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Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields

The Born probability is the integral of $|{\psi }_n{|}^2$ over the indicated interval. Changing variables to $\xi$ cancels the dimensional normalization factors and givesSince $\frac{d}{d\xi }(\xi e^{-{\xi }^2})=e^{-{\xi }^2}-2{\xi }^2{e}^{-{\xi }^2}$, integration givesAs $\ell \to \infty$, the correction term tends to zero and $P_0(\ell )\to 1$ by normalization. Hence $\boxed{P_1(\ell )\to 1}$. This is expected because the expanding interval exhausts the whole position space and the state is normalized: the particle is somewhere on the line with total probability one.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

For $x>0$, the geometric series givesEvery summand is nonnegative and has integral $(n+\alpha {)}^{-2}$ by part (a). This numerical series converges by comparison with ${\sum }_{n\ge 1}{n}^{-2}$. The series-integration rule proved above therefore yieldsThe value at the single endpoint zero does not affect this Lebesgue integral.

On $0<x<1$, the geometric expansion gives summands $h_n(x)=x^{n+\beta -1}-x^{n-\beta }$. Their integrals areBoth powers are integrable because $0<\beta <1$. If $\beta <1/2$, each $h_n$ is nonnegative, since its first exponent is smaller. Their integrals form a convergent nonnegative series: the $n=0$ term is finite and subsequent terms are bounded by a constant times $n^{-2}$. The proved series-integration rule applies.

If $\beta >1/2$, apply precisely that rule to $-h_n$, which are nonnegative, and negate the result. At $\beta =1/2$ all summands and the integrand are zero. Thus in every case the integral is absolutely integrable andThe common sign of the summands is the reason no unproved theorem for integrating arbitrary signed series is needed.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise for measurable functions, then $h$ is measurable and $\int h={\lim }_n\int h_n$, allowing infinity.

Apply it to the partial sums $S_N={\sum }_{n=0}^N{f}_n$. They are nonnegative, measurable, and increase pointwise to $f$. Finite additivity of the integral and monotone convergence yieldThe assumed finite sum of integrals makes $f$ integrable. This proves the series-integration rule needed below directly from the stated theorem, without assuming an additional interchange theorem.

Let $N$ be the vector subspace of integrable functions that vanish almost everywhere. The Lebesgue space is the quotient vector space $L^1(\mathbb{R})={\mathcal{L}}^1(\mathbb{R})/N$: its elements are equivalence classes of functions equal almost everywhere. Addition and scalar multiplication are $[f]+[g]=[f+g]$ and $c[f]=[cf]$.

Set $\Vert [f]{\Vert }_1=\int |f|$. This is well-defined because replacing $f$ by an equal-almost-everywhere representative does not change the integral. It is nonnegative, homogeneous, and obeys the triangle inequality by $|f+g|\le |f|+|g|$. Part (i) says its only zero element is the zero class. Hence it is a norm on $L^1$.

No such sequence exists. By Fatou's lemma on $(0,1)$,But the proposed nonnegative measurable integrand is positive at every point of $(0,1)$, so its integral cannot be zero. Explicitly, the sets where it is at least $1/k$ cover $(0,1)$; at least one has positive measure, forcing a positive integral. This contradicts the displayed bound.

Orient $\partial B$ counterclockwise, and define positive circulation by that orientation. The outward normal from the body satisfies $nds=(dy,-dx)$. Pressure acts inward, soThe steady Bernoulli equation for this constant-density irrotational flow is $p+\rho |v{|}^2/2=C$. The constant has zero integral around a closed curve. With complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, let $W=w^{\prime }=u-iv$. The no-penetration condition makes the boundary a streamline, so $dw=Wdz$ is real along it. Thus $Wdz=\overline{W}d\overline{z}$, and $|W{|}^{2}d\overline{z}=W^{2}dz$. Substitution proves Blasius force theorem:At infinity the Laurent expansion has the formThere is no source term because the impermeable body has zero net flux. The circulation term follows from $\oint Wdz=\Gamma$. The coefficient of $z^{-1}$ in $W^2$ is $\Gamma (U-iV)/(\pi i)$. Analyticity in the fluid lets the body contour be enlarged to a large counterclockwise circle, so the residue theorem gives $\oint W^{2}dz=2\Gamma (U-iV)$. HenceIts component along $(U,V)$ is zero: the drag is zero. The lift magnitude is $\boxed{\rho |\Gamma |\sqrt{U^2+V^2}}$. With the stated positive counterclockwise circulation, the signed lift along the left normal $(-V,U)/\sqrt{U^2+V^2}$ is $-\rho \Gamma \sqrt{U^2+V^2}$. Reversing the circulation convention reverses this signed formula. This is the Kutta–Joukowski theorem with its orientation made explicit.

