Let $I = \int_C {\frac{y}{{x^2 + y^2 }}dx} – \frac{x}{{x^2 + y^2 }}dy$ where $C$ is a circle oriented counterclockwise.
〔1〕Evaluate $I$ if $C$ is given by $(x-2016)^2+(y-2016)^2=1$
〔2〕Evaluate $I$ if $C$ is given by $x^2+y^2=1$.

〔1〕By Green’s theorem
$\begin{array}{cl} \Rightarrow I & =\iint_{\Omega}Q_x-P_ydA \\ & =\iint_{\Omega}\underbrace { – \frac{{1 \cdot (x^2 + y^2 ) – x \cdot (2x)}}{{(x^2 + y^2 )^2 }} – \frac{{1 \cdot (x^2 + y^2 ) – y(2y)}}{{(x^2 + y^2 )^2 }}}_0dA \\ & =0 ∎ \end{array}$

〔2〕 由於被積函數包含奇異點 (分母有 0)，因此不滿足 Green’s theorem
$C:\ x^2+y^2=1\Rightarrow \left\{ {\begin{array}{l} x = \cos t \\ y = \sin t \end{array}} \right.,\ 0\leq t\leq 2\pi$
$\begin{array}{cl} \Rightarrow I & =\int_0^{2\pi } {\frac{{\sin t}}{1}( – \sin t)} dt – \frac{{\cos t}}{1}(\cos t)dt \\ & =-\int_0^{2\pi } {dt} \\ & =-2\pi ∎\end{array}$

Find the maximum and minimum values of the function $f(x,y,z)=x^2-y^2$ on the surface $x^2+2y^2+3z^2=1$.

Let $g(x,y,z)=x^2+2y^2+3z^2-1$
Let $\nabla f + \lambda \nabla g = 0$
$\Rightarrow (2x, – 2y,0) + \lambda (2x,4y,6z) = (0,0,0)$
$\begin{array}{l} { \Rightarrow \left\{ {\begin{array}{l} {2x + 2\lambda x = 0} \\ { – 2y + 4\lambda y = 0} \\ {6\lambda z = 0}\end{array}} \right.} \\ {x^2 + 2y^2 + 3z^2 = 1} \end{array}\Rightarrow \left\{ {\begin{array}{l} {x(1 + \lambda ) = 0} \\ {y( – 1 + 2\lambda ) = 0} \\ {\lambda z = 0} \end{array}} \right.\Rightarrow \left\{ {\begin{array}{l} {x = 0} & {{\text{or}}} & {\lambda = – 1} \\ {y = 0} & {{\text{or}}} & {\lambda = \frac{1}{2}} \\ {z = 0} & {{\text{or}}} & {\lambda = 0} \end{array}} \right.$

$\begin{array}{cl} \because & f( \pm 1,0,0) = ( \pm 1)^2 – 0^2 = 1 \\ & f(0, \pm \frac{1}{{\sqrt 2 }},0) = 0^2 – ( \pm \frac{1}{{\sqrt 2 }})^2 = – \frac{1}{2} \\ & f(0,0, \pm \frac{1}{{\sqrt 3 }}) = 0^2 – 0^2 = 0\\ \therefore & \left\{ {\begin{array}{l} {{\text{Max}}{\rm{.}} = 1} \\ {{\text{min}}{\rm{.}} = – \frac{1}{2}}\end{array}} \right. ∎ \end{array}$

Compute $\mathop {\lim }\limits_{x \to \infty } (\sqrt {x + \sqrt {x + \sqrt x } } – \sqrt x )$.

