}^1|x|dx=1$. Zero net signed area therefore means balanced positive and negative contributions, rather than an empty region.

The short algebraic justification uses the positive part $f_{+}=\max (f,0)$ and negative part $f_{-}=\max (-f,0)$: $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$. Their integrals give $A_{+}$ and $A_{-}$, and linearity of integration gives the difference and sum formulas. For Lebesgue integration, require both contributions to be finite to avoid the undefined subtraction $\infty -\infty$.^[ref]

For integrable real-valued functions $f,g$ on a measure space $(X,\mu )$ and fixed real numbers $a,b$, their linear combination $af+bg$ is integrable, and the Lebesgue integral satisfiesThe same rule applies to any finite summation: ${\int }_X{\sum }_{j=1}^n{a}_j{f}_{j}d\mu ={\sum }_{j=1}^n{a}_j{\int }_X{f}_{j}d\mu$ for integrable functions $f_j$ and fixed scalars $a_j$. Finite signed values matter: integrability ensures all these integrals are finite, so no undefined $\infty -\infty$ occurs.

This is the distributive property of finite summation carried into integration: a fixed factor can be pulled out, and a finite sum can be integrated term by term. For simple functions, distribute the finite sums of value times measure; approximation extends additivity to nonnegative measurable functions, and splitting into a positive part and negative part gives the integrable signed case. See Lecture 12: Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem (Proposition 122: linearity for integrable functions). For example, ${\int }_0^1(2x-3)dx=2{\int }_0^{1}xdx-3{\int }_0^{1}1dx=-2$. With a probability measure, linearity of integration gives linearity of expectation.

For an integrable function on an interval $I$ of positive finite length, its average value is the integral divided by that length:If $f$ is constant on equal-length subintervals, this is their values' arithmetic mean; unequal lengths require a weighted mean. For example, values $2$ and $6$ on two equal halves give average value $4$, whereas value $2$ on the first quarter and $6$ on the remaining three quarters give $5$. The integral measures total contribution; the average value of a function divides out the interval length.

For the function $f(x)=2x$ on $[0,1]$, the average value of a function is ${\int }_0^{1}2xdx=1$. Replacing it by the constant $1$ preserves its integral on this interval. This is why the Haar projection uses cell averages to obtain a coarser description while retaining each cell's total contribution.

For a concrete comparison, take $f(x)=x^2$ on $[0,6]$. The integral is $I={\int }_0^6{x}^{2}dx=72$, whereas the unit-width left and right Riemann sums are $L=55$ and $R=91$. The exact integer power sum formula givesThe endpoint choice explains the opposite signs of the $18$ correction. The remaining correction is one in both cases.

The same method is useful far beyond squares of numbers. Substituting $B_0$, $B_1$, $B_2$ and $B_4$ into the integer power sum formula gives the exact fourth-power resultAt $N=10$, this isThus a finite polynomial calculation replaces ten additions, or a million additions if $N$ is a million. Although the individual corrections can be rational numbers, the result is the exact integer sum.

There is also a useful numerical integration interpretation. The average of the left and right Riemann sums is the composite trapezoidal rule. For $x^2$, the endpoint corrections cancel but its value is $73$, one above the true integral. Subtracting the $B_2$ correction makes this particular polynomial example exact. For $f(x)=x^4$, the two interior corrections above can be written $B_2(f^{\prime }(N)-f^{\prime }(0))/2!$ and $B_4(f^{\prime \prime \prime }(N)-f^{\prime \prime \prime }(0))/4!$. These are exact identities for these polynomials; extending such corrections to general functions requires a separate error analysis.

An insulin receptor substrate is a docking protein phosphorylated by the activated insulin receptor. Its phosphotyrosine residues recruit downstream signaling proteins, including SH2 domain-containing components of phosphoinositide 3-kinase. This connects reception to the pathway described under insulin, rather than making the insulin receptor substrate another extracellular ligand.^[ref]

A Euclidean bounce is a finite-action Euclidean saddle associated with decay of a false vacuum. It returns to the same metastable vacuum at the two ends of Euclidean time, instead of joining two degenerate stable vacua. Its characteristic negative fluctuation direction produces the imaginary part associated with a decay rate; the leading semiclassical suppression is $e^{-S_B/\hslash }$. Translation zero modes are treated as collective coordinates. It should not be confused with the stable double-well tunnelling instanton whose repeated crossings give a real level splitting.

