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Take the Fourier transform in $x$. The wave equation becomes the ordinary differential equationwith initial data $\tilde{u}(k,0)=\tilde{f}(k)$ and ${\partial }_t\tilde{u}(k,0)=\tilde{g}(k)$. Therefore

Part (c) and the Translation property of the Fourier transform invert the first term toFor the second term choose an antiderivative $p$ of $g$. By part (b), $\tilde{g}=ik\tilde{p}$, and henceIts inverse transform isCombining the terms gives the D'Alembert formula

The function $y_1=x$ solves the linear ordinary differential equation. Apply reduction of order by writing $y=xv(x)$. Substituting $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$ and cancelling the terms in $v$ givesFor $w=v^{\prime }$, this becomes a first-order linear ordinary differential equation:A partial fraction decomposition provides the required antiderivative:ThereforeThe logarithm has positive argument on the required interval. The Wronskian of $x$ and $1+x\log ((x-1)/x)$ is $1/(x-1)\ne 0$, proving their linear independence and hence completeness of this general solution.

Enforce the fixed-end Dirichlet boundary condition by taking the odd extension of the initial velocity to the whole line:This agrees with the specified initial condition for $s\ge 0$. To derive the relevant form of D'Alembert's formula, the zero initial displacement in b gives $G=-F$, up to a cancelling constant. The initial velocity then gives $-2c{F}^{\prime }=v_o$. HenceAn antiderivative, valid across $s=0$, isEvaluation therefore gives the solution throughout $x>0,t>0$:For $ct\le x$, replace $|x-ct|$ by $x-ct$; for $ct\ge x$, replace it by $ct-x$. Thus the same expression includes the reflected wave after it reaches the fixed end.

At $x=0$ the two terms agree, so $y(0,t)=0$. At $t=0$ they again agree, so $y(x,0)=0$. Differentiating the integral with respect to $t$ giveswhich recovers $y_t(x,0)=\kappa x{e}^{-x/L}$. Since $v_o$ is a continuously differentiable function, the formula is twice continuously differentiable even across $x=ct$, and satisfies the wave equation there as well.

The core contributes no change in $\phi$, since ${\phi }^{\prime }=0$ there. Integrating the radial gravitational field across the shell, with $u=r/r_0$, givesUsing the supplied antiderivative,HenceThe negative sign agrees with ${\phi }^{\prime }(r)<0$ in the material shell under the paper's gravitational potential sign conventions.

The printed arguments $x-t$ and $x+t$ assume unit wave speed. For the stated wave equation with general $c>0$, the correct arguments are $x-ct$ and $x+ct$, as in the paper's subsequent formula. Equivalently, use rescaled time $\tilde{t}=ct$ to obtain the printed unit-speed form.

Introduce coordinates $\xi =x-ct$, $\eta =x+ct$ along the characteristic lines for the one-dimensional wave equation, and write $u(x,t)=U(\xi ,\eta )$. The chain rule givesso the wave equation becomesThus $U_{\xi }$ is independent of $\eta$, and integration gives the countertravelling decomposition of the one-dimensional wave equationConversely, differentiating this expression twice verifies the wave equation for any sufficiently differentiable $F,G$.

The two initial conditions giveDifferentiating the first and solving these two equations for the derivatives givesFor a fixed reference point $x_0$, their antiderivatives can be writtenSubstituting the two travelling arguments cancels the constants and combines the integrals. This proves D'Alembert's formula with initial velocity:At $t=0$ the integral vanishes, giving $u=f$; differentiating by the fundamental theorem of calculus gives $u_t=g$ there. These checks also fix the sign and the factor $1/(2c)$.

We prove the symmetric derivative criterion for monotonicity using an antiderivative. Fix $a<u<v<b$ and defineThe fundamental theorem of calculus gives $F^{\prime }=f$ because $f$ is continuous. For each interior $x$,Since $f^s(x)>0$, the definition of symmetric derivative gives $\delta >0$ such that $f(x+t)-f(x-t)>t{f}^s(x)$ for $0<t<\delta$. Therefore, for $0<h<\delta$,

We next show that $F$ is a convex function. On any closed interval $[r,s]\subset (a,b)$ let $\ell$ be the affine function joining $(r,F(r))$ to $(s,F(s))$. If $F-\ell$ were positive anywhere, the extreme value theorem would give a positive maximum at an interior point $x$. For every sufficiently small $h>0$, that maximum would implyThe affine function contributes zero to this symmetric difference, contradicting the strict positive inequality above. Thus $F\le \ell$ on every such interval, exactly convexity.

