Integrals and Miscellaneous 17

\begin{eqnarray}\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3} &=& \frac{11}{32}\zeta(5)-\frac{5}{8}\zeta(2)\zeta(3) \\ \sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2} &=& \frac{21}{32}\zeta(5)-\frac{3}{4}\zeta(2)\zeta(3) \\ \sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3} &=& \frac{19}{32}\zeta(5)-4\Li_5\left(\frac{1}{2}\right)-4\Li_4\left(\frac{1}{2}\right)\ln2\\&&-\frac{7}{4}\zeta(3)\ln^22+\frac{2}{3}\zeta(2)\ln^32+\frac{11}{8}\zeta(2)\zeta(3)-\frac{2}{15}\ln^52\end{eqnarray}$$\int_0^1\frac{\ln^2 x\ln(1-x)\ln(1+x)}{x}dx=-\frac{27}{16}\zeta(5)+\frac{3}{4}\zeta(2)\zeta(3)$$

PROOF:

調和数を含んだ級数(Euler-sum)とゼータ関数 part16

$$I:=\int_0^1\frac{\ln x}{\sqrt{x(1-x^2)}}=-\frac{\sqrt{\pi} \G^2(\frac{1}{4})}{4\sqrt{2}}$$PROOF.
Putting $u=x^2$ yields\begin{eqnarray*}I&=&\frac{1}{4}\int_0^1 u^{-\frac{3}{4}}(1-u)^{-\frac{1}{2}}\ln udu\\&=&\frac{1}{4}\left.\dd{B(p,q)}{p}\right|_{p=1/4,q=1/2} \\&=& \frac{1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right)\right]\\&=&-\frac{\sqrt{\pi} \G^2(\frac{1}{4})}{4\sqrt{2}}\end{eqnarray*}

$$\sum_{n=1}^\infty\frac{(\frac{1}{2})_n}{n!(4n+1)}\left(1+\frac{1}{5}+\cdots+\frac{1}{4n-3}\right)=\frac{\G^2(\frac{1}{4})\ln2}{8\sqrt{2\pi}}$$PROOF.
Recall the formula\begin{equation}\sum_{n=1}^\infty\frac{(c)_n(\nu)_n}{(\lambda)_nn!}[\psi(c+n)-\psi(c)] =\frac{\G(\lambda)\G(\lambda-\nu-c)}{\G(\lambda-c)\G(\lambda-\nu)}\left(\psi(\lambda-c)-\psi(\lambda-\nu-c)\right)\tag{1}\end{equation}proved here. We set $c=\frac{1}{4}$ , $\lambda=\frac{5}{4}$ and $\nu=\frac{1}{2}$ to get$$\sum_{n=1}^\infty\frac{(\frac{1}{2})_n}{n!(4n+1)}\left(1+\frac{1}{5}+\cdots+\frac{1}{4n-3}\right)=\frac{\G^2(\frac{1}{4})\ln2}{8\sqrt{2\pi}}$$

$$I:=\int_0^1\frac{\ln(1-x)}{\sqrt{x-x^3}}dx=\frac{\G^2(\frac{1}{4})}{4\sqrt{2\pi}}\left(\ln2-\pi\right)$$\begin{eqnarray*}\sum_{n=1}^\infty\frac{(\frac{1}{2})_n}{n!(4n+1)}\left(1+\frac{1}{5}+\cdots+\frac{1}{4n-3}\right) &=& \frac{\G^2(\frac{1}{4})}{8\sqrt{2\pi}}\ln2 \\ \sum_{n=1}^\infty\frac{(\frac{1}{2})_n}{n!(4n+1)}\left(\frac{1}{3}+\frac{1}{7}+\cdots+\frac{1}{4n-1}\right) &=& \frac{\G^2(\frac{1}{4})}{16\sqrt{2\pi}}\ln2\\\sum_{n=1}^\infty\frac{(\frac{1}{2})_n}{n!(4n+1)}\left(1+\frac{1}{3}+\cdots+\frac{1}{4n-1}\right) &=& \frac{3\G^2(\frac{1}{4})}{16\sqrt{2\pi}}\ln2\end{eqnarray*}PROOF:

logを含む難しい積分12(超幾何関数の微分の応用)