The injective hull of a uniform module is uniform. An injective endomorphism of a uniform injective module splits its image, so its complementary summand must vanish and it is an automorphism. Consequently all nonunits are precisely the noninjective endomorphisms. Their nonzero kernels are essential; intersection of two such kernels is nonzero, making their sum noninjective. Composition on either side preserves noninjectivity, unless the other map is a unit, which merely transports the kernel. Thus the nonunits form the unique maximal ideal, and the quotient is a division ring.

For a domain satisfying the left Ore condition, its division ring of fractions $Q$ is an injective essential extension of the left regular module. For a homomorphism $f:I\to Q$ on a nonzero left ideal, choose $a\in I\setminus \{0\}$. A common left multiple $sx=ta$ gives $sf(x)=tf(a)$, hence $f(x)=x{a}^{-1}f(a)$; right multiplication by $a^{-1}f(a)$ extends it. The Baer criterion proves injectivity. Clearing a left denominator proves essentiality. The same calculation gives ${\mathrm{End}}_R({}_{R}Q)\cong Q^{\mathrm{op}}$, with the opposite ring required because right multiplications compose in reversed order.

An L2 initial condition specifies a square-integrable initial function up to equality almost everywhere. It need not have pointwise nodal values or classical derivatives. L2-compatible initialization of grid data uses bounded averaging or projection, and evolution estimates use the relevant L2 norm.

For an initial mass function proportional to $M^{-p}$ on $M\ge M_{\min }$ with $p>1$, integration and normalization give the displayed upper-tail fraction. The probability integral transform makes $X$ uniform on $[0,1]$ for a randomly sampled birth mass. Evolutionary lifetime regions can then be measured as ordinary areas when the independent age coordinate is also uniform.

If a cohort is periodically confirmed event-free but exact event reports update only those who fail, censoring the remaining subjects at their last scheduled confirmation creates informative censoring. Between confirmations the nominal risk set retains future observed failures and removes nonfailures. An apparent final failure may therefore empty the recorded risk set even though other cohort members remain event-free. Use a common verified cutoff or a complete event register with a common administrative censoring date.

Add one observation $z$ to the empirical sample, so $F_{n+1,z}=\frac{n}{n+1}{F}_n+\frac{1}{n+1}{\delta }_z$. The displayed finite difference is a finite-sample counterpart of the influence function. Its supremum magnitude measures worst-case one-observation sensitivity; unlike an infinitesimal derivative it records the actual response at contamination fraction $1/(n+1)$.

This plugs the empirical distribution into the population influence function. For an M-estimator it is $-[n^{-1}{\sum }_i{\partial }_{\theta }\psi (y_i,T(F_n)){]}^{-1}\psi (z,T(F_n))$, when this denominator is well-defined and nonzero. For functionals involving a density at a quantile, an unsmoothed empirical measure may not permit this derivative, so smoothing or an actual sensitivity curve is required.

Let $q$ be the unique $p$th quantile, with $0<p<1$, and suppose the density $f$ is continuous and positive at $q$. Define quantiles of contaminated distributions using the generalized inverse. For $z\ne q$, differentiating $(1-\epsilon )F(q_{\epsilon })+\epsilon 1_{z\le q_{\epsilon }}=p$ givesAt $z=q$, the generalized-inverse quantile stays exactly $q$ under this contamination, so its derivative is zero. The step formula thus holds $F$-almost everywhere and determines the asymptotic variance $p(1-p)/(nf(q{)}^2)$. Its gross-error sensitivity is $\max (p,1-p)/f(q)$, but its local-shift sensitivity is infinite because of the jump.

