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An isotropic tensor has unchanged components under every proper rotation of the Cartesian axes. Every rank-zero scalar is isotropic, and $c{\delta }_{ij}$ is a rank-two example. The zero rank-one tensor is isotropic, but no nonzero vector is: a half-turn about an axis perpendicular to it reverses its sign.

For the inertia tensor, write $u=x-G$ and $X=P-G$, so $x-P=u-X$. Expand its defining integrand. Every cross term has an integral containing ${\int }_{B}u\rho d^{3}x=0$ by the definition of the centre of mass. The remaining terms give the tensor parallel-axis theorem:For a uniform unit ball, rotational symmetry gives $\int \rho u_i{u}_j{d}^{3}u=c{\delta }_{ij}$. Its trace is $\int \rho |u{|}^2{d}^{3}u=3M/5$, so $c=M/5$ and $I(G)=2MI/5$. ThereforeAssuming positive mass, it is isotropic exactly when $P=G$. For $X\ne 0$, its eigenvalues and directions areTo give an explicit orthonormal eigenbasis when $X_1^2+X_2^2>0$, take $e_1=(-X_2,X_1,0)/\sqrt{X_1^2+X_2^2}$ and $e_2=e_{\parallel }\times e_1$. If $X$ is a nonzero vector along the third axis, take the first two coordinate unit vectors instead. At $X=0$, all three eigenvalues are $2M/5$ and every orthonormal basis works. The inertia tensor of a displaced uniform ball describes the same result for an arbitrary radius.

Treat the left side as a causal convolution with $t^{-\alpha }$. Under the usual Laplace-transform hypotheses, the convolution theorem for Laplace transforms givesFor an absolutely continuous $f$ with $f(0)=0$, the derivative transform is $\mathcal{L}[f^{\prime }]=s\hat{f}$. ConsequentlySince $\mathcal{L}[t^{\alpha -1}]=\Gamma (\alpha )s^{-\alpha }$, inverting the product provesHere $0<\alpha <1$ makes both power kernels locally integrable. For example assume exponential growth bounds sufficient for the transforms and a regularity class where the convolution is defined. This is the Abel integral equation of fractional order; the initial value removes the extra boundary term in the derivative transform.

For $f(t)=t$, the remaining integral is ${\int }_0^t(t-\tau {)}^{\alpha -1}d\tau =t^{\alpha }/\alpha$. ThusSubstitution into the original equation checks the result: scaling $\tau =tu$ gives $tB(\alpha +1,1-\alpha )/[\Gamma (\alpha +1)\Gamma (1-\alpha )]=t$ by the beta-gamma identity.

Cauchy's integral theorem states that a holomorphic function on a simply connected domain has zero contour integral around every closed piecewise smooth curve in that domain. One proof begins with the Cauchy theorem for a triangle. Subdivide a triangle into four equal triangles. Their oriented boundary integrals sum to the original integral because internal edges cancel, so choose a nested subtriangle carrying at least one quarter of the preceding integral's modulus. At stage $n$, its perimeter and diameter are $P/2^n,d/2^n$, and the nested triangles converge to a point $z_0$. Complex differentiability gives $f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$, with $ϵ(z)\to 0$. The affine terms have primitives and integrate to zero. Therefore $|I_n|\le \sup |ϵ|Pd/4^n$, while $|I_0|\le 4^n|I_n|$. Letting $n\to \infty$ proves $I_0=0$. Triangulation cancels internal edges for polygonal contours; local primitives and deformation through the domain extend the result to the stated piecewise smooth curves. Simple connectedness ensures the necessary contractions do not pass through a missing point.

Apply the theorem to $e^{-a{z}^2}$ on the rectangle between heights zero and $\rho$, with vertical sides at $\mathrm{Re}z=\pm R$. On either vertical side,The modulus of each side integral is at most $|\rho |e^{-a{R}^2+a{\rho }^2}\to 0$. Both horizontal integrals converge absolutely. The two horizontal orientations are opposite, so the limit gives $I(\rho ,a)=I(0,a)$, also for negative $\rho$. This is Gaussian contour translation.

Now complete the square:The substitution $z=x-i\nu /(2a)$ moves the path to height $-\nu /(2a)$. Its Gaussian integral equals the real-axis integral by the preceding result. HenceThis also gives the Fourier transform of a Gaussian in the convention with no prefactor in the forward transform.

The Klein four-group has multiplication tableSince it is abelian, every conjugacy class is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An irreducible representation is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a finite group, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.

A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices areThey are distinct irreducible representations, exhausting the four possibilities. The character table of the Klein four-group isThe specified diagonal matrices obey $D(a{)}^2=D(b{)}^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a group representation. Its coordinate axes are invariant lines carrying respectively ${\chi }_{01},{\chi }_{10},{\chi }_{11}$. ThereforeThe trivial representation ${\chi }_{00}$ is absent. If $H$ denotes the four-row character matrix above, direct row products give $H{H}^T=4I$. This verifies character orthogonality:The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.

Let $L=d^2/d{x}^2+x^{-1}d/dx-x^{-2}$. A Green's function $G(x,\xi )$ satisfies $L_{x}G=\delta (x-\xi )$ and the chosen homogeneous boundary conditions in $x$. Away from $x=\xi$, it is a linear combination of $x$ and $x^{-1}$. It must be continuous there and obey the unit derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$, because the leading differential coefficient is one. Thensolves $Ly=f$ whenever the integral and boundary limits are valid. The Wronskian of $x,x^{-1}$ is $-2/x$. The Green kernels for a radial dipole Euler equation implement the two different choices of boundary conditions below.

At a point of complex differentiability, the complex derivative is independent of the direction of approach. Taking increments along the real axis gives $f^{\prime }=u_x+i{v}_x$; taking imaginary increments gives $f^{\prime }=v_y-i{u}_y$. Equality of their real parts and imaginary parts proves the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

For the specified real part, observe that $z^2{e}^z$ has exactly that real part. Equivalently, integrating $v_y=u_x$ and checking $v_x=-u_y$ givesThusThe freedom is only a constant imaginary part: the difference of two such holomorphic functions has zero real part, so its Cauchy-Riemann equations make it constant on each connected domain.

For $s=|\mathbf{x}-{\mathbf{x}}_0|>0$, the radial Laplacian gives ${\nabla }^2(1/s)=s^{-2}{\partial }_s(s^2{\partial }_s(1/s))=0$. The outward flux of $\nabla [-1/(4\pi s)]$ through a small sphere centred at the source is $[1/(4\pi s^2)]4\pi s^2=1$. Applying Green's second identity to a smooth test function outside that small sphere and shrinking its radius therefore gives the distributional identityThe kernel tends to zero at infinity. Thus it is the fundamental solution with the sign convention $\Delta G=+\delta$. The sign will reverse when converting it into the electrostatic potential of a positive point charge through Poisson's equation.

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

For the normalized second-order equation $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, an ordinary point has analytic $P,Q$. A regular singular point is a singular point where $(x-x_0)P$ and $(x-x_0{)}^{2}Q$ extend analytically through $x_0$. The two finite singular points here are $0$ and $1/2$. At zero, the respective limiting coefficients are $1/2,0$; at $1/2$, they are $-1,0$. Thus both are regular singular points, with indicial roots $0,1/2$ and $0,2$, respectively.