Proposition16~30の解説

\begin{eqnarray} \int_0^1\frac{\ln x\ln(1-x^4)}{1+x^2}dx&=&\int_0^1\frac{1-x^2}{1-x^4}\ln x\ln(1-x^4)dx\\ &=&\frac{1}{16}\int_0^1(x^{-3/4}-x^{-1/4})\frac{\ln x\ln(1-x)}{1-x}dx\ \ \ \ (x^4\to x) \end{eqnarray}
\begin{eqnarray} \int_0^1x^{s-1}\frac{\ln x\ln(1-x)}{1-x}dx&=&\lim_{t\to0}\frac{\partial^2}{\partial s\partial t}\int_0^1x^{s-1}(1-x)^{t-1}dx\\ &=&\lim_{t\to0}\frac{\partial^2}{\partial s\partial t}\frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)}\\ &=&\lim_{t\to0}\frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)}\left((\psi(s)-\psi(s+t))(\psi(t)-\psi(s+t))-\psi'(s+t)\right) \end{eqnarray}
ここで,
\begin{eqnarray} \Gamma(t)&=&\frac{1}{t}+\psi(1)+O(t)\\ \psi(s)-\psi(s+t)&=&-\psi'(s)t-\psi''(s)\frac{t^2}{2}+O(t^3)\\ \psi(t)-\psi(s+t)&=&-\frac{1}{t}+\psi(1)-\psi(s)+O(t)\\ \psi'(s+t)&=&\psi'(s)+\psi''(s)t+O(t^2) \end{eqnarray}
なので,
\begin{eqnarray} (RHS)&=&\lim_{t\to0}\left(\frac{1}{t}+\psi(1)+O(t)\right)\left(\left(-\frac{1}{2}\psi''(s)-\psi'(s)(\psi(1)-\psi(s))\right)t+O(t^2)\right)\\ &=&-\frac{1}{2}\psi''(s)-\psi'(s)(\psi(1)-\psi(s)) \end{eqnarray}
となり,
\begin{eqnarray} \int_0^1\frac{\ln x\ln(1-x^4)}{1+x^2}&=&\frac{1}{16}\left(\frac{\psi''(\frac{3}{4})-\psi''(\frac{1}{{4}})}{2}+\psi'\left(\frac{3}{4}\right)\left(\psi(1)-\psi\left(\frac{3}{4}\right)\right)-\psi\left(\frac{1}{4}\right)\left(\psi(1)-\psi\left(\frac{1}{4}\right)\right)\right)\\ &=&\frac{\pi^3}{16}-3\beta(2)\ln 2 \end{eqnarray}
と計算できます.

\begin{eqnarray} \int_0^\infty\left(\frac{\sin\tan x}{x}\right)^kdx&=&\frac{1}{2}\int_{-\infty}^{\infty}\left(\frac{\sin\tan{x}}{x}\right)^kdx\\ &=&\frac{1}{2}\sum_{n\in\mathbb{Z}}\int_{n\pi-\frac{\pi}{2}}^{n\pi+\frac{\pi}{2}}\left(\frac{\sin\tan{x}}{x}\right)^kdx\\ &=&\frac{1}{2}\sum_{n\in\mathbb{Z}}\int_{-\pi/2}^{\pi/2}\frac{\sin^k\tan{x}}{(x+n\pi)^k}dx \end{eqnarray}
$k=1$のとき,Cを大半円の経路として
\begin{eqnarray} (RHS)&=&\frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin\tan{x}}{\tan{x}}dx\\ &=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin x}{x(1+x^2)}dx\\ &=&\mathrm{Im}\frac{1}{2}\oint_C\frac{e^{iz}}{z(1+z^2)}dz\\ &=&\frac{\pi}{2}+\mathrm{Im}\ \mathrm{Res}_{z=i}\frac{e^{iz}}{z(1+z^2)}\\ &=&\frac{\pi}{2}(1-e^{-1}) \end{eqnarray}
$k=2$のとき,
\begin{eqnarray} (RHS)&=&\frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin^2\tan{x}}{\sin^2{x}}dx\\ &=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}dx=\frac{\pi}{2} \end{eqnarray}
$k=3$のとき,
\begin{eqnarray} (RHS)&=&\frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin^3\tan{x}}{\tan{x}\sin^2{x}}dx\\ &=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^3x}{x^3}dx=\frac{3\pi}{8} \end{eqnarray}