空間内の曲面：パラメータ表示、接平面、法ベクトル、ガウス写像、面積

曲面の第１基本形式と第２基本形式

曲面の曲がり具合 ～法曲率～

From @infseriesbot\begin{eqnarray*}\int_0^1\frac{dx}{1+x\ln x} &=& \sum_{n=1}^\infty (-1)^{n-1}\int_0^1 x^{n-1}\ln^{n-1}xdx \\&=& \sum_{n=1}^\infty (-1)^{n-1}\cdot(-1)^1\frac{n-1}{n}\int_0^1 x^{n-1}\ln^{n-2}xdx \\&=& \sum_{n=1}^\infty (-1)^{n-1}\cdot(-1)^2\frac{(n-1)(n-2)}{n^2}\int_0^1 x^{n-1}\ln^{n-3}xdx \\&\vdots&\\ &=& \sum_{n=1}^\infty\frac{(n-1)!}{n^n}\end{eqnarray*}

\begin{eqnarray*}\int_0^1\frac{dx}{(1+x\ln x)^2} &=& \int_0^1\left(\sum_{n=1}^\infty(-1)^{n-1}x^{n-1}\ln^{n-1}x\right)^2dx \\&=& \sum_{n=1}^\infty n\int_0^1(-1)^{n-1}x^{n-1}\ln^{n-1}xdx\\&=&\sum_{n=1}^\infty n\frac{(n-1)!}{n^n}\\&=&\sum_{n=1}^\infty \frac{n!}{n^n}\end{eqnarray*}where we used Cauchy product and the result above.

From @infseriesbot\begin{eqnarray*}\int_0^1\frac{t^{a-1}(1-t)^{b-1}}{1-zt^p(1-t)^q}dt &=& \int_0^1 t^{a-1}(1-t)^{b-1}\sum_{n=0}^\infty z^n t^{pn}(1-t)^{qn}dt \\&=& \sum_{n=0}^\infty z^n \int_0^1 t^{pn+a-1}(1-t)^{q+b-1}dt \\&=& \sum_{n=0}^\infty B\Bigl(pn+a,qn+b\Bigr)z^n \end{eqnarray*}

$$\sum_{n=1}^\infty \frac{\zeta(2n)}{n}z^{2n}=\ln\frac{\pi z}{\sin\pi z}$$PROOF.
Recall the formula$$\sum_{n=1}^\infty \zeta(2n)z^{2n}=\frac{1}{2}(1-\pi z\cot\pi z)$$which we obtained here. After dividing it by $z$, we integrate from $\epsilon$ to $z$ to find \begin{eqnarray*}\sum_{n=1}^\infty \frac{\zeta(2n)}{n}z^{2n} &=& \int_\epsilon^z\left(\frac{1}{z}-\pi\cot\pi z\right)dz \\&=& \ln\frac{\pi z}{\sin\pi z}+\ln\frac{\sin\pi \epsilon}{\pi\epsilon} \\&=& \ln\frac{\pi z}{\sin\pi z}\quad( \mathrm{as}\;\epsilon\to0)\end{eqnarray*}

$$\sum_{n=1}^\infty \frac{\zeta(4n)}{n}z^{4n}=\ln\frac{\pi^2 z^2}{\sin\pi z\sinh\pi z}$$PROOF.
\begin{eqnarray*}\sum_{n=1}^\infty \frac{\zeta(4n)}{n}z^{4n} &=& \sum_{n=1}^\infty \frac{\zeta(2n)}{n}z^{2n} + \sum_{n=1}^\infty \frac{\zeta(2n)}{n}(iz)^{2n} \\&=& \ln\frac{\pi z}{\sin\pi z}+ \ln\frac{i\pi z}{\sin i\pi z}\\&=&\ln\frac{\pi z}{\sin\pi z}+ \ln\frac{\pi z}{\sinh \pi z} \\&=&\ln\frac{\pi^2 z^2}{\sin\pi z\sinh\pi z}\end{eqnarray*}

