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For $0<\epsilon <e^{-1}$, set $u=\log (1/x)$. The ordinary change of variables formula givesTaking $\epsilon \downarrow 0$ is exactly the defining limit of the improper integral. By the improper power integral, it converges for $\theta >1$, and

In the stated units, $\alpha =1/r$. The change of variables formula givesThus, with ${\alpha }_0=1/r_0$,

For a locally integrable $u$, the change of variables formula $y=Ax$ givesThis motivatesBecause pullback by an invertible linear map acts continuously on the Schwartz space, the right side is a continuous linear functional of $\phi$. It therefore defines a tempered distribution and agrees with ordinary pullback when $u$ is a function.

For every $\phi \in \mathcal{S}$, the change of variables formula gives the identityUsing this, the distributional Fourier transform, and the pullback formula from part b,Therefore

SetBecause the region lies in the positive quadrant, its four inequalities becomeThus this change of variables sends $D$ to a rectangle in the $uv$-plane. Its Jacobian determinant issoSince $x^2=uv$, the double integral is

Since ${\delta }_{1/t}\phi (x)=\phi (x/t)$ maps the Schwartz space continuously to itself, the formuladefines a continuous linear functional on $\mathcal{S}$, hence a tempered distribution. For a locally integrable function $u$, the change of variables formula $y=tx$ givesso this dilation of a distribution agrees with ordinary function dilation.

Because $0<\alpha <n$, $|x{|}^{-\alpha }$ is locally integrable at the origin and has only polynomial growth at infinity, so it defines a tempered distribution. It is a homogeneous distribution of degree $-\alpha$ and is radial. Its Fourier transform is therefore radial and homogeneous of degree $\alpha -n$, so it must have the form $c_{\alpha }|\lambda {|}^{\alpha -n}$. In particular, $\boxed{\beta =n-\alpha }$.

To determine the constant, use the stated Gamma integral representation, Fubini's theorem, and the Fourier transform of a Gaussian:The change of variables formula $s=|\lambda {|}^2/(4\tau )$ then givesThis is precisely the Fourier transform of the Riesz kernel.

For these increasing warps, the square-root velocity function is $Q(h)(t)=\sqrt{h^{\prime }(t)}$. The chain rule givesThereforewhere the last equality again uses the change of variables formula. Taking square roots proves invariance under common right composition by $\gamma$.

Because a strictly monotone map fixing both endpoints is increasing, the change of variables formula $t=h(s)$ givesAssuming $\mathbb{E}\Vert X{\Vert }^2>0$, defineThe Hilbert-space variance identity then gives

Under the regularity needed to interchange expectation and differentiation, $\mathbb{E}{h}^{\prime }(s)=1$ because $\mathbb{E}h(s)=s$. HenceThus $A=1$ exactly when the integrated covariance in the numerator vanishes. In particular, this holds when the amplitude $X$ and the time warp $h$ are independent random variables.

A phonon has zero chemical potential and energy $\hslash \omega$. The total internal energy is consequentlyWith the change of variable $x=\beta \hslash \omega$ and $\beta \hslash {\omega }_D=T_D/T$, this becomes

Set $\tilde{\Gamma }(x)=\Gamma (x)-\Gamma (b)$. Then $\tilde{\Gamma }(b)=0$, and subtraction of a constant leaves every edge difference unchanged:The linear maphas determinant $\pm 1$: its inverse is $\gamma (x)=\tilde{\gamma }(x)-\tilde{\gamma }(a)$, because $\gamma (a)=0$. The change of variables formula therefore shows that the new density is proportional toon functions vanishing at $b$. This is precisely the discrete Gaussian free field with Dirichlet boundary condition at $b$.

Because $x^{-2}$ is integrable on $[1,\infty )$, the Dominated convergence theorem applied to the defining integral proves that $v$ is continuous on $\mathbb{R}$.

For $\lambda >0$, the change of variables formula $t=\lambda x$ givesand for $\lambda <0$ the analogous formula is obtained with $e^{it}$. On either open half-line the lower endpoint stays away from zero locally, so repeated differentiation under the integral sign proves smoothness. Thus

The displayed map is a diffeomorphism of the open unit square onto itself. For fixed $u$, the second coordinate decreases strictly from one to zero as $v$ increases from zero to one. Its inverse is $u=1-x$, $v=(1-y)/[1-(1-x)y]$, and its positive Jacobian determinant is $[1-(1-x)y{]}^2/x$. Composing it with $(t,w)\mapsto ((1-t)/(1-wt),1-w)$ gives determinant $[1-(1-x)y][1-(1-x^2)y{]}^2/[x(1-y)]$. The reciprocal determinant therefore integrates to one over the target square by the change of variables formula, despite boundary singularities.

For the local flow of a smooth compactly supported vector field and any scalar BV space function $w$, the change of variables formula and polar decomposition of a vector measure giveThe two cofactor matrices are $I\pm tB+O(t^2)$ uniformly. On $|{\sigma }_w|=1$, the linear terms in their Euclidean norm expansions cancel. Integration bounds the remainder by $C{t}^2|Dw|(\Omega )$. The identity includes all components of the decomposition of a BV derivative; it does not require smoothness of $w$ or its jump interfaces.