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For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

The trapezoidal rule is exact for every affine function $f(x)=\alpha +\beta x$, equivalently every polynomial of polynomial degree at most one: the linear interpolant is the function itself. These functions have $f^{\prime \prime }=0$, so the preceding upper bound is also zero. There is no contradiction: zero error is precisely what the bound predicts.

The degree of precision of a quadrature rule is one, since the quadratic polynomial in part (i) is not integrated exactly. This does not exclude exactness for particular higher-degree or nonpolynomial functions. For example $f(x)=(x-m{)}^3$, $m=(a+b)/2$, has zero integral and opposite endpoint values, so the trapezoidal rule also gives zero. Its second derivative is generally nonzero, making the upper bound positive; an upper bound need not be attained. Signed quadrature error can cancel.

It suffices to assume $f\in C^2[a,b]$. The polynomial interpolation error for the endpoint linear interpolant gives, for every interior $x$, some ${\xi }_x\in (a,b)$ withLet $M={\max }_{[a,b]}|f^{\prime \prime }|$, which exists by continuity of the second derivative and the extreme value theorem. Since $(x-a)(b-x)\ge 0$,Integrating this pointwise inequality and using the triangle inequality for an integral yieldsThusThis derivation does not require choosing ${\xi }_x$ as a continuous function of $x$; the pointwise upper bound is sufficient.

The constant is sharp. Take the quadratic polynomial $f(x)=(x-a{)}^2$ and write $L=b-a$. Its exact integral is $L^3/3$, its trapezoidal rule value is $L^3/2$, and $\max |f^{\prime \prime }|=2$. ConsequentlyEquality holds, so no smaller universal constant than $1/12$ works under these assumptions. This is the sharp one-panel trapezoidal error bound.

Use the nonpolynomial function $f(x)=1+\sin x$ on $[-1,1]$. Its exact integral isbecause the sine is an odd function.

For every uniform partition of an interval with $N\ge 1$ panels, the quadrature nodes satisfy $x_{N-j}=-x_j$. Their composite trapezoidal rule quadrature weights agree, so the two sampled sine values cancel. If there is a node at zero its sine value is already zero. The constant term is integrated exactly, givingThis is symmetric trapezoidal exactness for odd functions, applied after adding a constant function so that the resulting definite integral is nonzero. It is exactness for this particular integrand, rather than a claim that every nonpolynomial function is integrated exactly.

A Newton-Cotes quadrature integrates a Lagrange interpolation polynomial at equally spaced quadrature nodes. For the Newton-Cotes closed quadrature, the endpoints are included: write $x_j=a+jh$, $h=(b-a)/n$, and useThese quadrature weights depend on the quadrature nodes, not on the sampled function. Open Newton-Cotes quadrature uses equally spaced interior quadrature nodes instead.

For Simpson's rule, use $n=2$ and set $x=a+ht$, so $0\le t\le 2$. The Lagrange cardinal polynomials becomeIntegrating these quadratic polynomials gives $1/3$, $4/3$, $1/3$, respectively. The change of variable contributes $h$, soIts degree of precision of a quadrature rule is three: it integrates every quadratic polynomial by construction, and also the cubic nodal polynomial $(x-a)(x-a-h)(x-b)$, whose integral vanishes by odd symmetry about $a+h$. Any cubic polynomial differs from its quadratic polynomial interpolant by a constant multiple of this nodal polynomial.

Fix $i$, write $\alpha =x_{i-1}$, $\beta =x_i$, and let $E=T-S$ on this interval. The Hermite interpolation conditions give bothTwo applications of integration by parts therefore giveOn this single cubic piece, the same calculation also givesHere $S^{(4)}=0$ inside the interval, and all four endpoint values of $E,E^{\prime }$ are zero. Since $T^{\prime \prime }=E^{\prime \prime }+S^{\prime \prime }$, we obtain the exact identityThis is an integral inequality, with no assertion that the integrand on the left has one sign pointwise. The stronger nodal derivative conditions make the boundary terms vanish on each interval separately; part (i) only needed their cancellation after summing.

The stated Hermite interpolating polynomial exists uniquely in ${\Pi }_{2n+1}$. Indeed, a polynomial in the kernel of a linear map of the value-and-derivative data map has a double root of a polynomial at each of $n+1$ distinct interpolation nodes. A nonzero such polynomial would have polynomial degree at least $2n+2$, which is impossible. The map is therefore an injective function and a linear map between vector spaces of the same finite dimension of a vector space, $2n+2$, and hence is a linear isomorphism. No higher smoothness of the original function $f$ is required to construct it.

