Integrals and Miscellaneous 2

\begin{eqnarray*}I&&=\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx\\&&=2\int_0^1\frac{\arctan y}{y}dy\quad(y=\tan\frac{x}{2})\\&&=2\int_0^1\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}y^{2n}dy\\&&=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\&&=2\beta(2)\end{eqnarray*}

分母に $\sin$ があるとWeierstrass置換で $1+y^2$ が消えることが分かった。

$$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}=?$$\begin{eqnarray*}\int_0^1\frac{dx}{1+x^3}&=&\int_0^1\sum^\infty_{n=0}(-x^3)^{n}dx\\&=&\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}\end{eqnarray*}Therefore,$$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}=\int_0^1\frac{dx}{1+x^3}$$This integral is evaluated below:\begin{eqnarray*}\int_0^1\frac{dx}{1+x^3}&=&\frac{1}{3}\int_0^1\left(\frac{1}{1+x}-\frac{x-2}{1-x+x^2}\right)dx\\&=&\frac{1}{3}\int_0^1\left(\frac{1}{1+x}-\frac{1}{2}\frac{2x-1}{1-x+x^2}+\frac{3}{2}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}\right)dx\\&=&\frac{1}{3}\left(\frac{\pi}{\sqrt{3}}+\log 2\right)\end{eqnarray*}Hence,\begin{equation}\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}=\frac{1}{3}\left(\frac{\pi}{\sqrt{3}}+\log 2\right)\end{equation}

Another solution:

$$\int_0^1\frac{\cos\log x}{1+x^2}dx=\frac{\pi}{4\cosh\frac{\pi}{2}}$$

証明(proof)はココ！

2022年3月30日 複素積分演習（cos(log x)/(1+x^2)）

\begin{eqnarray*}I&&=\int_0^\infty\frac{\arctan x^2}{x^2}dx\\&&=\left[-\frac{\arctan x^2}{x}\right]_0^\infty+\int_0^\infty\frac{2dx}{1+x^4}\\&&=\int_0^\infty\frac{2dx}{1+x^4}\\&&=\frac{1}{2\sqrt{2}}\int^\infty_0\Bigl(\frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1}+\frac{\sqrt{2}}{x^2+\sqrt{2}x+1}\\&&\quad-\frac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1}+\frac{\sqrt{2}}{x^2-\sqrt{2}x+1}\Bigr)dx\\&&=\frac{1}{2\sqrt{2}}\left[\log\left|\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right|\right]^\infty_0\\&&\quad+\frac{1}{2}\int^\infty_0\frac{dx}{(x+\frac{1}{\sqrt{2}})^2+\frac{1}{2}}+\frac{1}{2}\int^\infty_0\frac{dx}{(x-\frac{1}{\sqrt{2}})^2+\frac{1}{2}}\\&&=\frac{1}{2}\int^\infty_\frac{1}{\sqrt{2}}\frac{dx}{x^2+\frac{1}{2}}+\frac{1}{2}\int^\infty_{-\frac{1}{\sqrt{2}}}\frac{dx}{x^2+\frac{1}{2}}\\&&=\frac{1}{\sqrt{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}-\frac{\pi}{4}+\frac{\pi}{4}\right)\\&&=\frac{\pi}{\sqrt{2}}\end{eqnarray*}