$\begin{array}{cl} \mathop {\lim }\limits_{x \to \infty } (\sqrt {x + \sqrt {x + \sqrt x } } – \sqrt x ) &= \mathop {\lim }\limits_{x \to \infty } (\sqrt {x + \sqrt {x + \sqrt x } } – \sqrt x ) \cdot \frac{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{x + \sqrt {x + \sqrt x } – x}}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {x + \sqrt x } /\sqrt x }}{{(\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x )/\sqrt x }} \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + \sqrt {\frac{1}{x}} } }}{{\sqrt {1 + \sqrt {\frac{1}{x} + \frac{1}{{x^{3/2} }}} } + 1}} = \frac{1}{2} ∎ \end{array}$

For what positive $x$ does the following series converge?
$\sum\limits_{n = 1}^\infty {(\sqrt[n]{x} – 1)}$

For $x=1$, $\sum\limits_{n = 1}^\infty {(\sqrt[x]{x} – 1)} = 0$ converges.

For $x\ne 1$
$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{x^{\frac{1}{n}} – 1}}{{\frac{1}{n}}}\mathop = \limits^{{\text{Let }}t = \frac{1}{n}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{x^t – 1}}{t}\mathop = \limits^{\text{L}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{x^t \ln x}}{1} = \underbrace {\ln x}_{ \in \mathbb{R}} \ne 0\ (x\ne 1) \\ & \text{and }\sum\limits_{n = 1}^\infty {\frac{1}{n}}\text{ diverges. (by }p\text{-series test)} \\ \therefore & \sum\limits_{n = 1}^\infty {(x^{\frac{1}{n}} – 1)}\text{ diverges. (by limit comparison test)} \end{array}$

the series converges only for $x=1 ∎$

Let $B = \left\{ {(x,y,z) \in \mathbb{R}^3 :x^2 + y^2 + z^2 \le 1} \right\}$. Evaluate the integral
$\iiint_{B}\frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV$.

◎ 已知積分區域為一球，具有很好的對稱性
$\begin{array}{cl} \iiint_{B}\frac{{x^4 + 2y^4 }}{{z^4 + 4y^4 + x^4 }}dV & =\iiint_B \frac{{x^4 + 2y^4 + z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }} – \frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \\ & =\iiint_BdV-\iiint_B \frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \end{array}$

$\begin{array}{cl} \iiint_B \frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV & =\iiint_B 1-\frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \\ & =\iiint_B dV-\iiint_B \frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \end{array}$
$\Rightarrow 2\iiint_B \frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV=\frac{4}{3}\pi$
$\Rightarrow \iiint_B\frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV=\frac{2}{3}\pi ∎$

The $n$-th derivative of $\frac{1}{x^{2016}-1}$ has the form $\frac{P_n}{(x^{2016}-1)^{n+1}}$ where $P_n(x)$ is a polynomial. Find $P_n(1)$ for all $n\ge 0$.

$\begin{array}{cl} f'(x) & =\frac{{ – 1}}{{(x^{2016} – 1)^2 }} \cdot 2016x^{2015} \\ & =\frac{{ – 2016x^{2015} }}{{(x^{2016} – 1)^2 }},\ P_1 (x)=-2016x^{2016} \end{array}$

$\because\ f^n (x) = \frac{{P_n (x)}}{{(x^{2016} – 1)^{n + 1} }} = P_n (x)\frac{1}{{(x^{2016} – 1)^{n + 1} }}$
$\begin{array}{cl} \therefore\ f^{n+1}(x) & =P’_n (x) \cdot \frac{1}{{(x^{2016} – 1)^{n + 1} }} + P_n (x) \cdot \frac{{ – (n + 1)}}{{(x^{2016} – 1)^{n + 2} }} \cdot 2016x^{2015} \\ & = \frac{{P’_n (x)(x^{2016} – 1) – P_n (x) \cdot (n + 1) \cdot 2016x^{2015} }}{{(x^{2016} – 1)^{n + 2} }}\end{array}$
$\Rightarrow P_{n + 1} (x) = P’_n (x)(x^{2016} – 1) – P_n (x) \cdot (n + 1) \cdot 2016x^{2015}$

