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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

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.

First let $b>0$. Substitute $x=-y$ in the negative half of the integral. Since $g$ is an odd function,The two halves cancel, soFor $b=0$ both limits coincide. For $b<0$, reverse the limits and apply the result with $-b>0$. Thus the identity holds for every real $b$.

On $(-\pi ,\pi )$ the function is odd, so its mean and all its cosine Fourier coefficients vanish:The assigned endpoint value does not affect these integrals. The product of the two odd functions $x|x|$ and $\sin nx$ is even, hence the sine Fourier coefficients areTwo applications of integration by parts giveThereforeThe periodic extension has a jump discontinuity at every odd multiple of $\pi$: its left and right limits are ${\pi }^2$ and $-{\pi }^2$. Its assigned value at each such point is $-{\pi }^2$. Thus it is odd away from these points, but it is not a globally odd function pointwise. The Fourier series will take the midpoint value zero at these jumps.

Since the initial velocity vanishes, D'Alembert's formula reduces toEach translated copy has half the original amplitude. To identify each requested cross-section, track the translated support intervals and add the values where they overlap.

At $t=\pi$, $f(x-\pi )$ is supported on $[-\pi ,0]\cup [2\pi ,3\pi ]$ and $f(x+\pi )$ on $[-3\pi ,-2\pi ]\cup [0,\pi ]$. On these intervals $\sin (x\pm \pi )=-\sin x$, soThe pieces meet continuously at zero, where both contributions vanish.

At $t=3\pi /2$, the support of $f(x-3\pi /2)$ is $[-\pi /2,\pi /2]\cup [5\pi /2,7\pi /2]$; that of $f(x+3\pi /2)$ is $[-7\pi /2,-5\pi /2]\cup [-\pi /2,\pi /2]$. In the central overlap,so the two contributions cancel. The remaining wave profile isFor the time cross-section at the origin, the initial profile is an odd function, soFor the time cross-section at $x=\pi /2$, the left-travelling contribution is nonzero for $\pi /2\le t\le 3\pi /2$ and equals $\cos t/2$. The right-travelling contribution is nonzero for $3\pi /2\le t\le 5\pi /2$ and also equals $\cos t/2$. At the common endpoint both vanish. HenceThe following graphs show all four cross-sections. In particular, the last trace reaches $-1/2$ at $t=\pi$ and $1/2$ at $t=2\pi$.

Because $u(0,t)=0$, restriction to the half-line solves the following initial-boundary value problem with a homogeneous Dirichlet boundary condition:This is wave reflection by odd extension. If a condition at infinity is desired, take the solution with finite wave energy; here the profile has bounded support at every finite time.

The initial profile is continuous and piecewise continuously differentiable, but its first derivative jumps at the support endpoints. Thus the formula gives a weak solution globally and a piecewise classical solution away from the characteristic lines for the one-dimensional wave equation issuing from those endpoints. The initial velocity is zero in the square-integrable function sense and pointwise wherever the initial profile is differentiable; at the kink points a pointwise right time derivative need not equal zero. This regularity qualification is needed for the given data.

For a nonzero integer $k$, direct integration giveswhereas ${\int }_{-\pi }^{\pi }1dx=2\pi$. The product-to-sum formulastherefore establish Fourier orthogonality: for positive integers $m,n$,Here ${\delta }_{mn}$ is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by $\cos (mx)$ or $\sin (mx)$ and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral givesIf $f$ is an even function, then $f(x)\sin (nx)$ is an odd function, while $f(x)\cos (nx)$ is an even function. ThusThe Fourier series consequently reduces to a Fourier cosine series.

Absolute integrability now requiresFor $\alpha \le 0$, the integrand is bounded by $z{e}^{-z^2/2}$, whose integral is finite. For $\alpha >0$, once $z^2\ge 1/\alpha$, the exponent is at least $\alpha z^4/2$, so the integral diverges. HenceAgain the zero follows from the odd function and symmetry; outside the integrable range both signed parts diverge, so the expected value is undefined.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

Use $d(\ln \sec x)/dx=\tan x$ and $\ln \sec 0=0$. To obtain enough terms of tangent, write $t(x)=x+a{x}^3+b{x}^5+O(x^7)$; it is an odd function. The identity $t^{\prime }=1+t^2$ givesso $a=1/3$, $b=2/15$. Integrate this Taylor expansion term by term from zero to $x$:There is no constant term, and only even powers occur because $\ln \sec x$ is an even function near zero. Thus reaching the first three nonzero terms requires expansion through degree six.

