Integrals and Miscellaneous 14

Suppose $a>b>c\ge0$ and $F(x;k)$ denotes incomplete elliptic integral of the first kind as Legendre form then$$\int_{a-(a-b)x^2}^a\frac{ds}{\sqrt{(a-s)(b-s)(c-s)}}=\frac{2}{\sqrt{a-c}}F\left(x;\sqrt{\frac{a-b}{a-c}}\right)$$PROOF.
$$F\left(x;k\right):=\int_0^x\frac{du}{\sqrt{(1-u^2)(1-k^2u^2)}} $$Putting $k^2=\frac{a-b}{a-c}$ where $a>b>c\ge 0$ and $u=\frac{t}{\sqrt{a-b}}$ to find$$F\left(x;\sqrt{\frac{a-b}{a-c}}\right)=\sqrt{a-c}\int_0^{\sqrt{a-b}\:x}\frac{dt}{\sqrt{(a-b-t^2)(a-c-t^2)}}$$By setting $a-t^2=s$ we obtain$$\int_{a-(a-b)x^2}^a\frac{ds}{\sqrt{(a-s)(b-s)(c-s)}}=\frac{2}{\sqrt{a-c}}F\left(x;\sqrt{\frac{a-b}{a-c}}\right)$$

Reference:
藤原松三郎『微分積分学 第1巻-数学解析第一編』

リーマンのP方程式と24個の特殊解

$$F\left[\begin{matrix}a,b\\a+b+\frac{1}{2}\end{matrix};z\right]=F\left[\begin{matrix}2a,2b\\a+b+\frac{1}{2}\end{matrix};\frac{1-\sqrt{1-z}}{2}\right]$$$$F\left[\begin{matrix}a,a+\frac{1}{2}\\c\end{matrix};z^2\right]=(1+z)^{-2a}F\left[\begin{matrix}2a,c-\frac{1}{2}\\2c-1\end{matrix};\frac{2z}{1+z}\right]$$

リーマンP方程式から超幾何関数の二次変換を導出

\begin{eqnarray}F\left[\begin{matrix}\frac{1}{2},1-u\\2u+\frac{1}{2}\end{matrix};\frac{1}{4}\right]&=&\frac{2\sqrt{2}\G(u+\frac{3}{4})\G(u+\frac{1}{4})}{3\G^2(u+\frac{1}{2})} \\ F\left[\begin{matrix}\frac{1}{2},1-u\\2u-\frac{1}{2}\end{matrix};\frac{1}{4}\right]&=&\frac{2\sqrt{2}\G(u-\frac{1}{4})\G(u+\frac{1}{4})}{3\G(u-\frac{1}{6})\G(u+\frac{1}{6})}\end{eqnarray}Proof here.

From @integralsbot:$$\int_0^1\frac{dx}{(1-x+x^2)\sqrt[3]{x(1-x)}}=\frac{4\pi}{3\sqrt{3}}$$PROOF.
We easily see that the integrand has a symmetry about $x=1/2$ then\begin{eqnarray*}LHS &=& 2\int_0^\frac{1}{2}\frac{dx}{(1-x+x^2)\sqrt[3]{x(1-x)}} \\&=& 2\int_0^\frac{1}{2}\frac{dy}{(y^2+\frac{3}{4})\sqrt[3]{\frac{1}{4}-y^2}}\quad(\because y=\frac{1}{2}-x)\\&=& 4^\frac{4}{3}\int_0^1\frac{du}{(u^2+3)\sqrt[3]{1-u^2}}\quad(\because u=2y) \\&=& 2^{-\frac{1}{3}}\int_0^1 t^{-\frac{1}{3}}(1-t)^{-\frac{1}{2}}\left(1-\frac{t}{4}\right)^{-1}dt\quad(\because t=1-u^2) \\&=& 2^{-\frac{1}{3}}\frac{\sqrt{\pi}\G(\frac{2}{3})}{\G(\frac{7}{6})} F\left[\begin{matrix}\frac{2}{3},1\\\frac{7}{6}\end{matrix};\frac{1}{4}\right] \\&=& \frac{2^\frac{2}{3}\sqrt{\pi}\G(\frac{2}{3})}{\sqrt{3}\G(\frac{7}{6})}F\left[\begin{matrix}\frac{1}{2},\frac{1}{6}\\\frac{7}{6}\end{matrix};\frac{1}{4}\right]\end{eqnarray*}We used Euler transformation in the last line. Recalling a formula$$F\left[\begin{matrix}1-a,\frac{1}{2}\\2a-\frac{1}{2}\end{matrix};\frac{1}{4}\right]=\frac{2\sqrt{2}}{3}\frac{\G(a-\frac{1}{4})\G(a+\frac{1}{4})}{\G(a-\frac{1}{6})\G(a+\frac{1}{6})}$$which was deduced here.