\begin{eqnarray} \int_0^1\frac{\mathrm{Li}_2(1-x^2)}{1-x^2}dx&=&\int_0^1\frac{\mathrm{Li}_2(x^2)}{x\sqrt{1-x^2}}dx\\ &=&\sum_{n\gt0}\frac{1}{n^2}\int_0^1\frac{x^{2n-1}}{\sqrt{1-x^2}}dx\\ &=&\sum_{n\gt0}\frac{1}{2n^3\beta_n}\\ &=&2\sum_{n\gt0}\int_0^1\frac{x^{2n-1}}{2n^2\beta_n}dx\\ &=&2\int_0^1\frac{(\sin^{-1}x)^2}{x}dx\\ &=&2\int_0^{\pi/2}\frac{x^2}{\tan x}dx=\frac{\pi^2}{2}\ln2-\frac{7}{4}\zeta(3) \end{eqnarray}

\begin{eqnarray} \int_0^1\frac{\tan^{-1}x\ln(1+x^2)}{1+x^2}dx&=&\int_0^{\pi/4}x\ln\frac{1}{\cos^2x}dx\\ &=&-2\int_0^{\pi/4}x\left(-\ln2+\sum_{n\gt0}\frac{(-1)^{n-1}}{n}\cos(2nx)\right)dx\\ &=&\frac{\pi^2}{16}\ln2-\sum_{n\gt0}\left(\frac{\pi}{4}\frac{(-1)^{n-1}}{n^2}\sin\frac{\pi n}{2}+\frac{(-1)^{n-1}}{2n^3}\left(\cos\frac{\pi n}{2}-1\right)\right)\\ &=&\frac{\pi^2}{16}\ln2-\frac{\pi}{4}\beta(2)-\frac{1}{2}\sum_{n\gt0}\frac{(-1)^{n-1}}{8n^3}+\frac{1}{2}\sum_{n\gt0}\frac{(-1)^{n-1}}{n^3}\\ &=&\frac{\pi^2}{16}\ln2-\frac{\pi}{4}\beta(2)+\frac{21}{64}\zeta(3) \end{eqnarray}

\begin{eqnarray} p_{\pm}(s,t;a;x)&=&\frac{e^{a\pi ix^2}}{\sinh^2\pi x}\frac{e^{s\pi x}}{e^{t\pi x}\pm1}\\ p_{\pm}(s,t;a;x+ni)&=&\frac{e^{a\pi ix^2}}{\sinh^2\pi(x+ni)}\frac{e^{\pi((s-an)x+sni-an^2i)}}{e^{t\pi(x+ni)}\pm1} \end{eqnarray}
ここで$n\in\mathbb{N}$とすると,
\begin{eqnarray} p_{\pm}(s,t;a;x+ni)=\frac{e^{a\pi ix^2}}{\sinh^2\pi x}\frac{e^{\pi((s-2an)x+sni-an^2i)}}{e^{t\pi(x+ni)}\pm1} \end{eqnarray}
となる.ここで,
\begin{eqnarray} p_{\pm}(s,t;a;x)-p_{\pm}(s,t;a;x+ni)=\frac{e^{a\pi ix^2}}{\sinh^2\pi x} \end{eqnarray}
となる$s,t,a,n$の条件を考える.
$s=2an=t,tn\in2\mathbb{N}$とすれば,
\begin{eqnarray} p_{\pm}(2an,2an;a;x)-p_{\pm}(2an,2an;a;x)&=&\frac{e^{a\pi ix^2}}{\sinh^2\pi x}\left(\frac{e^{t\pi x}-e^{an^2\pi i}}{e^{t\pi x}\pm1}\right) \end{eqnarray}
よって$an^2$の偶奇により$\pm$を決定すればよく,
偶数のとき$-$,奇数のとき$+$にすればよい.
そして実軸と実軸から高さ$n$の平行な線の長方形の経路$D$でコーシーの主値と考えれば,
\begin{eqnarray} \int_{-\infty}^{\infty}p_{\pm}(x)-p_{\pm}(x+ni)dx=\oint_Dp_{\pm}(z)dz \end{eqnarray}
とできるので,留数計算により積分が計算できます.