$P_1 (1)=-2016$
$\Rightarrow P_{n + 1} (1) = P'(1) \cdot (1^{2016} – 1) – P_n (1) \cdot (n + 1) \cdot 2016$
$\Rightarrow P_2 (1) = ( – 2016) \times ( – 1) \times 2 \times 2016 = 2016^2 \times 2$
$\Rightarrow P_3 (1) = 2016^2 \times 2 \times ( – 1) \times 3 \times 2016 = 2016^3 \times ( – 1) \times 2 \times 3$
$\vdots$
$\Rightarrow P_n (1) = ( – 1)^n \cdot (2016)^n \cdot (n!) ∎$

〔1〕Prove that $\int_0^\infty {(\frac{{\sin x}}{x})^2 } dx = \int_0^\infty {\frac{{\sin x}}{x}} dx$. (10 分)
〔2〕Evaluate the improper integral $\int_0^\infty {\frac{{\sin x}}{x}} dx$. (10 分)

〔1〕
$\begin{array}{cl} \int_0^\infty {(\frac{{\sin x}}{x})^2 } dx & =\int_0^\infty {x^{ – 2} } \sin ^2 xdx \\ & =\left. { – x^{ – 1} \sin ^2 x} \right|_0^\infty + \int_0^\infty {x^{ – 1} } \cdot 2\sin x\cos xdx \\ & =\underbrace {\left. { – \frac{{\sin ^2 x}}{x}} \right|_0^\infty }_0 + \int_0^\infty {\frac{{\sin 2x}}{x}} dx\\&=\int_0^\infty {\frac{{\sin 2x}}{x}} dx\ (\text{Let }u = 2x \Rightarrow du = 2dx)\\&=\int_0^\infty {\frac{{\sin u}}{{\frac{u}{2}}} \cdot } \frac{{du}}{2} = \int_0^\infty {\frac{{\sin x}}{x}} dx ∎\end{array}$

〔2〕
$\begin{array}{cl} \int_0^\infty {\frac{{\sin x}}{x}} dx & =\int_0^\infty {\frac{{\sin x}}{x}} (\int_0^\infty {e^{ – u} } du)dx \\ & =\int_0^\infty {\sin x} \int_0^\infty {e^{ – u} } \frac{{du}}{x}dx\ (\text{Let }u = xy \Rightarrow du = xdy) \\ & =\int_0^\infty {\int_0^\infty {(\sin x)e^{ – xy} } } dydx\\&=\int_0^\infty {\int_0^\infty {e^{ – yx} \sin x} } dxdy \end{array}$

$\begin{array}{cl} \because\ \int_0^\infty {e^{ – yx} \sin x} dx & =\left. { – e^{ – yx} \cos x} \right|_{x = 0}^{x = \infty } – \left. {ye^{ – yx} \sin x}\right|_{x = 0}^{x = \infty } – \int_0^\infty {y^2 e^{ – yx} \sin x} dx \\ & =0 – \underbrace {( – e^0 \cos 0)}_{ – 1} – y^2 \int_0^\infty {e^{ – yx} \sin x} dx \\ & =1 – y^2 \int_0^\infty {e^{ – yx} \sin x} dx \end{array}$
$\begin{array}{cl} & \Rightarrow (1 + y^2 )\int_0^\infty {e^{ – yx} \sin x} dx = 1 \Rightarrow \int_0^\infty {e^{ – yx} \sin x} dx = \frac{1}{{1 + y^2 }} \\ \therefore & \int_0^\infty {\frac{{\sin x}}{x}} dx=\int_0^\infty {\frac{1}{{1 + y^2 }}} dy = \left. {\tan ^{ – 1} y} \right|_{y = 0}^{y = \infty } = \frac{\pi }{2} – 0 = \frac{\pi }{2} ∎\end{array}$

For each continuous function $f:[0,1] \to\mathbb{R}$, let $I(f) = \int_0^1 {xf(x)(x – f(x))} dx$. Find the maximum value of $I(f)$ over all such function $f$.