Suppose $n\ge 2$ , $n\in \NN$ then$$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{n}{2^{n-1}}$$PROOF.
We define $\a:=e^{\frac{2\pi}{n}i}$. From $\a$ is a primitive $n$-th root of 1, it follows that$$z^n-1=(z-1)\prod_{k=1}^{n-1}(z-\a^k)$$Differentiating by $z$ yields$$nz^{n-1}=\prod_{k=1}^{n-1}(z-\a^k)+(z-1)\frac{d}{dz}\prod_{k=1}^{n-1}(z-\a^k)$$Substituting $z=1$, we have \begin{eqnarray*}n &=& \prod_{k=1}^{n-1}(1-\a^k) \\&=& \prod_{k=1}^{n-1}\left(1-e^{\frac{2\pi k}{n}i}\right) \\&=& \prod_{k=1}^{n-1}e^{\frac{\pi k}{n}i}\left(e^{-\frac{\pi k}{n}i}-e^{\frac{\pi k}{n}i}\right) \\&=& (-2i)^{n-1}e^{i\frac{\pi}{n}(1+2+\cdots +(n-1))}\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}\\&=& 2^{n-1}\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}\end{eqnarray*}Hence,$$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{n}{2^{n-1}}$$

Reference:
Rainville, E.D. (1960) Special Functions. The Macmillan Company, New York.

$\mathfrak{R}s>1$ , $m\le n\in\NN$$$\zeta\left(1-s,\frac{m}{n}\right)=\frac{2\G(s)}{(2n\pi)^s}\sum_{l=1}^n\cos\left(\frac{\pi s}{2}-\frac{2\pi ml}{n}\right)\zeta\left(s,\frac{l}{n}\right)$$
Proof here:

【ζ4】フルヴィッツゼータ関数のフーリエ展開・複素積分・Critical Stripへの拡張(ゼータ関数の基礎4)

From @infseriesbot$$I:=\int_0^1 \left(\frac{1-x}{x}\right)^{ax}\sin\pi axdx =\frac{\pi a}{2}e^{-a}$$PROOF(not so rigorous).$$f(z):=\left(\frac{1-z}{z}\right)^{az}e^{-i\pi az}=e^{az\{\ln(1-z)-\ln z-i\pi\}}$$We also define the contour $C$ encircling the two branch points $0,1$ counter-clockwise, composed of two small arcs and two lines along the real axis. And we suppose$$-\pi\le\arg z<\pi\;,\; 0\le \arg(1-z)<2\pi$$ The branch cut exists just between $0$ and $1$, for we find f(z) is continuous when crossing the real axis at $-\epsilon$ and $1+\epsilon$.

\begin{eqnarray*}\oint_Cf(z)dz &=& \int_1^0 e^{ax\{\ln(1-x)-\ln x+i\pi\}}dx+\int_0^1 e^{ax\{\ln(1-x)-\ln x-i\pi\}}dx\\&=& -2i\int_0^1 \left(\frac{1-x}{x}\right)^{ax}\sin\pi axdx \\&=& -2iI\end{eqnarray*}Then we calculate the residue:\begin{eqnarray*}\oint_Cf(z)dz &=& -2\pi i\mathrm{Res}_{z=\infty}f(z) \\&=& 2\pi i\mathrm{Res}_{z=0}\frac{1}{z^2}f\left(\frac{1}{z}\right)\\&=& 2\pi i\mathrm{Res}_{z=0}\left[\frac{1}{z^2}e^{\frac{a}{z}\ln(1-z)}\right]\end{eqnarray*}Expanding RHS\begin{eqnarray*}\frac{1}{z^2}e^{\frac{a}{z}\ln(1-z)} &=& \frac{1}{z^2}\sum_{n=0}^\infty\frac{a^n\ln^n(1-z)}{n!z^n}\\&=&\frac{1}{z^2}\left[1+\frac{a}{z}\ln(1-z)+\frac{a^2}{2!z^2}\ln^2(1-z)+\frac{a^3}{3!z^3}\ln^3(1-z)+\cdots\right]\\&=& \frac{1}{z^2}\Biggl[1-\frac{a}{z}\left(z+\frac{z^2}{2}+\frac{z^3}{3}\cdots\right)\\&&+\frac{a^2}{2!z^2}\left(z+\frac{z^2}{2}+\frac{z^3}{3}\cdots\right)^2\\&&-\frac{a^3}{3!z^3}\left(z+\frac{z^2}{2}+\frac{z^3}{3}\cdots\right)^3+\cdots\Biggr]\end{eqnarray*}Hence,$$a_{-1}=-\frac{a}{2}e^{-a}$$$$\therefore\quad\oint_Cf(z)dz=-2\pi i\frac{a}{2}e^{-a}$$Therefore,$$I=\frac{\pi a}{2}e^{-a}$$