The local-shift sensitivity of a scalar statistical functional isIt measures the sensitivity of the first-order contamination effect to moving the contaminating observation. If the influence function is continuously differentiable with bounded derivative, the mean value theorem gives ${\lambda }^{\ast }\le {\sup }_x|{\mathrm{IF}}^{\prime }(x)|$, and taking $y\to x$ proves equality. A jump in the influence function makes ${\lambda }^{\ast }$ infinite, even when gross-error sensitivity is finite.

For observations expressed in centered, standardized coordinates, the rejection point iswith value $\infty$ if the set is empty. A finite rejection point means sufficiently distant contamination has zero first-order effect. Boundedness of the influence function does not imply a finite rejection point: the influence function of a quantile has nonzero constant tails. Nor does tending to zero at infinity guarantee a finite rejection point unless those tails become identically zero.

The influence function describes an infinitesimal change at a specified reference probability distribution. The replacement breakdown point asks how many observations in a particular finite sample can be replaced before an estimator becomes unbounded. The finite-sample maximum bias records the intervening changes. These are different questions: a bounded influence function at one baseline is not a general bound over all epsilon-contamination neighborhoods, nor a proof of a positive replacement breakdown point.^[ref]

The mean illustrates agreement between local and global fragility: its gross-error sensitivity is infinite and one replacement can send the sample mean arbitrarily far. The median has finite gross-error sensitivity under the stated positive-density assumptions, while its sample median withstands fewer than about half the observations being replaced under ordinary tie conventions. Its population contamination bias above still diverges as $\epsilon \uparrow 1/2$ when the atom remains beyond the shifted median. Here $x$ must also move outward; taking that limit with a fixed finite atom would invalidate the displayed quantile formula.

For scalar location with a fixed positive scale, the Huber score has a direct global argument: if strictly more than half of the sample stays fixed, then beyond every retained observation by more than $ks$, their saturated scores outweigh all possible replaced scores. Every root therefore stays in a bounded interval. This is a separate finite-replacement argument; estimating scale changes its premises. A smooth bounded influence function is valuable local protection, but global robustness also depends on the full estimating rule, scale choice, and root selection.

Consider the population statistical model $F_{\epsilon ,x}=(1-\epsilon )N(0,1)+\epsilon {\delta }_x$. The following are deterministic statistical functional calculations, not observed contamination experiments. Compare the actual location bias with the linear approximation $\epsilon \mathrm{IF}(x;T,F)$. The mean bias is exactly $\epsilon x$. For positive $x$ exceeding the shifted median and $0\le \epsilon <1/2$, the generalized-inverse median bias is exactlyThis formula applies to both $x=1$ and $x=8$ throughout the plotted interval $0\le \epsilon \le 0.1$. If the contaminating atom lies inside the quantile jump containing probability $1/2$, the generalized-inverse median instead equals that atom.

For the Huber location estimator with $s=1$ and $k=1.345$, solve its actual population equation, rather than substituting the influence function:Writing $a=t-k$ and $b=t+k$, integration of the normal distribution giveswhere $\varphi$ is the probability density function of $N(0,1)$. Its derivative is $-[\Phi (t+k)-\Phi (t-k)]<0$, so the mixture score is strictly decreasing for $\epsilon <1$. For $\epsilon <1/2$, its limits at negative and positive infinity have opposite signs even with arbitrary distant point contamination. The displayed examples use the unique root, evaluated by the bisection method to numerical precision.

At $x=8$ and $\epsilon =0.01$, the exact mean, median, and Huber location estimator biases are $0.080000$, $0.012660$, and $0.016541$; the influence function predictions are $0.080000$, $0.012533$, and $0.016375$. At $\epsilon =0.05$, they become $0.400000$, $0.066012$, and $0.086241$, versus predictions $0.400000$, $0.062666$, and $0.081875$. The bounded alternatives control the effect of the distant atom, but their actual response curves bend away from the tangent.

The nearby atom $x=1$ gives a useful comparison: at $\epsilon =0.05$, the Huber location estimator bias is $0.060237$, slightly below its linear approximation $0.060874$, whereas the median bias is still $0.066012$. The direction and size of approximation error depend on the contaminating location. Even accurate local influence calculations must specify $F$, the scale, the cutoff, and the actual contamination fraction.