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With the stated sign convention, write $t=y-x$ to obtainThe Cauchy principal value of the cosine part vanishes by oddness, while the Dirichlet integral givesThus the Hilbert transform has Fourier multiplier $-i\mathrm{sgn}(\omega )$. Taking the real part yields

Direct evaluation of the integral givesThe continuous value at $k=0$ is zero, in agreement with the vanishing integral of the odd function $f$. Thus

Consider the meromorphic functionClose the contour with a large semicircle in the upper half-plane, indenting above the three real pole $-1,0,1$. The large-arc integral vanishes by Jordan's lemma, while the upper-half-plane indentation rule givesThe three residue areTheir sum is $\cos 1-1$, so the residue theorem yields

The real part of the integrand is an odd function, so its full-line principal value is zero. Its imaginary part,is an even function. Taking imaginary parts and halving the full-line integral therefore gives

Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integralvanish because its integrand is odd in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its inertia tensor at the centre has the isotropic form

Forpart (a) and $\lambda =M{\ell }^2/6$ giveThe displayed symmetric matrix has orthogonal eigenvectorswith corresponding eigenvaluesFor a unit angular velocity, $|\mathbf{L}|=|I\mathbf{\omega }|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is

Apply the formula toThe upper-half-plane poles areand their residues are $e^{i{z}_k}/(4z_k^2)$. Writing $c=1/\sqrt{2}$, their sum isSince the cosine part of $x{e}^{ix}/(1+x^4)$ is odd, its integral vanishes, while the sine part is even. Henceso

Near $t=0$, take $\Omega (t)=\log F(t)$ to be the continuous local matrix logarithm with $\Omega (0)=0$. The symmetry identity gives $F(-t)=F(t{)}^{-1}$, and uniqueness of this logarithm yieldsThus $\Omega$ is an odd function. The local-logarithm qualification is necessary because the matrix exponential is not globally injective; the statement is naturally understood either near $t=0$ or as an identity of formal power series.

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

For two zero-body-force Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\hat{\mathbf{u}},\hat{\mathbf{\sigma }})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow statesApply it first with the auxiliary translating-sphere solution and then with the auxiliary rotating-sphere solution. The swimmer is force-free and torque-free, while the auxiliary surface tractions are known. The resulting surface slip velocity formulas are

On $r=a$, the first part of the prescribed slip isIts surface average is $-2\mathbf{A}/3$, whereas the $\mathbf{B}$ term has zero average by oddness. ThusThe $\mathbf{A}$ term contributes no rotation. For the other term, the isotropic second and fourth surface moments giveand hence