We set $a=5/6$ to find $$\int_0^1\frac{dx}{(1-x+x^2)\sqrt[3]{x(1-x)}}=\frac{4\pi}{3\sqrt{3}}$$

\begin{eqnarray*}\int_0^\frac{\pi}{2}x\cot x\ln(\sin x)dx &=& \int_0^1\frac{\arcsin y \ln y}{y}dy\quad(y=\sin x) \\&=& -\frac{1}{2}\int_0^1\frac{\ln^2y}{\sqrt{1-y^2}}dy \\&=& -\frac{1}{16}\int_0^1 u^{-\frac{1}{2}}(1-u)^{-\frac{1}{2}}\ln^2udu \quad(y^2=u)\\&=& -\frac{1}{16}\left.\dd{B}{x}(x,y)\right|_{x=y=1/2}\\&=& -\frac{\pi}{16}\left[\left\{\psi(1/2)-\psi(1)\right\}^2+\psi'(1/2)-\psi'(1)\right]\\&=&-\frac{\pi\ln^22}{4}-\frac{\pi^3}{48}\end{eqnarray*}

The derivatives of the beta function here. Special values of digamma function here and here.

From @infseriesbot.\begin{eqnarray*}\sqrt{e} &=& \sum_{n=0}^\infty\frac{1}{n!}\frac{1}{2^n}\\ &=& \sum_{n=0}^\infty\frac{1}{(2n)!!}\end{eqnarray*}

From @infseriesbot.\begin{eqnarray*}\sum_{n=0}^\infty\frac{(4n)!}{2^{8n}n!^4} &=& {}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{4},\frac{3}{4}\\1,1\end{matrix};1\right] \\&=& {}_2F_1\left[\begin{matrix}\frac{1}{8},\frac{3}{8}\\1\end{matrix};1\right]^2\quad(\because \mathrm{Clausen's\:formula})\\&=& \frac{\pi}{\G(\frac{7}{8})^2\G(\frac{5}{8})^2}\end{eqnarray*}

From @infseriesbot.\begin{eqnarray*}\sum_{n=0}^\infty\frac{(2n)!(3n)!}{108^{n}n!^5} &=& {}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{3},\frac{2}{3}\\1,1\end{matrix};1\right] \\&=& {}_2F_1\left[\begin{matrix}\frac{1}{6},\frac{1}{3}\\1\end{matrix};1\right]^2\quad(\because \mathrm{Clausen's\:formula})\\&=& \frac{\pi}{\G(\frac{5}{6})^2\G(\frac{2}{3})^2}\end{eqnarray*}

We proved Clausen's formula here.

More general case of 2023/1/9\begin{eqnarray*}\sum_{n=0}^\infty\frac{(4n)!}{n!^4}\left(\frac{x}{256}\right)^n &=& {}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{4},\frac{3}{4}\\1,1\end{matrix};x\right] \\&=& {}_2F_1\left[\begin{matrix}\frac{1}{8},\frac{3}{8}\\1\end{matrix};x\right]^2\quad(\because \mathrm{Clausen's\:formula})\\&=& \frac{2^\frac{5}{2}}{\pi^2\left(\sqrt{2}+\sqrt{1-\sqrt{1-x}}\right)}K\left(\frac{\sqrt{2}\left(1-\sqrt{1-x}\right)^\frac{1}{4}}{\left(\sqrt{2}+\sqrt{1-\sqrt{1-x}}\right)^\frac{1}{2}}\right)^2\end{eqnarray*}

Putting $x=32/81$ gives$$\sum_{n=0}^\infty\frac{(4n)!}{648^nn!^4}=\frac{3\pi}{4\G(\frac{3}{4})^4}$$

We proved Clausen's formula here.