\begin{eqnarray*}I&&=\int_0^1\frac{dx}{x^4+1}\\&=&\int_0^1\frac{dx}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}\\&=&\int_0^1\left(\frac{-\frac{1}{2\sqrt{2}}x+\frac{1}{2}}{x^2-\sqrt{2}x+1}+\frac{\frac{1}{2\sqrt{2}}x+\frac{1}{2}}{x^2+\sqrt{2}x+1}\right)dx\\&=&\frac{1}{4\sqrt{2}}\int^1_0\Bigl(\frac{2x+\sqrt{2}}{x^2+\sqrt{2}x+1}+\frac{\sqrt{2}}{x^2+\sqrt{2}x+1}\\&&-\frac{2x-\sqrt{2}}{x^2-\sqrt{2}x+1}+\frac{\sqrt{2}}{x^2-\sqrt{2}x+1}\Bigr)dx\\&=&\frac{1}{4\sqrt{2}}\left[\log\left|\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right|\right]^1_0\\&&+\frac{1}{4}\int^1_0\left(\frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1}\right)dx\\&=&\frac{\log(\sqrt{2}+1)}{2\sqrt{2}}+\frac{1}{4}\int^1_0\frac{dx}{(x+\frac{1}{\sqrt{2}})^2+\frac{1}{2}}+\frac{1}{4}\int^1_0\frac{dx}{(x-\frac{1}{\sqrt{2}})^2+\frac{1}{2}}\end{eqnarray*}Substitute $x+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\tan\theta$ in the second term, $x-\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\tan\theta$ in the third term.We have\begin{eqnarray*}\int_0^1\frac{dx}{x^4+1}&=&\frac{\log(\sqrt{2}+1)}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\int^{\theta_1}_\frac{\pi}{4}d\theta+\frac{1}{2\sqrt{2}}\int^{\theta_2}_{-\frac{\pi}{4}}d\theta\\&=&\frac{\log(\sqrt{2}+1)}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}(\theta_1+\theta_2)\end{eqnarray*} $\tan\theta_1=\sqrt{2}+1$ , $\tan\theta_2=\sqrt{2}-1$ and we can see$$\tan(\theta_1+\theta_2)\longrightarrow\infty$$So we get $\theta_1+\theta_2=\frac{\pi}{2}$ then\begin{equation}\int_0^1\frac{dx}{x^4+1}=\frac{\log(\sqrt{2}+1)}{2\sqrt{2}}+\frac{\pi}{4\sqrt{2}}\end{equation}

$$I=\int_0^\infty\frac{xdx}{(x^2+4\pi^2)^2(e^x-1)}$$Let $x=2\pi y$, then$$I=\frac{1}{4\pi^2}\int_0^\infty\frac{ydy}{(1+y^2)^2(e^{2\pi y}-1)}$$

Binet's second formula:$$\log\G(z)=\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log2\pi+2\int_0^\infty\frac{\arctan\frac{y}{z}}{e^{2\pi y}-1}dy$$The second derivative of this is$$\psi'(z)=\frac{1}{z}+\frac{1}{2z^2}+4\int_0^\infty\frac{yzdy}{(z^2+y^2)^2(e^{2\pi y}-1)}$$Substitute $z=1$, we obtain$$\psi'(1)=\frac{3}{2}+16\pi^2I$$Use $\psi'(1)=\frac{\pi^2}{6}$,$$I=\frac{1}{96}-\frac{3}{32\pi^2}$$

Binet's second formula の導出はこちら

2022年3月6日 【γ19】対数ガンマ関数におけるビネの第2公式の導出（アベル・プラナの和公式,ポリガンマ関数）（ガンマ関数の基礎19）

$$I=\int_0^\infty\frac{\cos x}{(1+x^2)^3}dx$$Consider$$\oint_C\frac{e^{iz}}{(1+z^2)^3}dz$$$C$:Semicircle of large radius $R$ which includes a pole $z=i$. $\G$ is the big arc of $C$

$$\oint_C=\left.2\pi i\frac{1}{2!}\frac{d^2}{dz^2}\frac{e^{iz}}{(z+i)^3}\right|_{z=i}=\frac{7\pi}{8e}$$

\begin{eqnarray*}\left|\int_\G\right|&&\le\int_0^\pi\left|\frac{e^{iRe^{i\t}}}{(1+R^2e^{2i\t})^3}R\right|d\t\\&&=R\int_0^\pi\frac{e^{-R\sin\t}}{|1+R^2e^{2i\t}|^3}d\t\\&&\le R\int_0^\pi\frac{e^{-R\sin\t}}{\Bigl||1|-|R^2e^{2i\t}|\Bigr|^3}d\t\\&&\le\frac{R}{(R^2-1)^3}\int_0^\pi e^{-R\sin\t} d\t\\&&=\frac{2R}{(R^2-1)^3}\int_0^\frac{\pi}{2} e^{-R\frac{2}{\pi}\t} d\t\longrightarrow 0\end{eqnarray*}Hence,$$I=\frac{1}{2}\frac{7\pi}{8e}=\frac{7\pi}{16e}$$