$a=4$のとき$n=1$であれば十分なので
\begin{eqnarray} f(z):&=&\frac{e^{4\pi iz^2}}{\sinh^2\pi z}\frac{e^{8\pi z}}{e^{8\pi z}-1}\\ \mathrm{Im}\oint_Df(z)dz&\Rightarrow&\int_{-\infty}^{\infty}\frac{\sin4\pi x^2}{\sinh^2\pi x}dx \\ &=&\mathrm{Im}\ \ 2\pi i\left(\frac{1}{2}\mathrm{Res} (f(z),0)+\frac{1}{2}\mathrm{Res}(f(z),i)+\sum_{k=1}^{7}\mathrm{Res}\left(f(z),\frac{ki}{8}\right)\right)\\ &=&\frac{5}{4}+\frac{1-2\sqrt2}{4}=\frac{3-\sqrt2}{2} \end{eqnarray}
虚部を取ったのは実部の積分が発散するためなのですが,実際は
\begin{eqnarray} \int_0^{\pi}f(\varepsilon e^{ix})\varepsilon ie^{ix}dx&=&i\int_0^\pi\frac{1}{8\pi^3\varepsilon^2e^{2ix}}+\frac{1}{2\pi^2\varepsilon e^{ix}}+\frac{5\pi+4i}{8\pi^2}+O(\varepsilon)dx\\ &=&\frac{1}{\pi^2\varepsilon}+\frac{5i}{8}-\frac{1}{2\pi}+O(\varepsilon)\\ \int_0^{\pi}f(\varepsilon e^{-ix}+i)\varepsilon ie^{-ix}dx&=&i\int_0^{\pi}\frac{1}{8\pi^3\varepsilon^2e^{-2ix}}-\frac{1}{2\pi^2\varepsilon e^{-ix}}+\frac{5\pi+4i}{8\pi^2}+O(\varepsilon)dx\\ &=&\frac{1}{\pi^2\varepsilon}+\frac{5i}{8}-\frac{1}{2\pi}+O(\varepsilon) \end{eqnarray}
のように計算すべきです.これが実部の発散の原因なのでそれと打ち消すような積分があれば実部もいい感じに表せます.今回は,この主値計算は虚部を考えれば同じ値を返すので,以後prop23,29,30も同様に留数計算で値を出させてもらいます.

\begin{eqnarray} \int_0^{\infty}\frac{dx}{1+(x+\tan x)^2}&=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{1+(x+\tan x)^2}\\ &=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{1+\left(x-\sum_{n\in\mathbb{Z}}\frac{1}{z+\frac{\pi}{2}+\pi n}\right)^2}\\ &=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{1+x^2}\\ &=&\frac{\pi}{2} \end{eqnarray}

$a=\frac12$のとき$n=2$で十分です.
\begin{eqnarray} f(z):&=&\frac{e^{\frac{\pi i}{2}z^2}}{\sinh^2\pi z}\frac{e^{2\pi z}}{e^{2\pi z}-1}\\ \mathrm{Im}\oint_Df(z)dx&\Rightarrow&\int_{-\infty}^{\infty}\frac{\sin\frac{\pi x^2}{2}}{\sinh^2\pi x}dx\\ &=&\mathrm{Im}\ 2\pi i\left(\frac{1}{2}\mathrm{Res}(f(z),0)+\frac12\mathrm{Res}(f(z),2i)+\mathrm{Res}\left(f(z),i\right)\right)\\ &=&\mathrm{Im}\ 2\pi i\left(\frac12\left(\frac{i}{4\pi^2}+\frac{i}{4\pi^2}\right)+\frac{1+i\pi}{4\pi^2}\right)\\ &=&\frac{1}{2\pi}\\ \therefore\int_0^{\infty}\frac{\sin\frac{\pi x^2}{2}}{\sinh^2\pi x}dx&=&\frac{1}{4\pi} \end{eqnarray}