Application of Abel-Plana summation formula variation:\begin{eqnarray*}\psi'\left(z+\frac{1}{2}\right)&=&\frac{1}{z}-4\int_0^\infty\frac{yzdy}{(y^2+z^2)^2(e^{2\pi y}+1)}\\\psi\left(z+\frac{1}{2}\right)&=&\ln z+2\int_0^\infty\frac{ydy}{(y^2+z^2)(e^{2\pi y}+1)}\\\ln\G\left(z+\frac{1}{2}\right)&=&z\ln z-z+\frac{\ln2\pi}{2}-2\int_0^\infty\frac{\arctan\frac{y}{z}}{e^{2\pi y}+1}dy\end{eqnarray*}Proof here:

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）

$$F\left[\begin{matrix}a,b\\c\end{matrix};z\right]=F\left[\begin{matrix}a+1,b\\c\end{matrix};z\right]-\frac{bz}{c}F\left[\begin{matrix}a+1,b+1\\c+1\end{matrix};z\right]$$$$F\left[\begin{matrix}a,b\\c\end{matrix};z\right]=F\left[\begin{matrix}a+1,b\\c+1\end{matrix};z\right]+\frac{b(a-c)z}{c(c+1)}F\left[\begin{matrix}a+1,b+1\\c+2\end{matrix};z\right]$$$$F\left[\begin{matrix}a,b\\c\end{matrix};z\right]=F\left[\begin{matrix}a,b\\c+1\end{matrix};z\right]+\frac{abz}{c(c+1)}F\left[\begin{matrix}a+1,b+1\\c+2\end{matrix};z\right]$$Proof here:

超幾何関数に関するガウスの隣接関係式

From @infseriesbot.$${}_3F_2\left[\begin{matrix}2a,2b,a+b\\2a+2b,a+b+\frac{1}{2}\end{matrix};z\right]={}_2F_1\left[\begin{matrix}a,b\\a+b+\frac{1}{2}\end{matrix};z\right]^2$$Proof here.

From @SrinivasR1729\begin{eqnarray*}\int_0^1&&x^s\arctan\frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}dx\\&&= \frac{\arctan(\sqrt{2}-1)}{s+1}+\frac{1}{4(s+1)}\int_0^1\frac{x^{s+1}dx}{\sqrt{1-x^2}}\quad(\because IBP)\\&&=\frac{\pi}{8(s+1)}+\frac{1}{8(s+1)}\int_0^1u^\frac{s}{2}(1-u)^{-\frac{1}{2}}du\quad(x^2=u)\\&&=\frac{\pi}{8(s+1)}+\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{8(s+1) \G(\frac{s+3}{2})}\end{eqnarray*}

Suppose $a<1$ ,$$I:=\int_0^1\frac{(\ln x)^n}{1-a+ax}dx=\frac{(-1)^{n-1}n!}{a}\Li_{n+1}\left(\frac{a}{a-1}\right)$$PROOF.
When $a<0$, from $0<\frac{a}{a-1}<1$ it follows that \begin{eqnarray*}I&=&\frac{1}{1-a}\sum_{k=0}^\infty\left(\frac{a}{a-1}\right)^k\int_0^1 x^k\ln^nxdx \\&=& \frac{1}{1-a}\sum_{k=0}^\infty\left(\frac{a}{a-1}\right)^k\frac{(-1)^nn!}{(k+1)^{n+1}}\\&=&\frac{(-1)^{n-1}n!}{a}\Li_{n+1}\left(\frac{a}{a-1}\right)\end{eqnarray*}where we find that it holds also for $a=0$.
When $0<a<1$, substituting $u=\frac{a}{a-1}x$ yields\begin{eqnarray*}I&=& -\frac{1}{a}\int_0^\frac{a}{a-1}\frac{\ln^n\frac{a-1}{a}u}{1-u}du \\&=& -\frac{n}{a}\int_0^\frac{a}{a-1}\frac{\ln(1-u)\ln^{n-1}\frac{a-1}{a}u}{u}du\quad(\because IBP)\\&=&-\frac{n(n-1)}{a}\int_0^\frac{a}{a-1}\frac{\Li_2(u)\ln^{n-2}\frac{a-1}{a}u}{u}du\quad(\because IBP)\\&\vdots&\\&=&\frac{(-1)^{n-1}n!}{a}\int_0^\frac{a}{a-1}\frac{\Li_n(u)}{u}du\quad(\because IBP)\\&=&\frac{(-1)^{n-1}n!}{a}\Li_{n+1}\left(\frac{a}{a-1}\right)\end{eqnarray*}Or$$\int_0^1\frac{(\ln x)^n}{1-ax}dx=(-1)^n\frac{\Li_{n+1}(a)}{a}$$

