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For a scalar smooth map $\Phi :{\mathbb{R}}^n\to \mathbb{R}$, integration can be decomposed into its level surfaces with the gradient weight displayed in the title. Near a regular level, using $\Phi$ as the normal coordinate gives the volume element $dx=dSds/|\nabla \Phi |$, which explains the formula by integration by substitution. In particular, $\int W(x)\delta (\Phi (x))dx={\int }_{\Phi =0}W/|\nabla \Phi |dS$ when the zero level is regular and the integrals are well-defined. General versions allow Lipschitz functions and maps into higher-dimensional spaces.

Set $u=Br/A$, retaining $r_0$ in $B$. Replacing only the lower integration limit by zero givesSubstitution of the constants yieldsTo quantify the approximation, write $ϵ=\delta M/M\ll 1$. The true lower limit is $u_0=B{r}_0/A=ϵ$, and an elementary integration by substitution, $u={\sin }^2\chi$, givesSince $\arcsin \sqrt{ϵ}-\sqrt{ϵ(1-ϵ)}=\frac{2}{3}{ϵ}^{3/2}+O({ϵ}^{5/2})$, the approximate time is too large by a relative fraction $\frac{4}{3\pi }{ϵ}^{3/2}+O({ϵ}^{5/2})$. This explains why changing the lower limit introduces little error for a small excess mass. The cosmological turnaround time diverges as $\delta M\to 0^{+}$, consistently with the absence of finite cosmological turnaround in an exactly critical Einstein-de Sitter universe.

Put $u=x^2$, so $xdx=du/2$. Then integration by substitution givesThe last two integrals are elementary exponential moments, or values of the Gamma function.

Use integration by substitution with $\theta =\pi /4-\phi /2$, so $\sin \phi =\cos 2\theta$ and $d\phi =-2d\theta$. Reversing the transformed limits givesThe original integrand is finite at both endpoints.

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 $-1/2$.

Use the integration by substitution $x=\sin \theta$, with $-\pi /6\le \theta \le \pi /6$. On this interval $\cos \theta >0$, so it cancels the positive denominator square root. The transformed integral isIts integrand is even. Therefore the answer in the PDF's modulus-amplitude convention for the elliptic integral of the second kind isThe modulus is $2/3$, not the squared parameter $4/9$.

Set $s=\sin u$, so $ds=\cos udu$, and write $J_n=I_{n,1}(\pi /2,0)$. The integration by substitution gives $J_n={\int }_0^1(1-s^3{)}^{n}ds$. For $n>0$, integration by parts yieldsBoth endpoint terms vanish. Use $s^3=1-(1-s^3)$ inside the remaining integral:ConsequentlyThis is the $p=3$ instance of the power-complement integral recurrence, with initial value $J_0=1$.

Subtract the integrands first, using $\cos u-{\cos }^{3}u=\cos u{\sin }^{2}u$. Then set $s=\sin u$:The further integration by substitution $w=s^3$ makes $dw=3s^{2}ds$, soThis also holds at $n=0$.

The indicated integration by substitution givesPointwise, the argument of $\phi$ tends to $x$. Since the Gaussian integral is $\int e^{-{\eta }^2}d\eta =\sqrt{\pi }$, the expected limit is $\phi (x)$. To justify it for unbounded data, use the approximate identity property rather than assuming a uniform bound on $\phi$.

In the original variable, normalization givesFor $|\xi -x|<\delta$, continuity makes the difference smaller than any prescribed $\epsilon$ if $\delta$ is small enough, and the kernel mass is at most one. On the complement the Gaussian mass tends to zero; the data-weighted tail tends to zero by heat-kernel convergence for Gaussian-integrable initial data. This provesFor heat-kernel convergence from a conditionally convergent Gaussian integral, the same tail conclusion follows using the bounded primitive of $(\phi (\xi )-\phi (x))e^{-{\xi }^2/(4T)}$. The factor $q=K(x-\xi ,t)e^{{\xi }^2/(4T)}$ has its maximum at $\xi =xT/(T-t)$, approaching $x$. For sufficiently small $t$, it is monotone on each of the two tails outside $|\xi -x|<\delta$, and its values at the tail endpoints tend to zero exponentially. Integration by parts bounds each tail by a constant times its endpoint value, proving the same limit without an absolute-convergence assumption. The approach $t\downarrow 0$ is from positive time.

For $m=-1$, set $u=\ln x$, $du=dx/x$. This integration by substitution givesThis case is treated separately because the denominator $m+1$ in part (a) vanishes.

Use integration by substitution with $x=2u$ and the Gaussian being an even function:The factor two from the differential exactly cancels the half-line factor in the given full-line Gaussian integral.

The integral converges absolutely: near zero it behaves like $\log x$, and at infinity like $(\log x)/x^2$. Let its value be $J$. The integration by substitution $x=1/t$ givesThereforeAbsolute convergence makes this cancellation between the two ends legitimate.

Use integration by substitution with the reciprocal variable $t=1/x$, so $dx=-dt/t^2$ and $\log x=-\log t$. The endpoints $x=1,\infty$ become $t=1,0$. ThusRenaming the dummy integration variable proves the identity without evaluating either side.

Set $u=\ln x$, so $du=dx/x$. The real domain is $x>1$, ensuring $u>0$. The integration by substitution followed by integration by parts givesThus

Set $x=2\sin \theta$ with $-\pi /2<\theta <\pi /2$, so $dx=2\cos \theta d\theta$ and $\sqrt{4-x^2}=2\cos \theta$. The integration by substitution givesThis uses the principal inverse sine; the chosen angle interval keeps the square root positive.

Apply the recurrence repeatedly down to $m=2$:Including the base factor $1/(n-1)$ and rewriting the products as factorials givesThe formula also includes $m=2$ because $0!=1$. It is symmetric in $m,n$, consistent with the integration by substitution $x=1/t$. Equivalently it is the Euler beta function $B(m-1,n-1)$, although that identification is not needed for the recurrence derivation.

Since $1+x^2>0$ for all real $x$, divide the equation by that factor. Its linear coefficient can be integrated by polynomial division:Thus a positive integrating factor is $M=e^{3x^2/2}(1+x^2{)}^{-3/2}$. Multiplying the divided equation gives the product rule formThe integration by substitution $u=1+x^2$ givesThe initial condition $y(0)=1$ fixes $C-2=1$. The required solution is thereforeNo real singularity was introduced by the division or the integrating factor, so the solution extends over the entire real line.

The same integration by substitution gives the expected valueThe remaining integrand is an odd function. Its integral is absolutely convergent, since ${\mathrm{sech}}^{2}t\le 4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. ConsequentlyEquivalently, the logistic distribution is symmetric about $\mu$, and its exponentially decaying tails ensure that its expected value exists.

Separate one factor of the hyperbolic sine and use the hyperbolic Pythagorean identity ${\sinh }^{2}x={\cosh }^{2}x-1$. With $u=\cosh x$ and $du=\sinh xdx$, integration by substitution givesFor real $x$, the hyperbolic cosine never vanishes. Differentiating the result gives $\sinh x(1-{\mathrm{sech}}^{2}x)={\sinh }^{3}x/{\cosh }^{2}x$, checking the sign of the reciprocal term.

Use integration by substitution with $u=\ln x$, so $du=dx/x$. For $x>1$, also $u>0$, andThe domain condition explains why no absolute value is needed around the inner logarithm.