\begin{eqnarray*}I&&=\int_0^\infty\frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}}dx\\&&=\frac{1}{\pi}\int_0^\infty\frac{y^{-\frac{4}{5}}}{1+y^2}dy\\&&=\frac{1}{\pi}\int_0^\frac{\pi}{2}(\tan\t)^{-\frac{4}{5}}d\t\\&&=\frac{1}{2\pi}B\left(\frac{1}{10},\frac{9}{10}\right)\\&&=\frac{1}{2\pi}\G\left(\frac{1}{10}\right)\G\left(\frac{9}{10}\right)\\&&=\frac{1}{2\sin\frac{\pi}{10}}\\&&=\frac{\sqrt{5}+1}{2}\end{eqnarray*}

\begin{eqnarray*}I&&=\int_0^\infty\arctan e^{-x} dx\\&&=\int_0^1\frac{\arctan y}{y}dy\quad(e^{-x}=y)\\&&=\int_0^1\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}y^{2n}dy\\&&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\&&=\beta(2)\end{eqnarray*}

\begin{eqnarray*}I&&=\int_0^\infty\frac{x}{(1+x^2)^2}\frac{dx}{\tanh\frac{\pi x}{2}}\\&&=\int_0^\infty\frac{x}{(1+x^2)^2}dx+2\int_0^\infty\frac{xdx}{(1+x^2)^2(e^{\pi x}-1)}\\&&=\frac{1}{2}+\int_0^\infty\frac{\frac{1}{2}ydy}{(y^2+\frac{1}{4})^2(e^{2\pi y}-1)}\quad(x=2y)\end{eqnarray*}

Binet's second formula:$$\log\G(z)=\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log2\pi+2\int_0^\infty\frac{\arctan\frac{y}{z}}{e^{2\pi y}-1}dy$$The second derivative of this is$$\psi'(z)=\frac{1}{z}+\frac{1}{2z^2}+4\int_0^\infty\frac{yzdy}{(z^2+y^2)^2(e^{2\pi y}-1)}$$Substitute $z=1/2$ , we obtain$$\int_0^\infty\frac{\frac{1}{2}ydy}{(y^2+\frac{1}{4})^2(e^{2\pi y}-1)}=\frac{1}{4}\left(\psi'\left(\frac{1}{2}\right)-2-2\right)=\frac{\pi^2}{8}-1$$Hence,$$I=\frac{\pi^2}{8}-\frac{1}{2}$$

Binet's second formula の導出はこちら

2022年3月6日 【γ19】対数ガンマ関数におけるビネの第2公式の導出（アベル・プラナの和公式,ポリガンマ関数）（ガンマ関数の基礎19）

\begin{eqnarray*}I&&=\int_0^\frac{\pi}{2}\sqrt{\tan x}dx\\&&=\int_0^\frac{\pi}{2}\cos^{-\frac{1}{2}} x\sin^{\frac{1}{2}}xdx\\&&=\frac{1}{2}B\left(\frac{1}{4},\frac{3}{4}\right)=\frac{\pi}{\sqrt{2}}\end{eqnarray*}

\begin{eqnarray*}J&&=\int_0^\frac{\pi}{2}\frac{dx}{\sqrt{\tan x}}\\&&=\frac{1}{2}B\left(\frac{1}{4},\frac{3}{4}\right)=\frac{\pi}{\sqrt{2}}\end{eqnarray*}

\begin{eqnarray*}K&&=\int_0^\frac{\pi}{2}\sqrt[3]{\tan x}dx\\&&=\int_0^\frac{\pi}{2}\cos^{-\frac{1}{3}} x\sin^{\frac{1}{3}}xdx\\&&=\frac{1}{2}B\left(\frac{1}{3},\frac{2}{3}\right)=\frac{\pi}{\sqrt{3}}\end{eqnarray*}