An explicit illustration uses interpolation nodes $0,1,2$, values $0,1,0$, and zero derivatives at every interpolation node. The clamped cubic spline is $S(x)=3x^2-2x^3$ on $[0,1]$ and its reflection on $[1,2]$. Its Hermite interpolating polynomial is the quartic polynomial $T(x)=x^2(2-x{)}^2$. On $[0,1]$, $T-S=x^2(1-x{)}^2$, and the corresponding reflected formula holds on $[1,2]$. Direct integration gives $J(S)=24$, $J(T)=128/5$, and $J(T-S)=8/5$, exactly as the energy identity predicts.

Let $R$ be a second cubic spline matching the same values and endpoint derivatives. The minimum roughness property of a clamped cubic spline applied with $S$ as the minimizing cubic spline gives $J(R)\ge J(S)$, where $J(u)={\int }_{x_0}^{x_n}(u^{\prime \prime }{)}^{2}dx$. Applying the same argument with their roles exchanged gives $J(S)\ge J(R)$. Hence equality holds, and the exact identity from part (i) yieldsThe integrand is continuous and nonnegative. If it were positive somewhere, it would have a positive integral over a small interval. It follows that $R^{\prime \prime }=S^{\prime \prime }$ everywhere. Consequently $R-S$ is an affine function. Its value and derivative vanish at $x_0$, so it is identically zero. Thus $R=S$, proving uniqueness of the clamped cubic spline with the prescribed data. Existence was assumed in the question; this argument establishes the remaining uniqueness claim.

Put $E=T-S$. Matching interpolation data and endpoint derivatives givesThe endpoint slope constraints make $S$ a clamped cubic spline. The decisive step in its minimum roughness property of a clamped cubic spline is the orthogonality identityProve it separately on each interval. Two applications of integration by parts, using the vanishing fourth derivative of the cubic piece, giveThe second boundary term is zero because $E$ vanishes at both endpoints of this interval. Sum over $i$. At each interior spline knot, the two values of $S^{\prime \prime }{E}^{\prime }$ agree because both factors are continuous, so the contributions cancel. The remaining outer boundary terms vanish because $E^{\prime }$ is zero at both outer endpoints. This proves the identity.

Expanding the square now gives the stronger exact equalityThe last integral is nonnegative, proving the required inequality. It measures the excess second derivative roughness penalty of $T$ over $S$.

The argument applies directly to twice continuously differentiable competitors. More generally it holds when $T^{\prime }$ is an absolutely continuous function and $T^{\prime \prime }$ is square-integrable, the usual finite-energy class for this second derivative roughness penalty. Mere existence of pointwise second derivatives does not by itself guarantee that all displayed integrals exist; the variational statement is understood for admissible competitors with this regularity.

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

For $x>0$ the geometric ratio $e^{-2x}$ has modulus less than one, andEvery summand is nonnegative. The monotone convergence theorem therefore permits integration of the sum:The integral is finite because its integrand is bounded above by $e^{-x}$, whose integral is one. The displayed paired sums are the even partial sums of the alternating harmonic series; its odd partial sums differ by a term tending to zero. Thus the required equality holds with the usual series ordering. One can additionally evaluate it by $t=e^{-x}$:Grouping into nonnegative pairs supplies the justified interchange; absolute convergence of the ungrouped integrated alternating series is not needed and would be false.

The inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-x_i{x}_j)dm$. The mixed moments vanish by reflection in $x$ or $y$. Every diagonal integrand is even in $z$, so its integral over the upper hemisphere is half its integral over the corresponding full ball of mass $2M$. A uniform ball's coordinate moments are each $(2M)a^2/5$, obtained by spherical symmetry and $⟨r^2⟩=3a^2/5$. Hence the hemisphere has each coordinate second moment $M{a}^2/5$ andThe tensor parallel axis theorem gives $I_O=I_C+M(|d{|}^{2}I-d{d}^t)$ with $d=(0,0,3a/8)$. ThereforeTranslation along the symmetry axis changes the two transverse moments but leaves the axial one unchanged.