Let $f_0 (x)$ be the function such that $I(x_0)$ reaches its max.
Let $\eta :\ [0,1]\to\mathbb{R}$ be continuous such that $\left\{ {\begin{array}{l} {\eta (0) = 0} \\ {\eta (1) = 0} \end{array}} \right.$
Let $f_\alpha (x) = f_0 (x) + \alpha \eta (x)$
$\because\ I(f_0)$ reaches its max.
$\therefore\ \frac{d}{{d\alpha }}\left. {I(f_\alpha )} \right|_{\alpha = 0} = 0$
$\begin{array}{cl} I(f_\alpha) & =\int_0^1 {xf_\alpha (x) \cdot (x – f_\alpha (x))} dx \\ & =\int_0^1 {x^2 f_\alpha (x) – x\left[ {f_\alpha (x)} \right]^2 } dx\end{array}$
$\begin{array}{cl} \Rightarrow \frac{d}{{d\alpha }}I(f_\alpha ) & =\frac{d}{{d\alpha }}\int_0^1 {x^2 f_\alpha (x) – x\left[ {f_\alpha (x)} \right]^2 } dx \\ & =\int_0^1 {x^2 \frac{d}{{d\alpha }}f_\alpha (x) – x \cdot 2f_\alpha (x) \cdot } \frac{d}{{d\alpha }}f(x)dx \\ & =\int_0^1 {x^2 \eta (x) – 2x(f_\alpha (x) + \alpha \eta (x))} \cdot \eta (x)dx \end{array}$
$\begin{array}{cl} \underbrace {\frac{d}{{d\alpha }}\left. {I(f_\alpha )} \right|_{\alpha = 0} }_0 & =\int_0^1 {x^2 \eta (x) – 2xf_0 (x)\eta (x)} dx \\ & =\int_0^1 {\eta (x)(x^2 – 2xf_0 (x))} dx\end{array}$
$\because\ \eta$ is arbitrary.
$\therefore\ x^2 – 2xf_0 (x) = x(x – 2f_0 (x)) = 0 \Rightarrow f_0 (x) = \frac{\pi }{2}$
$\begin{array}{cl} \text{So max. }I(f) & =I(\frac{x}{2}) \\ & =\int_0^1 {x \cdot \frac{x}{2} \cdot (x – \frac{x}{2})} dx \\ & =\int_0^1 {\frac{{x^2 }}{2} \cdot \frac{x}{2}} dx = \frac{1}{4}\left[ {\frac{{x^4 }}{4}} \right]_{x = 0}^{x = 1} = \frac{1}{{16}} ∎\end{array}$

Evaluate $\int_0^\infty {(x – \frac{{x^3 }}{2} + \frac{{x^5 }}{{2 \cdot 4}} – \frac{{x^7 }}{{2 \cdot 4 \cdot 6}} + \cdots )(1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 4^2 }} + \frac{{x^6 }}{{2^2 \cdot 4^2 \cdot 6^2 }} + \cdots )} dx$.

Let $f(x)={(x – \frac{{x^3 }}{2} + \frac{{x^5 }}{{2 \cdot 4}} – \frac{{x^7 }}{{2 \cdot 4 \cdot 6}} + \cdots )(1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 4^2 }} + \frac{{x^6 }}{{2^2 \cdot 4^2 \cdot 6^2 }} + \cdots )}$

$\begin{array}{cl} x – \frac{{x^3 }}{2} + \frac{{x^5 }}{{2 \cdot 4}} – \frac{{x^7 }}{{2 \cdot 4 \cdot 6}} + \cdots & =x(1 – \frac{{x^2 }}{2} + \frac{{x^4 }}{{2 \cdot 4}} + \cdots ) \\ & =x(1 – \frac{{(x^2 )^1 }}{{2^1 }} + \frac{{(x^2 )^2 }}{{2^2 \cdot 2!}} – \frac{{(x^2 )^3 }}{{2^3 \cdot 3!}} + \cdots ) \\ & =x(1 + ( – 1)^1 (\frac{{x^2 }}{2})^1 + \frac{{( – 1)^2 (\frac{{x^2 }}{2})^2 }}{{2^2 }} + \cdots )\\&=x(1 + \frac{{( – \frac{{x^2 }}{2})^1 }}{{1!}} + \frac{{( – \frac{{x^2 }}{2})^2 }}{{2!}} + \frac{{( – \frac{{x^2 }}{2})^3 }}{{3!}} + \cdots )\\&=xe^{ – \frac{{x^2 }}{2}} \end{array}$