For a differentiable convex function, secant slopes bound endpoint derivatives. Explicitly, the chord inequality at $u+h$ and at $v-h$, followed by $h\downarrow 0$, givesSo $f$ is nondecreasing. If $f(u)=f(v)$ for some $u<v$, monotonicity would make $f$ constant on $[u,v]$, forcing $f^s(x)=0$ for $u<x<v$, a contradiction. Consequently $\boxed{f(u)<f(v)\text{ for every }u<v}$: $f$ is strictly increasing. This proof requires no continuity of $f^s$ and never applies the ordinary mean value theorem to a merely symmetrically differentiable function.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is

Divide by $x>1$. The product rule turns the linear ordinary differential equation intoEquivalently, $\ln x$ is an integrating factor after division by $x\ln x$. An antiderivative gives $\ln xy=\frac{1}{2}(\ln x{)}^2+C$. The initial condition at $e$ gives $C=-1/2$, henceIts derivative is $y^{\prime }=[1+(\ln x{)}^{-2}]/(2x)$, so substitution gives $x\ln x{y}^{\prime }+y=\ln x$, and $y(e)=0$. Every solution of a differential equation has the integrated form above, and the initial condition fixes its only constant. This agrees with the uniqueness theorem for ordinary differential equations on the stated interval.

Let $I={\int }_0^{\infty }{e}^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is $r$, soUsing integration by parts, with the boundary term $x{e}^{-x^2}$ vanishing at both endpoints,The substitution $u=x^2$ also gives the two antiderivativesThese values also follow from the Gaussian moment recurrence and can be checked by differentiation.

The upper boundary follows $y=x$ until its intersection with $y=1/x$ at $(1,1)$, then follows the hyperbola to $(2,1/2)$. The lower boundary is $y=0$. Therefore the two vertical-slice descriptions are $0\le x\le 1$, $0\le y\le x$, and $1\le x\le 2$, $0\le y\le 1/x$.

Integrating with respect to $y$ first reduces the double integral toUsing the antiderivatives from part (a),

The region lies in the first quadrant, inside the unit circle and above the positive branch of $xy=1/4$. In polar coordinates, $xy=(r^2/2)\sin 2\theta$, so its radial bounds are $(2\sin 2\theta {)}^{-1/2}\le r\le 1$. Such an interval exists when $\sin 2\theta \ge 1/2$, henceAt the intersections with the circle, $r=1$ and $\sin 2\theta =1/2$, giving exactly the two stated polar angles. Their Cartesian coordinates are $(\cos (\pi /12),\sin (\pi /12))$ and the swapped pair.

The Jacobian determinant contributes $dxdy=rdrd\theta$, while $xy(x^2+y^2)=r^4\sin 2\theta /2$. Thus the double integral isIts antiderivative is $-\cos 2\theta /24+\cot 2\theta /192$. Evaluating the bounds gives

Two applications of integration by parts give $I=\cosh x\sin x-\int \sinh x\sin xdx$ and $\int \sinh x\sin xdx=-\sinh x\cos x+I$. Thus an antiderivative is $[\cosh x\sin x+\sinh x\cos x]/2$. Evaluating its endpoints gives

With the change of variables $u=3+2x$, $du=2dx$, an antiderivative isHere $C$ is the constant of integration.

For $x\in (n,n+1)$, strict decrease gives $f(x)<f(n)$, while for $x\in (n-1,n)$ it gives $f(x)>f(n)$. Integrating over these unit intervals provesSumming the rectangle bounds givesThe partial sums and the improper integral are nondecreasing. These bounds show that one is bounded exactly when the other is bounded, proving the integral test for convergence. In the convergent case, sum the same interval inequalities for $n=N+1,N+2,\dots$ to obtainFor $f(x)=1/(x\ln x)$, the antiderivative is $\ln (\ln x)$, which grows without bound, so the first series diverges. For $f(x)=1/[x(\ln x{)}^2]$, the antiderivative is $-1/\ln x$, whose tail integral is finite, so the second series converges, withBoth functions are positive and strictly decreasing on $[2,\infty )$, as required.

Use the change of variables $u=x^2+1$, $du=2xdx$. ThenDifferentiating verifies this antiderivative on the whole real line.

The change of variables $u=x^2$, $du=2xdx$, givesDifferentiation verifies the antiderivative for every real $x$.

Again the function is even, so all sine coefficients vanish. Its constant term is $a_0/2=\sinh 1$. With $k=n\pi$, an antiderivative isThus $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. In the requested factored notation,This is the period-two specialization of the Fourier series of periodic hyperbolic sine and cosine. The function values match at the endpoints, although its derivative has a jump in the periodic extension.

An antiderivative of $\sin x$ is $-\cos x$. The fundamental theorem of calculus gives

Set $u=3-2x$, so $du=-2dx$. This integration by substitution givesThe antiderivative is taken on either interval separated by $x=3/2$. Differentiating verifies the sign and factor $-1/2$.

The integrand is a continuous function, so the fundamental theorem of calculus differentiates the upper limit directly:No explicit antiderivative is needed.