For sigma-finite measures, Tonelli's theorem states that the double integral of a nonnegative measurable function equals either iterated integral, allowing $+\infty$. Fubini's theorem states that if a measurable function is absolutely integrable, its sections are integrable almost everywhere, both iterated integrals exist, and they equal the double integral.

Assign any value at $(0,0)$, which does not affect Lebesgue integrability. Near that corner set $s=x+y$ and $t=(x-y)/(x+y)$. Then $x=s(1+t)/2$, $y=s(1-t)/2$, and the absolute Jacobian determinant is $s/2$. On $0<s<1$, the full range is $-1<t<1$, soThe remaining part of the unit square has $x+y\ge 1$ and is harmless. The improper power integral is finite precisely when $\alpha <3$. Under the given restriction,For those values, the change $(x,y)\mapsto (y,x)$ preserves the square and changes the sign of $F$. Absolute integrability justifies this change of variables, givingFor $\alpha \ge 3$ the positive and negative parts each diverge, so symmetry does not define a Lebesgue integral.

Set $x=\cos \theta$ with $0\le \theta \le \pi$. The cosine addition identity giveshence $T_0=1$, $T_1=x$, and $T_{n+1}=2x{T}_n-T_{n-1}$. By mathematical induction, $T_n$ is a polynomial of degree $n$: the initial cases hold, and the leading term of $2x{T}_n$ has degree $n+1$ while the subtraction has lower degree. Its leading coefficient is $2^{n-1}$ for $n\ge 1$.

The same change of variables gives the weighted integral

For a nonnegative measurable function, the Lebesgue integral is defined through nonnegative simple functions:The value may be infinite. A real measurable $g$ is Lebesgue integrable when $\int |g|dx<\infty$; then $\int g=\int g^{+}-\int g^{-}$, where $g^{+}=\max (g,0)$ and $g^{-}=\max (-g,0)$.

If $|g|\le h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves monotonicity of the Lebesgue integral:Hence $g$ is integrable, and both its positive and negative parts have finite integrals.

The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\le \min (1,|u|)$ give the integrable majorantIndeed, $\int H=2+2{\int }_1^{\infty }{x}^{-3}dx=3$. The comparison just proved gives $\int |k|\le 3$, so $k$ is Lebesgue integrable over the whole real line, despite its oscillations near zero. Its odd symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.

Fubini's theorem states that an absolutely integrable measurable function on a product of two $\sigma$-finite measure spaces has integrable sections almost everywhere, the corresponding iterated integrals exist, and both orders equal its product integral. In particular, for $f\in L^1({\mathbb{R}}^2)$,To establish integrability of the nonnegative function here, first use Tonelli's theorem: for a nonnegative measurable function the product integral equals either iterated integral with infinity allowed. This avoids assuming the conclusion needed for Fubini's theorem.

If $\alpha \ge -1$, ${\int }_0^{\infty }(1+x+y{)}^{\alpha }dx=\infty$ for every fixed $y$. If $\alpha <-1$, it isThe outer integral is finite exactly when $\alpha +1<-1$, that is, $\alpha <-2$. In that caseFor $-2\le \alpha <-1$ the outer integral diverges, including logarithmic divergence at $\alpha =-2$. Nonnegativity means finiteness of this integral is exactly absolute integrability; after proving it, Fubini's theorem allows either order.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

Integrate the Lagrange interpolation polynomial and exchange the integral with the finite sum:Therefore the interpolatory quadrature weights areThe resulting quadrature rule is exact for every polynomial of degree at most $n$, since its interpolant is itself. For a general continuous $f$, replacing $f$ by its interpolant supplies the approximation.

For a nonnegative measurable function, the Lebesgue integral is defined through nonnegative simple functions:The value may be infinite. A real measurable $g$ is Lebesgue integrable when $\int |g|dx<\infty$; then $\int g=\int g^{+}-\int g^{-}$, where $g^{+}=\max (g,0)$ and $g^{-}=\max (-g,0)$.

If $|g|\le h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves monotonicity of the Lebesgue integral:Hence $g$ is integrable, and both its positive and negative parts have finite integrals.

The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\le \min (1,|u|)$ give the integrable majorantIndeed, $\int H=2+2{\int }_1^{\infty }{x}^{-3}dx=3$. The comparison just proved gives $\int |k|\le 3$, so $k$ is Lebesgue integrable over the whole real line, despite its oscillations near zero. Its odd symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.