REFERENCE:
「複素線積分」Wikipedia(2023/4/7)
Residue at ∞ ,The LibreTexts libraries(2023/4/7)

Solve $$y''(z)+\frac{1}{z^3}y(z)=0$$SOLUTION.
The equation has an irregular singular point at $z=0$ then we cannot use Frobenius method there. Substituting $z=1/w$ yields$$\frac{d^2u}{dw^2}+\frac{2}{w}\frac{du}{dw}+\frac{1}{w}u=0\;,\; u(w):=y(z)$$This equation has a regular singular point at $w=0$ then the Frobenius series exists.

However, instead of Frobenius method, we substitute $4w=\zeta^2$ and $u=\phi(\zeta)/\zeta$ to find$$\zeta^2\phi''+\zeta\phi'+(\zeta^2-1)\phi=0$$which is the Bessel equation of $\nu=1$. Hence,$$\phi(\zeta)=2A J_1(\zeta)+2B Y_1(\zeta)$$where $A,B$ are arbitrary constants.$$\therefore\quad y(z)=\sqrt{z}\left[AJ_1\left(\frac{2}{\sqrt{z}}\right)+BY_1\left(\frac{2}{\sqrt{z}}\right)\right]$$

$$\lim_{x\to 0}\frac{F\left[\begin{matrix}a,b\\a+b\end{matrix};1-x\right]}{\ln\frac{1}{x}}=\frac{1}{B(a,b)}$$PROOF.
Recall the linear combination of 2F1, namely\begin{eqnarray}F\left[\begin{matrix}a,b\\c\end{matrix};z\right] &=& \frac{\G(c)\G(c-a-b)}{\G(c-a)\G(c-b)}F\left[\begin{matrix}a,b\\a+b-c+1\end{matrix};1-z\right]\\&&\quad+\frac{\G(c)\G(a+b-c)}{\G(a)\G(b)}(1-z)^{c-a-b}F\left[\begin{matrix}c-a,c-b\\c-a-b+1\end{matrix};1-z\right]\end{eqnarray}which we obtained here. Substituting $c=a+b+\epsilon$ we find\begin{eqnarray*}F\left[\begin{matrix}a,b\\a+b+\epsilon\end{matrix};z\right] &=& \frac{\G(a+b+\epsilon)\G(1+\epsilon)\G(1-\epsilon)}{\epsilon\G(a+\epsilon)\G(b+\epsilon)\G(a)\G(b)}\\&&\times\sum_{n=0}^\infty\frac{(1-z)^n}{n!}\left(\frac{\G(a+n)\G(b+n)}{\G(1+n-\epsilon)}-\frac{\G(a+n+\epsilon)\G(b+n+\epsilon)}{\G(1+n+\epsilon)}(1-z)^\epsilon\right)\\&=&\frac{\G(a+b+\epsilon)\G(1+\epsilon)\G(1-\epsilon)}{\G(a+\epsilon)\G(b+\epsilon)\G(a)\G(b)}\\&&\times\sum_{n=0}^\infty\frac{(1-z)^n}{n!}\left(\frac{f(\epsilon)-f(0)}{\epsilon}-\frac{g(\epsilon)-g(0)}{\epsilon}\right)\end{eqnarray*}where we define$$f(\epsilon)=\frac{\G(a+n)\G(b+n)}{\G(1+n-\epsilon)}\;,\;g(\epsilon)=\frac{\G(a+n+\epsilon)\G(b+n+\epsilon)}{\G(1+n+\epsilon)}(1-z)^\epsilon$$Taking the limit of $\epsilon\to0$ yields$$F\left[\begin{matrix}a,b\\a+b\end{matrix};z\right]=\frac{1}{B(a,b)}\sum_{n=0}^\infty\frac{(a)_n(b)_n}{n!^2}(1-z)^n\left[2\psi(n+1)-\psi(a+n)-\psi(b+n)-\ln(1-z)\right]$$Hence, when $\delta<<1$, $$F\left[\begin{matrix}a,b\\a+b\end{matrix};1-\delta\right]=\frac{1}{B(a,b)}\left[\ln\frac{1}{\delta}+O(1)\right]$$Therefore,$$\lim_{x\to 0}\frac{F\left[\begin{matrix}a,b\\a+b\end{matrix};1-x\right]}{\ln\frac{1}{x}}=\frac{1}{B(a,b)}$$Application here.