With the weighted inner product $⟨p,q{⟩}_w={\int }_{-1}^{1}p(x)q(x)w(x)dx$, a family of orthogonal polynomials has $\mathrm{deg}{p}_j=j$, each $p_j$ nonzero, andContinuity and positivity of the weight make the integral finite and give $⟨p_j,p_j{⟩}_w>0$. Unit norm or monic leading coefficient is an additional optional normalization.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

For $x>0$, the geometric series givesEvery summand is nonnegative and has integral $(n+\alpha {)}^{-2}$ by part (a). This numerical series converges by comparison with ${\sum }_{n\ge 1}{n}^{-2}$. The series-integration rule proved above therefore yieldsThe value at the single endpoint zero does not affect this Lebesgue integral.

On $0<x<1$, the geometric expansion gives summands $h_n(x)=x^{n+\beta -1}-x^{n-\beta }$. Their integrals areBoth powers are integrable because $0<\beta <1$. If $\beta <1/2$, each $h_n$ is nonnegative, since its first exponent is smaller. Their integrals form a convergent nonnegative series: the $n=0$ term is finite and subsequent terms are bounded by a constant times $n^{-2}$. The proved series-integration rule applies.

If $\beta >1/2$, apply precisely that rule to $-h_n$, which are nonnegative, and negate the result. At $\beta =1/2$ all summands and the integrand are zero. Thus in every case the integral is absolutely integrable andThe common sign of the summands is the reason no unproved theorem for integrating arbitrary signed series is needed.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise for measurable functions, then $h$ is measurable and $\int h={\lim }_n\int h_n$, allowing infinity.

Apply it to the partial sums $S_N={\sum }_{n=0}^N{f}_n$. They are nonnegative, measurable, and increase pointwise to $f$. Finite additivity of the integral and monotone convergence yieldThe assumed finite sum of integrals makes $f$ integrable. This proves the series-integration rule needed below directly from the stated theorem, without assuming an additional interchange theorem.

Let $N$ be the vector subspace of integrable functions that vanish almost everywhere. The Lebesgue space is the quotient vector space $L^1(\mathbb{R})={\mathcal{L}}^1(\mathbb{R})/N$: its elements are equivalence classes of functions equal almost everywhere. Addition and scalar multiplication are $[f]+[g]=[f+g]$ and $c[f]=[cf]$.

Set $\Vert [f]{\Vert }_1=\int |f|$. This is well-defined because replacing $f$ by an equal-almost-everywhere representative does not change the integral. It is nonnegative, homogeneous, and obeys the triangle inequality by $|f+g|\le |f|+|g|$. Part (i) says its only zero element is the zero class. Hence it is a norm on $L^1$.

No such sequence exists. By Fatou's lemma on $(0,1)$,But the proposed nonnegative measurable integrand is positive at every point of $(0,1)$, so its integral cannot be zero. Explicitly, the sets where it is at least $1/k$ cover $(0,1)$; at least one has positive measure, forcing a positive integral. This contradicts the displayed bound.

Away from the boundary, derivatives of a smoothly varying solution are much smaller than $k^{2}f$ unless the outer solution vanishes. Nonzero boundary data therefore require a rapidly varying layer. If its thickness is $\delta$, the radial second derivative is of order $f/{\delta }^2$; balancing it with $k^{2}f$ gives $\delta =k^{-1}$.

Put $R=k(1-r)$ and write $f(r,\theta )=F(R,\theta )$. The equation becomesFor fixed $R$ and boundary data with angular variation independent of $k$, the leading equation is $F_{RR}=F$. Matching to the zero outer solution removes the growing exponential, and the boundary value at $R=0$ fixes its coefficient:This approximation describes the layer; it is not a uniform expansion in polar coordinates at the origin.

Laplace's method localizes an exponential integral near maxima of its real phase. At an isolated smooth interior maximum $t_0$ with $G^{\prime \prime }(t_0)<0$ and $F(t_0)\ne 0$, Taylor expansion and Gaussian integral giveAn endpoint maximum with nonzero inward phase slope instead gives an inverse-$k$ factor. Degenerate maxima require their first nonzero local phase term; continuous differentiability alone does not guarantee the quadratic formula.

Here $\cos (ix\sin \varphi )=\cosh (x\sin \varphi )$. In its two exponentials, the positive exponential dominates at the interior maximum $\varphi =\pi /2$ of $\sin \varphi$. Since $\sin (\pi /2+s)=1-s^2/2+O(s^4)$,The negative exponential has no comparable positive maximum and is negligible. Consequently, where $kr\to \infty$,In the boundary layer $r=1-R/k$, $r^{-1/2}=1+O(k^{-1})$, so this reduces to $e^{-R}$, agreeing with the formal result when $g=1$.

