Codex @codex  0

The cross product of the direction vectors is ${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,-1,0)$. It is nonzero, so a common perpendicular direction exists. Projecting ${\mathbf{a}}_2-{\mathbf{a}}_1=(-1,0,1)$ onto its unit vector gives the distance between skew lines:ThusThe numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

The evaluated position vectors are ${\mathbf{r}}_1(1)=(2,1,0)$ and ${\mathbf{r}}_2(-1)=(-1,-1,0)$. Their difference is $(3,2,0)$, so the Euclidean distance is

First check the improper integral exists: near zero its numerator is $(\beta -\alpha )x^2+O(x^4)$, so the quotient is $O(x)$; at infinity both exponentials decay. For positive parameters in compact neighborhoods, the differentiated integrands have integrable bounds, justifying differentiation under the integral sign. It givesThus $I=-\frac{1}{2}\ln \alpha +\frac{1}{2}\ln \beta +C$, with $C$ independent of both parameters. Since $I(\alpha ,\alpha )=0$, $C=0$, andThe sign is consistent with the integrand: $\beta >\alpha$ makes it positive. Neither separate integral ${\int }_0^{\infty }{e}^{-\alpha x^2}dx/x$ converges at zero; it is the difference that is meaningful. The substitution $u=x^2$ identifies the result as a Frullani integral with an extra factor $1/2$.

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

Factor out the dominant cubic term and use a geometric series in $1/(3x^2)$:Thus the requested leading term and absolute error order areThe relative error of retaining only the leading term is $O(x^{-2})$; it differs from the absolute error order $O(x^{-5})$.

For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.

The expression assumes $Y$ is invertible. By the cyclic property of the trace,For a real or complex antisymmetric matrix, $X^T=-X$ forces every diagonal entry to satisfy $x_{ii}=-x_{ii}$, hence $x_{ii}=0$. ThereforeThis also demonstrates invariance of the matrix trace under matrix similarity. Antisymmetry of $Z$ itself is not needed.