REFERENCE:
E.G.C.Poole,Introduction to the Theory of Linear Differential Equations(1936), chapter 6 Ex12.
Bateman.H, Eldélyi.A, Higher Transcendental Functions vol.1 (1953), eq(2.10)

From @MAGNA81407795Evaluate that $$I(u,s,n):=\int_0^\infty\frac{x^{s-1}dx}{\sqrt{x^2+u^2}(x+\sqrt{x^2+u^2})^n}$$SOLUTION.\begin{eqnarray*}I(u,s,n) &=& u^{s-n-1}\int_0^\infty \frac{y^{s-1}}{\sqrt{y^2+1}(y+\sqrt{y^2+1})^n}dy\quad(x=uy) \\&=& \frac{u^{s-n-1}}{2^{s-1}}\int_1^\infty z^{-n-s}(z^2-1)^{s-1}dx\quad\left(z=y+\sqrt{y^2+1}\right) \\&=& \frac{u^{s-n-1}}{2^s}\int_0^1 t^{\frac{n-s+1}{2}-1}(1-t)^{s-1}dt\quad(t=z^{-2}) \\&=& \frac{u^{s-n-1}\G(s)\G(\frac{n-s+1}{2})}{2^s\G(\frac{n+s+1}{2})} \end{eqnarray*}

From @fhdalalwi1412
$$I:=\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}dx=\frac{\pi}{2}\ln(1+\sqrt{2})$$PROOF.
$$I(s):=\int_0^1\frac{\arctan sx}{x\sqrt{1-x^2}}dx$$then we easily see $I(0)=0$ , $I(1)=I$.\begin{eqnarray*}I'(s) &=& \int_0^1\frac{dx}{(1+s^2x^2)\sqrt{1-x^2}} \\&=& \int_0^\frac{\pi}{2}\frac{d\t}{(1+s^2\sin^2\t)}\quad(x:=\sin\t) \\&=& \int_0^\infty\frac{du}{1+s^2+u^2}\quad(u:=\cot\t) \\&=& \frac{\pi}{2\sqrt{1+s^2}}\end{eqnarray*}Hence,$$I(s)=\frac{\pi}{2}\mathrm{arsinh}s = \frac{\pi}{2}\ln(s+\sqrt{s^2+1})$$$$\therefore\quad I=\frac{\pi}{2}\ln(1+\sqrt{2})$$

From @EylemGercek$$I:=\int_0^a\frac{dx}{1+b^{\frac{x}{a}-1}}$$SOLUTION.\begin{eqnarray*} I &=& a\int_0^1\frac{dx}{1+b^{x-1}} \\&=& \frac{a}{\ln b}\int_\frac{1}{b}^1\frac{dy}{y(1+y)}\quad(b^{x-1}=y)\\ &=& \frac{a}{\ln b}\ln\frac{1+b}{2}\end{eqnarray*}