以下,偏角を$(-\pi,\pi]$とする.
\begin{eqnarray} \kappa(z):=\frac1\pi\int_0^1\frac{dt}{\sqrt{t(1-t)(1-zt)}} \end{eqnarray}
の$z\in\mathbb{R}_{\gt1}$において,
\begin{eqnarray} \kappa(z\pm0i)&=&\frac{1}{\pi}\int_0^1\frac{dt}{\sqrt{t(1-t)(1-(z\pm0i)t)}}\\ &=&\frac{1}{\pi}\left(\int_0^{\frac{1}{z}}+\int_{\frac{1}{z}}^1\right)\frac{dt}{\sqrt{t(1-t)(1-(z\pm0i)t)}}\\ &=&\frac{1}{\pi}\left(\int_0^1\frac{dx}{\sqrt{zx\left(1-\frac{x}{z}\right)\left(1-(z\pm0i)\frac{x}{z}\right)}}+\int_0^{1-\frac{1}{z}}\frac{dy}{\sqrt{(1-y)y(1-z\mp0i+(z\pm0i)y)}}\right)\ \ \ \left(t\to\frac{x}{z},t\to1-y\right)\\ &=&\frac{\kappa(\frac{1}{z})}{\sqrt{z}}+\frac{1}{\pi}\int_0^1\frac{\sqrt{1-\frac{1}{z}}}{\sqrt{\left(1-(1-\frac{1}{z})t\right)t(1-z\mp0i+(z-1)t)}}dt\ \ \ \left(y\to\left(1-\frac{1}{z}\right)t\right)\\ &=&\frac{\kappa(\frac{1}{z})}{\sqrt{z}}+\frac{1}{\pi\sqrt{z}}\int_0^1\frac{dt}{\sqrt{\left(1-(1-\frac1z)t\right)t(-1\mp0i+t)}}\\ &=&\frac{\kappa(\frac{1}z)\pm i\kappa(1-\frac{1}z)}{\sqrt{z}} \end{eqnarray}
ですから,$\kappa(\frac{1}{x}\pm i)=\sqrt{x}(\kappa(x)\pm i\kappa(1-x))\ \ (0\lt x\lt1)$となることがわかりました.
そして,$a,b\gt0$のとき,
\begin{eqnarray} (a+ib)^{1/2}-(a-ib)^{1/2}&=&e^{\frac{1}{4}\ln(a^2+b^2)}(e^{\frac{1}{2}\tan^{-1}\frac{b}{a}}-e^{-\frac{1}{2}\tan^{-1}\frac{b}{a}})\\ &=&2i(a^2+b^2)^{1/4}\sin\left(\frac{1}{2}\tan^{-1}\frac{b}{a}\right)\\ &=&\sqrt{2}i\frac{b}{\sqrt{a+\sqrt{a^2+b^2}}} \end{eqnarray}
ですから,$f(z):=\kappa(z)^{1/2}z^{-2}$において次のような経路Cを考えれば,

$R\to\infty,\varepsilon\to0$で
\begin{eqnarray} \oint_Cf(z)dz&=&\int_{1+\varepsilon}^{R}f(z+0i)-f(z-0i)dz+\int_{-\pi}^{\pi}f(Re^{ix})Rie^{ix}dx-\int_{-\pi}^{\pi}f(1+\varepsilon e^{ix})\varepsilon ie^{ix}dx\\ &=&2\pi i\lim_{z\to0}\frac{d}{dx}\kappa(z)^{1/2}=2\pi i\lim_{z\to0}\frac{\frac{d}{dz}\kappa(z)}{2\kappa(z)^{1/2}}=\frac{\pi i}{4} \end{eqnarray}
となることと,
\begin{eqnarray} \int_1^{\infty}f(z+0i)-f(z-0i)dz&=&\int_0^1f\left(\frac{1}{z}+0i\right)-f\left(\frac{1}{z}-0i\right)\frac{dz}{z^2}\\ &=&\int_0^1\kappa\left(\frac{1}{x}+0i\right)^{1/2}-\kappa\left(\frac{1}{x}-0i\right)^{1/2}dx\\ &=&\int_0^1x^{1/4}\left((\kappa(x)+i\kappa(1-x))^{1/2}-(\kappa(x)-i\kappa(1-x))^{1/2}\right)dx\\ &=&\sqrt{2}i\int_0^1\frac{x^{1/4}\kappa(1-x)}{\sqrt{\kappa(x)+\sqrt{\kappa(x)^2+\kappa(1-x)^2}}}dx\\ &=&\frac{4i}{\sqrt{\pi}}\int_0^1x^{{3/2}}\frac{K'(x)}{\sqrt{K(x)+\sqrt{K(x)^2+K'(x)^2}}}dx \end{eqnarray}
と,
\begin{eqnarray} \left|\int_{-\pi}^{\pi}f(Re^{ix})Rie^{ix}dx\right|&\leq&\int_{-\pi}^{\pi}\left|\kappa(Re^{ix})^{1/2}\right|\frac{dx}{R}=O(R^{-3/2})\to0\\ \left|\int_{-\pi}^{\pi}f(1+\varepsilon e^{ix})\varepsilon ie^{ix}dx\right|&\leq&\int_{-\pi}^{\pi}|\kappa(1+\varepsilon e^{ix})|\frac{\varepsilon}{(1+|\varepsilon|)^2}dx=O(\varepsilon^{1/2})\to0 \end{eqnarray}
であることから,題意が示されました.