Define $y=k^{-1}\dot{x}+x^3/3-x$. Differentiation and the original equation give $\dot{y}=(\lambda -x)/k$, while rearrangement gives $\dot{x}=k(y-x^3/3+x)$. Write $h(x)=x^3/3-x$.

On the fast scale $\tau =kt$,Fast trajectories are horizontal. Their equilibria lie on the cubic $y=h(x)$; the transverse linearization is $-h^{\prime }(x)=1-x^2$, so the outer branches $|x|>1$ attract and the middle branch repels. On a horizontal line, arrows point right when $y>h(x)$ and left when $y<h(x)$.

The source prints the slow time as $T=k/t$. The slow variable needed here is $T=t/k$. With this correction,At leading order $y=h(x)$ andThe reciprocal printed variable would give $-T^2{y}_T=\lambda -x$, rather than the autonomous slow equations used in the stated period formula.

For $-1<\lambda <1$, the left attracting branch runs from $x=-2$ to the fold $x=-1$, where $y=2/3$. The horizontal fast jump at this height lands at $x=2$. The right branch then runs from $2$ to the fold $1$, where $y=-2/3$, followed by a fast jump to $-2$. The equilibrium at $x=\lambda$ is on the repelling middle branch. The two slow segments dominate the period:The jump times are negligible on this $O(k)$ scale. Substituting $u=-x$ in the first integral also gives the compact valueFor $\lambda =1/2$, the negative slow segment takes $k[1-\frac{3}{4}\log (5/3)]$, and the positive one takes $k[2-\frac{3}{4}\log 3]$. Thus $x(t)$ rises slowly from $-2$ to $-1$, jumps to $2$, decreases more slowly to $1$, then jumps back to $-2$; the jumps are nearly vertical in the time trace.

For $\lambda >1$ the fast horizontal field and attracting branches are unchanged. The right slow branch has arrows toward $x=\lambda$ from both directions; the left branch still moves toward $-1$ and jumps right. The equilibrium $(\lambda ,h(\lambda ))$ is stable, so the large relaxation cycle is replaced by approach to this equilibrium. If $\lambda >2$, a landing at $2$ moves right toward $\lambda$, rather than left. The sketches display the representative case $\lambda =3/2$.

For fixed nonzero $x$, expand the denominator geometrically:The effective expansion parameter is ${ϵ}^2/x^2$, so the expansion fails when $x=O(ϵ)$. In the inner variable $X=x/ϵ$,for fixed $X$. Expanding the leading outer term toward $x=0$ gives $x^{-2}$ at leading order; expanding the leading inner term for large $X$ and returning to $x$ gives ${ϵ}^{-2}{X}^{-2}=x^{-2}$. Thus the two one-term re-expansions match in $ϵ\ll x\ll 1$. The next common contribution is $-x^{-1}$ when the corresponding next inner order is retained.

On $x\ge \delta$, the relative denominator error is bounded by ${ϵ}^2/{\delta }^2$, henceIntegration by parts yieldsFor the last step split the remaining integral at one: ${\int }_{\delta }^{1}dx/x=-\log \delta$, and replacing $e^{-x}$ by one on that interval has a bounded error.

In the inner interval, $|e^{-x}-1|\le x$ gives an explicit error estimate:The $1/\delta$ terms cancel between regions. Every displayed remainder is $o({ϵ}^{-1})$ whenever $ϵ/\delta \to 0$ and $\delta \to 0$; for example $ϵ({ϵ}^2/{\delta }^3)=(ϵ/\delta {)}^3$. Therefore the leading answer is independent of the arbitrary intermediate cutoff:

On the two independent variables, the total derivative is $D_x={\partial }_x+{ϵ}^{-1}{\partial }_X$. Multiplication of $D_x(a{D}_{x}y)=f$ by ${ϵ}^2$ gives exactly the stated multiple-scale equation.

At order one, ${\partial }_X(a{\partial }_X{y}_0)=0$. Integrating twice givesThe periodic reciprocal coefficient has strictly positive mean. Unless $A=0$, the integral has a nonzero linear drift in $X$, violating the periodic cell ansatz and creating secular growth when $X=x/ϵ$. Thus $A=0$ and $y_0=y_0(x)$.

At order $ϵ$,Periodicity requires the derivative to have zero cell average. ConsequentlyAt order ${ϵ}^2$, the equation isThe second flux inside braces is $C(x)$, independent of $X$. Integrating over one period annihilates the first derivative, by periodicity, and yields $C^{\prime }(x)={\int }_0^{1}f(x,X)dX$. HenceThe effective transport coefficient is the harmonic mean of $a$ over the fast period; the source is averaged arithmetically. Smoothness sufficient for these formal differentiations and uniform positivity on the cells are understood.