数式の部分は，残念ならがら答えそのものを書いたシケ対プリントが出
回ったようだ．採点の比重を小さくすることにする．解答例を下に示す．
(1)
{\rm div}X=\nabla\cdot X=\sum_{k=1}^n{\partial\xi_k\over\partial x_k}
(2)
{(-1)^{n+1}\over\cos^{n+1}x}\det(B_{ik})
(3)
(1-x)\sum_{n=1}^\infty{x^n\over1-x^n}-\log{1\over1-x}
(4)
\int{x\over(px+q)(ax+b)}dx={1\over aq-bp}\left[{q\over p}\log|px+q|-{b\over a}\log|ax+b|\right]
(5)
\int{dx\over x^6-1}={1\over12}\log\left|{x^2-x+1\over x^2+x+1}\right|+{1\over6}\log\left|{x-1\over x+1}\right|+{1\over2\sqrt3}\arctan{\sqrt3x\over1-x^2}
(6)
\int{dx\over\sqrt{(x-a)^2+b^2}}=\log\left(x-a+\sqrt{(x-a)^2+b^2}\right)
(7)
{1\over\sqrt{1-a^2}}\log\left|{\sqrt{(1+a)(1+x)}-\sqrt{(1-a)(1-x)}\over\sqrt{(1+a)(1+x)}+\sqrt{(1-a)(1-x)}}\right|
(8)
{\cos^2x\over2}+\log|\sin x|
(9)
\int_{-1}^1{x\over(x-a)\sqrt{1-x^2}}dx=\pi
(10)
\left(e^{-ax}-{1\over1+bx}\right){1\over x}
(11)
\int_0^{\pi\over2}{\sin x\over\sqrt{1-k^2\sin^2x}}dx={1\over2k}\log{1+k\over1-k}
(12)
\sum_{n=1}^\infty{\sinh nx\over n!}=e^{\cosh x}\sinh(\sinh x)
(13)
\sum_{r=0}^{[n/2]}(-1)^r{n\over n-r}{n-r\choose r}(2\cosh x)^{n-2r}=2\cosh nx
(14)
\prod_{n=1}^\infty\left(1+(-1)^n{x\over n}\right)=\prod_{n=1}^\infty\left(1-{x\over2n-1}\right)\left(1+{x\over2n}\right)=\sqrt\pi\bigg/\Gamma\left(1+{x\over2}\right)\Gamma\left({1-x\over2}\right)
(15)
\cot z=\lim_{m\to\infty}\sum_{n=-m}^m{1\over z+n\pi}
(16)
g(p)=\int_0^\infty e^{-pt}f(t)dt
(17)
\sqrt{{\pi\over2}{\sqrt{p^2+\alpha^2}-p\over p^2+\alpha^2}}
(18)
\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty\exp\left[-\sum_{r=1}^n(x_r-y_r)^2\right]f(x_1,x_2,\cdots,x_n)dx_1dx_2\cdots dx_n
(19)
{1\over\Gamma(z)}\int_0^\infty{t^{z-1}e^{-at}\over1-e^{-t}}dt
(20)
\exp\left(\int\psi dz\right)Z_\nu(z)
(21)
S_n''(t)+S_n(t)=O_{n-1}(t)+O_{n+1}(t)
(22)
{}_pF_q(\alpha_1,\cdots,\alpha_p;\beta_1,\cdots,\beta_q;z)=\sum_{n=0}^\infty{(\alpha_1)_n\cdots(\alpha_p)_n\over(\beta_1)_n\cdots(\beta_q)_n}{z^n\over n!}
(23)
{\partial\ \over\partial z}\left[{z^{1-r}F(\alpha-\gamma+1,\beta-\gamma+1,2-\gamma;z)\over F(\alpha,\beta,\gamma;z)}\right]={(1-\gamma)z^{-\gamma}(1-z)^{\gamma-\alpha-\beta-1}\over[F(\alpha,\beta,\gamma;z)]^2}
(24)
N_{2p-k-m,k,m}=\sum_{r=0}^p(-1)^r{k\choose p-r}{m\choose r}
(25)
(z-e_1)^{\kappa_1}(z-e_2)^{\kappa_2}(z-e_3)^{\kappa_3}\prod_{r=1}^m(z-\xi_r)