\begin{eqnarray*}I&&=\int_0^\infty\frac{xdx}{(1+4x^2)^2\sinh\pi x}\\&&=\frac{4}{64}\int_0^\infty\frac{\frac{1}{4}ydy}{(y^2+\frac{1}{16})^2\sinh2\pi y}\end{eqnarray*}Abel-Plana formula:$$\sum_{k=0}^\infty f(k)=\frac{f(0)}{2}+\int_0^\infty f(\a)d\a+i\int_0^\infty\frac{f(iy)-f(-iy)}{e^{2\pi y}-1}dy$$Let $f(\a)=\frac{1}{(z+\a)^2}$ in Abel-Plana formula to obtain\begin{equation}\psi'(z)=\frac{1}{z}+\frac{1}{2z^2}+4\int_0^\infty\frac{yzdz}{(y^2+z^2)^2(e^{2\pi y}-1)}\tag{1}\end{equation}Abel-Plana formula's variation:$$\sum_{k=0}^{\infty}f\left(k+\frac{1}{2}\right)=\int_0^\infty f(\a)d\a+i\int_0^\infty\frac{f(-iy)-f(iy)}{e^{2\pi y}+1}dy$$Let $f(\a)=\frac{1}{(z+\a)^2}$ to obtain\begin{equation}\psi'\left(z+\frac{1}{2}\right)=\frac{1}{z}-4\int_0^\infty\frac{yzdz}{(y^2+z^2)^2(e^{2\pi y}+1)}\tag{2}\end{equation}By (1)-(2), we get$$\psi'(z)-\psi'\left(z+\frac{1}{2}\right)=\frac{1}{2z^2}+4\int_0^\infty\frac{yzdz}{(y^2+z^2)^2\sinh2\pi y}$$Let $z=1/4$,\begin{eqnarray*}I&&=\frac{1}{64}\left[\psi'\left(\frac{1}{4}\right)-\psi'\left(\frac{3}{4}\right)-8\right]\\&&=\frac{G}{4}-\frac{1}{8}\end{eqnarray*}

Abel-Plana formula:

2022年3月5日 アベル・プラナの和公式　複素積分を応用して

And its variation:

2022年4月5日 アベル・プラナの和公式のバリエーション（複素積分・cotでなくtanで導出する）

About polygamma function here:

2022年2月19日 【γ10】ポリガンマ関数の値、極、級数表示、ゼータ関数との関係（ガンマ関数の基礎シリーズ10）

\begin{eqnarray*}I&&=\int_0^\frac{\pi}{2}\frac{\cos x}{\sin^2x}\left(1-\frac{x}{\tan x}\right)dx\quad(t=\tan\frac{x}{2})\\&&=\frac{1}{2}\int_0^1\left(\frac{1}{t^2}-1-\frac{(1-t^2)^2}{t^3}\arctan t\right)dt\\&&=-1+\frac{1}{2\epsilon}-\frac{1}{2}\int_0^1\left(\frac{1}{t^3}-\frac{2}{t}+t\right)\arctan tdt\end{eqnarray*}Evaluating the integral termwise below:\begin{eqnarray*}\int_0^1\frac{\arctan t}{t^3}dt&&=\left[-\frac{\arctan t}{2t^2}\right]_0^1+\frac{1}{2}\int_0^1\frac{dt}{t^2(1+t^2)}\\&&=-\frac{\pi}{4}-\frac{1}{2}+\frac{1}{\epsilon}\end{eqnarray*}\begin{eqnarray*}\int_0^1\frac{\arctan t}{t}dt&&=\int_0^1\frac{1}{t}\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}t^{2n+1}dt\\&&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\&&=G\end{eqnarray*}\begin{eqnarray*}\int_0^1 t\arctan tdt&&=\left[\frac{t^2}{2}\arctan t\right]_0^1-\frac{1}{2}\int_0^1\frac{t^2}{1+t^2}dt\\&&=\frac{\pi}{4}-\frac{1}{2}\end{eqnarray*}Therefore,$$I=G-\frac{1}{2}$$

\begin{eqnarray*}I&&=\int_0^\infty\frac{dx}{\cosh^sx}\\&&=2^s\int_0^\infty\frac{e^{-sx}}{(1+e^{-2x})^s}\\&&=2^s\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^\infty e^{-(s+2n)x}(s)_ndx\quad(\mathrm{binomial \;theorem})\\&&=2^{s-1}\sum_{n=0}^\infty\frac{(-1)^n}{n!}(s)_n\frac{1}{n+\frac{s}{2}}\\&&=\frac{2^s}{s}F\left(s,\frac{s}{2},\frac{s}{2}+1;-1\right)\\&&=\frac{2^s}{s}\frac{\G^2(1+\frac{s}{2})}{\G(1+s)}\quad(\mathrm{Kummer's\;theorem})\\&&=2^{s-2}\frac{\G^2(\frac{s}{2})}{\G(s)}\\&&=\frac{\sqrt{\pi}\G(\frac{s}{2})}{2\G(\frac{s+1}{2})}\quad(\mathrm{duplication\;formula})\end{eqnarray*}