The integrand is even and decays at both infinity and zero, since $a,b>0$. On $x>0$ set $u=\sqrt{a}x-\sqrt{b}/x$, a strictly increasing bijection onto the whole real line. ThenUsing evenness for the negative half-axis givesThe second summand is an integrable odd function, so its integral vanishes. The remaining Gaussian integral givesFor the second integral use $x=1/t$ separately on each half-axis. The factor $x^{-2}dx$ becomes $-dt$, and reversing the transformed limits yields $J(a,b)=I(b,a)$. ThusThese identities are instances of the Gaussian integral with an inverse-square term. Splitting at zero before either change of variables avoids treating a discontinuous transformation as a single bijection.

The periodic ramp is an odd function, so $a_0=a_n=0$. Integration by parts givesThus the Fourier series of a periodic sawtooth isThe periodic function is smooth at $x=\pi /2$, so the Dirichlet-Jordan convergence theorem permits evaluation there. Even-index terms vanish and odd-index sines alternate in sign, giving

The integrand is an odd function, because $x$ is odd and $\cos x$ is even. Its integral over the symmetric interval is therefore

Write $y=\frac{1}{2}\sin (2x)$. It is an odd function with amplitude $1/2$ and period $\pi$. Its zeros on the specified interval are $-\pi ,-\pi /2,0,\pi /2,\pi$. The maxima are at $x=-3\pi /4,\pi /4$ and the minima at $x=-\pi /4,3\pi /4$, with values $1/2$ and $-1/2$. The middle panel of the preceding figure shows the two full oscillations.

Take $\sigma >0$ and substitute $u=(x-\mu )/\sigma$. The density becomes the standard Gaussian distribution density $\phi (u)=e^{-u^2/2}/\sqrt{2\pi }$. Its normalization follows from the supplied Gaussian integral after scaling by $\sqrt{2}$. Since $u\phi (u)$ is an integrable odd function, its integral is zero, and therefore $\mathbb{E}X=\mu$.

For the second moment, ${\phi }^{\prime }=-u\phi$. Integration by parts givesThe endpoint term vanishes by Gaussian decay. Consequently

The function $x^2\sin x$ is an odd function, since $x^2$ is even and the sine is odd. Its integral on the symmetric interval therefore cancels:

The Gaussian function $e^{-x^2}$ is even and $\sin x$ is odd, so their product is an odd function. Moreover $|e^{-x^2}\sin x|\le e^{-x^2}$, whose integral on the real line is finite. Thus the improper integral is absolutely convergent, not merely a symmetric principal value, and the contributions of its two half-lines cancel:

The integrand is an even function, since both $x$ and $\sin x$ are odd functions. Using integration by parts,The endpoint cosine terms vanish, leaving the sine contribution.

The flat function $f$ is an even function, has its unique zero and global minimum at the origin, and approaches the horizontal asymptote $1$ as $x\to \pm \infty$. It decreases for negative $x$ and increases for positive $x$, meeting the origin with a horizontal tangent line.

Its derivative is an odd function: $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$ off zero, with $f^{\prime }(0)=0$. It is negative to the left and positive to the right, tends to zero both at the origin and at infinity, and has one negative trough and one positive crest. Indeed $f^{\prime \prime }(x)=2e^{-1/x^2}(2-3x^2)/x^6$ away from zero, so the change in slope of $f^{\prime }$ occurs once on each side. The exact crest and trough coordinates are not needed for the sketch.

Insert the linear initial profile into the transformed integral from (b):The constant part integrates to $ax$. The other part is an absolutely integrable odd function, so its integral vanishes. HenceAs a direct check, $u_t=u_{xx}=0$ and $u(x,0)=ax$. Thus a linear temperature profile is stationary on the whole line. More generally, every affine function is preserved by this heat-kernel solution.