$1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 4^2 }} + \frac{{x^6 }}{{2^2 \cdot 4^2 \cdot 6^2 }} + \cdots$
$= 1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 1^2 \cdot 2^2 }} + \frac{{x^6 }}{{2^3 \cdot 1^2 \cdot 2^2 \cdot 3^2 }} + \cdots$
$= 1 + (\frac{x}{2})^2 + \frac{{(\frac{x}{2})^4 }}{{(2!)^2 }} + \frac{{(\frac{x}{2})^6 }}{{(3!)^2 }} + \cdots = \sum\limits_{n = 0}^\infty {\frac{{(\frac{x}{2})^{2n} }}{{(n!)^2 }}}$

$\begin{array}{cl} \int_0^\infty f(x)dx & =\int_0^\infty {xe^{ – \frac{{x^2 }}{2}} } \sum\limits_{n = 0}^\infty {\frac{{x^{2n} }}{{2^{2n} (n!)^2 }}} dx \\ & =\sum\limits_{n = 0}^\infty {\int_0^\infty {xe^{ – \frac{{x^2 }}{2}} \cdot \frac{{x^{2n} }}{{2^{2n} (n!)^2 }}} } dx \\ & =\sum\limits_{n = 0}^\infty {\frac{1}{{2^{2n} (n!)^2 }}\int_0^\infty {x^{2n + 1} \cdot e^{ – \frac{{x^2 }}{2}} } } dx \end{array}$

$\begin{array}{cl} \int_0^\infty {x^{2n + 1} \cdot e^{ – \frac{{x^2 }}{2}} } dx & =\int_0^\infty {x^{2n} e^{ – \frac{{x^2 }}{2}} \cdot x} dx\ (\text{Let }u = \frac{{x^2 }}{2} \Rightarrow du = 2xdx) \\ & =\int_0^\infty {2^n u^n e^{ – u} } du \\ & =2^n \underbrace {\int_0^\infty {u^n e^{ – u} } du}_{f(n)}\\&=2^n \left[ {\left. { – u^n e^{ – u} } \right|_{u = 0}^{u = \infty } + \int_0^\infty {nu^{n – 1} e^{ – u} } du} \right]\\&=2^n \cdot n\underbrace {\int_0^\infty {u^{n – 1} e^{ – u} } du}_{f(n – 1)} \end{array}$

$\Rightarrow\ f(1) = 1 \cdot f(0) = 1,\ f(2) = 2 \cdot f(1) = 2 \cdot 1 = 2!,\ f(3) = 3 \cdot f(2) = 3!$
$\cdots,\ f(n) = n \cdot f(n – 1) = n!$

$\begin{array}{cl} \Rightarrow \int_0^\infty f(x)dx & =\sum\limits_{n = 0}^\infty {\frac{1}{{2^{2n} (n!)^2 }}} \underbrace {\int_0^\infty {x^{2n + 1} \cdot e^{ – \frac{{x^2 }}{2}} } dx}_{2^n n!} \\ & =\sum\limits_{n = 0}^\infty {\frac{{2^n n!}}{{2^{2n} (n!)^2 }}} \\ & =\sum\limits_{n = 0}^\infty {\frac{1}{{2^n (n!)}}} = \sum\limits_{n = 0}^\infty {\frac{{(\frac{1}{2})^n }}{{n!}}} = e^{\frac{1}{2}} = \sqrt e ∎\end{array}$