\begin{eqnarray} \int_0^{\infty}\frac{\ln(1+x)}{\ln^2x+\pi^2}\frac{dx}{x^2}&=&\int_0^{\infty}\frac{\ln(1+x)-\ln x}{\ln^2x+\pi^2}dx\ \ \left(x\to\frac{1}{x}\right)\\ &=&\int_0^1\int_0^{\infty}\frac{1}{\ln^2x+\pi^2}\frac{dxdt}{x+t}\\ &=&\frac{1}{2\pi i}\int_0^1\int_0^{\infty}\left(\frac{1}{\ln x-\pi i}-\frac{1}{\ln x+\pi i}\right)\frac{dxdt}{x+t} \\ &=&\frac{1}{2\pi i}\int_0^1\oint_{\Gamma}\frac{1}{\ln z-\pi i}\frac{dzdt}{z+t}\\ &=&\int_0^1\mathrm{Res}('',-t)+\mathrm{Res}('',-1)dt\\ &=&\int_0^1\frac{1}{\ln t}+\frac{1}{1-t}dt\\ &=&\gamma \end{eqnarray}

\begin{eqnarray} \int_0^\pi x\ln^2(2\sin x)dx &=&\mathrm{Im}\ \frac{1}{3}\int_0^{\pi}\left(ix+\ln(2\sin x)\right)^3+ix^3dx\\ &=&\mathrm{Im}\frac{1}{3}\int_0^{\pi}\ln^3(e^{2ix}-1)dx+\frac{\pi^4i}{24}\\ &=&\frac{\pi^4}{24}+\mathrm{Im}\ \frac{1}{3}\oint_{|z|=1}\ln^3(z-1)\frac{dz}{2iz}\ \ \ (e^{2ix}\to z)\\ &=&\frac{\pi^4}{24}+\mathrm{Im}\ \frac{\pi}{3}\mathrm{Res}_{z=0}\frac{\ln^3(z-1)}{z}\\ &=&\frac{\pi^4}{24}+\mathrm{Im}\ \frac{\pi}{3}(\pi i)^3\\ &=&\frac{\pi^4}{24} \end{eqnarray}

$a,b\gt0$において,
\begin{eqnarray} \int_0^{\pi/2}\frac{\cos^a x\cos(ax)}{\cosh2b-\cos2x}dx&=&-\int_{-\pi/2}^{\pi/2}\left(\frac{e^{2ix}+1}{2}\right)^a\frac{dx}{e^{4ix}-2e^{2ix}\cosh2b+1}\\ &=&-\frac{1}{2i}\oint_{|z|=1}\left(\frac{z+1}{2}\right)^a\frac{dz}{(z-e^{2b})(z-e^{-2b})}\ \ (e^{2ix}\to z)\\ &=&-\pi\mathrm{Res}_{z=e^{-2b}}\frac{(\frac{z+1}{2})^a}{(z-e^{2b})(z-e^{-2b})}\\ &=&\pi\frac{(\frac{1+e^{-2b}}{2})^a}{e^{2b}-e^{-2b}}\\ &=&\frac{\pi}{2}\frac{\cosh^ab\ e^{-ab}}{\sinh2b} \end{eqnarray}
また,
\begin{eqnarray} \int_0^{\pi/2}\frac{\cos^ax\cos(ax)}{\cosh2b-\cos2x}dx&=&\mathrm{Re}\int_0^{\pi/2}\frac{e^{-a(ix-\ln\cos x)}}{\cosh2b-\cos2x}dx\\ \end{eqnarray}
ですから,両辺$a^{s-1}$をかけて$a$での$(0,\infty)$の積分を考えれば,
\begin{eqnarray} \mathrm{Re}\int_0^{\pi/2}\frac{(ix-\ln\cos x)^{-s}}{\cosh2b-\cos2x}dx=\frac{\pi}{2\sinh2b}(b-\ln\cosh b)^{-s} \end{eqnarray}
がわかり,両辺$s$での偏微分と$s\to0^{+}$を考えれば,
\begin{eqnarray} \mathrm{Re}\int_0^{\pi/2}\frac{\ln(ix-\ln\cos x)}{\cosh2b-\cos2x}dx&=&\frac{\pi}{2\sinh2b}\ln(b-\ln\cosh b)\\ \frac{1}{2}\int_0^{\pi/2}\frac{\ln(x^2+\ln^2\cos x)}{\cosh2b-\cos2x}dx&=& \end{eqnarray}
がわかります.最後に$\frac{2}{1+e^{-2b}}=\frac{1+\sqrt5}{2}$となる$b$,つまり$b=\frac{\ln(\sqrt5+2)}{2}$を代入すれば題意が成り立つことがわかりました.

