Integrals and Miscellaneous 20

$f(z):=(1-z^3)^{-1/2}$ , $C$ is the contour which consists of two lines and two arcs. One of the lines is indented at $z=1$ by an arc of radius $1/R$. The radius of the other arc is $R$.

From the Cauthy's theorem we find $\oint_C f(z)dz=0$. The integral round the arcs tend to zero.\begin{eqnarray*}\oint f(z)dz &=&\int_0^1(1-x^3)^{-\frac{1}{2}}dx+\int_1^\infty\left\{(x-1)e^{-\pi i}\right\}^{-\frac{1}{2}}(1+x+x^2)^{-\frac{1}{2}}dx+i\int_\infty^0(1+ix^3)^{-\frac{1}{2}}dx\\&=&\int_0^1(1-x^3)^{-\frac{1}{2}}dx+i\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx\\&&\quad-i\int_0^\infty(1+ix^3)^{-\frac{1}{2}}dx\\&=&\int_0^1(1-x^3)^{-\frac{1}{2}}dx+i\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx\\&&\quad-ie^{-\frac{\pi}{6}i}\int_0^\infty(1+y^3)^{-\frac{1}{2}}dx\end{eqnarray*}Hence,\begin{eqnarray*}\int_0^1(1-x^3)^{-\frac{1}{2}}dx&=&\frac{1}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx=\frac{1}{2}\int_{-\infty}^0(1-x^3)^{-\frac{1}{2}}dx\\\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx&=&\frac{\sqrt{3}}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx\end{eqnarray*}Therefore,\begin{eqnarray*}\int_{-\infty}^1(1-x^3)^{-\frac{1}{2}}dx&=&\frac{3}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx\\\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx&=&\frac{\sqrt{3}}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx\end{eqnarray*}$$\therefore\quad\int_{-\infty}^1(1-x^3)^{-\frac{1}{2}}dx=\sqrt{3}\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx$$

$$\int_0^1\int_0^1\frac{dxdy}{1+ixy}=G-i\frac{\pi^2}{48}$$PROOF:
Substituting $y=u/x$ gives\begin{align}LHS &= \int_0^1\frac{dx}{x}\int_0^x\frac{du}{1+iu} \\&=\int_0^1\frac{\arctan x}{x}dx-\frac{i}{2}\int_0^1\frac{\ln(1+x^2)}{x}dx\end{align}We obtained $$\int_0^1\frac{\arctan x}{x}dx=G$$ in "2022/4/3" and \begin{align} \int_0^1\frac{\ln(1+x^2)}{x}dx &=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 x^{2n-1}dx\\&=-\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\\&=\frac{1}{4}\zeta(2)\end{align}

※そもそも最初から級数にすればよかった。$$\int_0^1\int_0^1\frac{dxdy}{1+ixy}=\sum_{n=1}^\infty\int_0^1\int_0^1(-ixy)^{n-1}dxdy$$

$$\int_0^1\int_0^1\frac{(-\ln xy)^a}{1-xy}(1-x)dxdy=\G(a+2)\left\{\zeta(a+2)-\frac{1}{a+1}\right\}$$PROOF.
Substituting $y=u/x$ yields\begin{align} LHS &= \int_0^1dx\int_0^xdu \frac{(-\ln u)^a}{1-u}\frac{1-x}{x} \\&= \int_0^1du\int_u^1dx\frac{(-\ln u)^a}{1-u}\frac{1-x}{x}\\ &= \int_0^1\frac{(-\ln u)^a}{1-u}(-\ln u+u-1)du \\&=\int_0^\infty \left(\frac{t^{a+1}}{e^t-1}-t^ze^{-t}\right)dt\quad(u=e^{-t})\\ &= \G(a+2)\left\{\zeta(a+2)-\frac{1}{a+1}\right\}\end{align}where we used$$\zeta(s)=\frac{1}{\G(s)}\int_0^\infty\frac{x^{s-1}}{e^{x}-1}dx$$ obtained here.

$a\to -1$ yields $$\int_0^1\int_0^1\frac{1}{(1-xy)\ln xy}(1-x)dxdy=-\g$$ from$$\displaystyle\lim_{s\to 1}\left(\zeta(s,a)-\frac{1}{s-1}\right)=-\psi(a)$$ we found here.

REFERENCE:
Chapman, R., A proof of Hadjicostas's conjecture.(2004)

$\Phi(z,s,a)$ denotes Lerch transcendent.$$\dd{\Phi}{s}(-1,-1,1)=\ln\frac{A^3}{2^{1/3}e^{1/4}}$$PROOF.
\begin{align}LHS &= \left.\frac{d}{ds}\left(1-\frac{1}{2^{s-1}}\right)\zeta(s)\right|_{s=-1}\\&= -3\zeta'(-1)+4\zeta(-1)\ln 2\\&=-\frac{\ln 2}{3}-\frac{1}{4}+3\ln A\end{align}

REFERENCE:
Guillera, J., Sondow, J. Double integrals and infinite products for some classical constants via analytic continuations of Lerch’s transcendent. Ramanujan J 16, 247–270 (2008).

$$I_n:=\int_0^1\left(1-\sqrt{x}\right)^ndx=\frac{2}{(n+1)(n+2)}$$where we substitute $y=x^2$ and use the beta function.

ex)$$\int_0^1\left(1-\sqrt{x}\right)^{19}dx=\frac{1}{210}$$

REFERENCE:

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$$y'=x+y^2$$

１階非線型微分方程式の例

By termwise integration we obtain$$(m-1)\int_0^1\sum_{n=1}^\infty x^{m^n-1}dx=1$$