From P.J.Nahin(2015)$$\int_{-1}^1\frac{\cos x}{e^{1/x}+1}dx=\sin1$$PROOF.\begin{eqnarray*}LHS &=&\frac{1}{2}\int_{-1}^1\biggl[\biggl(\underbrace{\frac{\cos x}{e^{1/x}+1}+\frac{\cos (-x)}{e^{-1/x}+1}}_{even}\biggr)+\biggl(\underbrace{\frac{\cos x}{e^{1/x}+1}-\frac{\cos (-x)}{e^{-1/x}+1}}_{odd}\biggr)\biggr]dx \\&=&\int_0^1\cos x dx \\&=& \sin1\end{eqnarray*}

P.J.Nahin(2015)は面白い積分テクニックがたくさん載った人気書籍です！

Inside Interesting Integrals【楽天】

Inside Interesting Integrals【amazon】

$$\frac{22}{7}-\frac{1}{630}<\pi<\frac{22}{7}-\frac{1}{1260}$$PROOF.\begin{eqnarray*}\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx &=& \int_0^1\left(x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right)dx \\&=& \frac{22}{7}-\pi\end{eqnarray*}Since the integrand is positive, $\frac{22}{7}>\pi$ holds. Besides,$$\int_0^1\frac{x^4(1-x)^4}{2}dx<\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx<\int_0^1 x^4(1-x)^4dx$$We can easily evaluate the left and right integrals. Hence,$$\frac{22}{7}-\frac{1}{630}<\pi<\frac{22}{7}-\frac{1}{1260}$$

Reference:
Dalzell, D.P.(1944), "On 22/7", Journal of the London Mathematical Society, 19 (75 Part 3): 133–134

$n\in\NN$,\begin{eqnarray}I&:=&\int_0^1\frac{\ln^nx\Li_{n+1}(x)}{1+x}dx\\&=&\frac{(-1)^nn!}{2}\left[\zeta(2n+2)-\left(1-\frac{1}{2^n}\right)^2\zeta(n+1)^2\right]\tag{1}\end{eqnarray}

$$\int_0^1\frac{t\ln^ny}{1-ty}dy=(-1)^nn!\Li_{n+1}(t)\quad(t<1)$$$$\int_0^1dx\int_0^1dy\frac{\ln^nx\ln^ny}{1-xy}=n!^2\zeta(2n+2)$$$$\int_0^1\frac{\ln^nx}{1+x}dx=(-1)^nn!\left(1-\frac{1}{2^n}\right)\zeta(n+1)$$

\begin{eqnarray*}\int_0^1\frac{\ln x\Li_{2}(x)}{1+x}dx &=& -\frac{\pi^4}{480} \\\int_0^1\frac{\ln^2 x\Li_{3}(x)}{1+x}dx &=&\frac{\pi^6}{945}-\frac{9}{16}\zeta(3)^2\\\int_0^1\frac{\ln^3 x\Li_{4}(x)}{1+x}dx &=&-\frac{41\pi^8}{1209600}\\\int_0^1\frac{\ln^4 x\Li_{5}(x)}{1+x}dx &=& 12\zeta(10)-\frac{675}{64}\zeta(5)^2\\ \int_0^1\frac{\ln^5 x\Li_{6}(x)}{1+x}dx &=&-\frac{6823\pi^{12}}{3632428800}\end{eqnarray*}

Proof :

ポリログを含む積分1（重積分・級数展開）