\begin{eqnarray} q_{\pm}(s;a;x):&=&\frac{e^{a\pi ix^2}}{\cosh\pi x}\frac{e^{s\pi x}}{e^{s\pi x}\pm1}\\ q_{\pm}(s;a;x+ni)&=&\frac{e^{a\pi ix^2}}{\cosh\pi (x+ni)}\frac{e^{\pi((s-2an)x+sni-an^2i)}}{e^{s\pi(x+ni)}\pm1} \end{eqnarray}
$n\in\mathbb{N},s=2an,sn\in2\mathbb{N}$とすると,
\begin{eqnarray} q_{\pm}(2an;a;x+ni)=(-1)^n\frac{e^{a\pi ix^2}}{\cosh\pi x}\frac{e^{an^2\pi i}}{e^{s\pi x}\pm1} \end{eqnarray}
よって$an^2+n$の偶奇によって$\pm$を決めればよく,
偶数のとき$-$,奇数のとき$+$とすれば,
\begin{eqnarray} q_{\pm}(2an;a;x)-q_{\pm}(2an;a;x+ni)=\frac{e^{a\pi ix^2}}{\cosh\pi x} \end{eqnarray}
とできる.

$a=\frac{1}{2}$のとき$n=2$で十分
\begin{eqnarray} f_1(z):&=&\frac{e^{\frac{\pi i}{2}z^2}}{\cosh\pi z}\frac{e^{2\pi z}}{e^{2\pi z}-1}\\ \oint_{D}f_1(z)dz&\Rightarrow&\int_{-\infty}^{\infty}\frac{e^{\frac{\pi i}{2}x^2}}{\cosh\pi x}dx\\ &=&2\pi i\left(\frac{1}{2}\mathrm{Res}(f_1(z),0)+\frac{1}{2}\mathrm{Res}(f_1(z),2i)+\sum_{k=1}^{3}\mathrm{Res}\left(f_1(z),\frac{ki}{2}\right)\right)\\ &=&2\pi i\left(\frac{1}{4\pi}+\frac{1}{4\pi}-\frac{e^{3\pi i/8}}{2\pi}+\frac{i}{2\pi}-\frac{e^{3\pi i/8}}{2\pi}\right)\\ &=&2i\left(\frac{1+i}{2}-\cos\frac{3\pi}{8}-i\sin\frac{3\pi}{8}\right)\\ &=&\sqrt{2+\sqrt2}-1+i\left(1-\sqrt{2-\sqrt2}\right) \end{eqnarray}
$a=2$のとき$n=1$で十分
\begin{eqnarray} f_2(z):&=&\frac{e^{2\pi iz^2}}{\cosh\pi z}\frac{e^{4\pi z}}{e^{4\pi z}+1}\\ \oint_Df_2(z)dz&\Rightarrow&\int_{-\infty}^{\infty}\frac{e^{2\pi ix^2}}{\cosh\pi x}dx\\ &=&2\pi i\sum_{k=1}^{3}\mathrm{Res}\left(f_2(z),\frac{ki}{4}\right)\\ &=&2\pi i\left(-\frac{e^{7\pi i/8}}{2\sqrt2\pi}-\frac{1}{2\pi}-\frac{e^{7\pi i/8}}{2\sqrt2\pi}\right)\\ &=&\sqrt{2}\sin\frac{7\pi}{8}-i\left(1+\sqrt2\cos\frac{7\pi}{8}\right)\\ &=&\sqrt{1-\frac{1}{\sqrt2}}+i\left(\sqrt{1+\frac{1}{\sqrt2}}-1\right) \end{eqnarray}