Continued fractions II

Continued fractions for confluent hypergeometric functions

Today we show the proof of this nice continued fraction proposed by @nntaleb $$\frac{x}{e^x -1}= 1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3x}{6 + \cdots}}}}}$$ The proof will relly on the continued fractions theorem for confluent hypergeometric functions

Proof

Note that $$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{(1)_{n}}{(1)_{n}}\frac{x^n}{n!} = {}_{1}F_{1} (1;1;x)$$ where $$(x)_{n} = x(x+1)(x+2)\cdots(x+n-1)$$ is the Pochhhammer polynomial or rising factorial of degree $n$ and ${}_{1}F_{1}(a;b;z)$ is the confluent hypergeometric function, also denoted by $M(a;b;z)$. There is a continued fractions theorem for confluent hypergeometric functions:

Thorem Consider the function:

$${}_{1}F_{1}(b;c;z)= \sum_{n=0}^{\infty} \frac{(b)_{n}}{(c)_{n}} \frac{z^n}{n!}$$ Let $\left\{a_{n}\right\}$ be a sequence of complex numbers defined by $$a_{2n} = \frac{b+n}{(c+2n-1)(c+2n)} \quad n=1,2,3,...$$ $$a_{2n+1} = -\frac{c-b+n}{(c+2n)(c+2n+1)} \quad n=0,1,2,3,...$$ Then $$1+K_{n=1}^{\infty}\left(\frac{a_{n}z}{1}\right) = \frac{{}_{1}F_{1} (b;c;z)}{{}_{1}F_{1} (b+1;c+1;z)} \quad \forall z\in \mathbb{C}$$

Corollary If we let $b\to 0$ and making the transformation $c+1 \mapsto c$

$$1-\cfrac{\frac{1}{c}z}{1+\cfrac{\frac{1}{c(c+1)}z}{1-\cfrac{\frac{c}{(c+1)(c+2)}z}{1+\cfrac{\frac{2}{(c+2)(c+3)}z}{1-\cfrac{\frac{c+1}{(c+3)(c+4)}z}{1+ \cfrac{\frac{3}{(c+4)(c+5)}z}{1-\cdots}}}}}}= \cfrac{1}{{}_{1}F_{1} (1;c;z) }$$ $$\begin{align*} \Longrightarrow {}_{1}F_{1} (1;c;z) =&\cfrac{1}{1-\cfrac{\frac{1}{c}z}{1+\cfrac{\frac{1}{c(c+1)}z}{1-\cfrac{\frac{c}{(c+1)(c+2)}z}{1+\cfrac{\frac{2}{(c+2)(c+3)}z}{1-\cfrac{\frac{c+1}{(c+3)(c+4)}z}{1+ \cfrac{\frac{3}{(c+4)(c+5)}z}{1-\cdots}}}}}}}\\ =& \cfrac{1}{1-\cfrac{z}{c+\cfrac{z}{c+1 -\cfrac{cz}{c+1+\cfrac{2z}{c+3-\cfrac{(c+1)z}{c+4+\cfrac{3z}{c+5 - \cdots}}}}}}} \tag{*} \end{align*}$$ Applying $(*)$ to $e^{x}$: $$e^x = {}_{1}F_{1}(1;1;x) = \cfrac{1}{1-\cfrac{x}{1+\cfrac{x}{2 -\cfrac{x}{3+\cfrac{2x}{4-\cfrac{2x}{5+\cfrac{3z}{6 - \cdots}}}}}}}$$ Making the transformation $x\mapsto -x$: $$\Longrightarrow e^{-x} = \frac{1}{e^x} = \cfrac{1}{1+\cfrac{x}{1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}}}}$$ $$\Longrightarrow e^x = 1+\cfrac{x}{1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}}}$$ $$\Longrightarrow e^x -1 = \cfrac{x}{1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}}}$$ Finally, taking the reciprocal: $$\boxed{ \Longrightarrow \frac{x}{e^x -1}= 1-\cfrac{x}{2 +\cfrac{x}{3-\cfrac{2x}{4+\cfrac{2x}{5-\cfrac{3z}{6 + \cdots}}}}} }$$

Confluent hypergeometric function II

Kummer's transformation for ${}_{1}F_{1}$

Here is the proof of this transformation posted by @infseriesbot $$\sum_{n=0}^{\infty} \frac{z^n}{(a)_{n+1}} = e^{z}\sum_{n=0}^{\infty} \frac{(-z)^n}{(a+n)}$$

Proof:

Recall the definition of the Pochhammer polynomial (rising factorial) $$(a)_{n} = a(a+1)\cdots(a+n-1)$$ Hence $$\begin{align*} S = \sum_{n=0}^{\infty} \frac{z^n}{(a)_{n+1}} =& \sum_{n=0}^{\infty} \frac{z^n}{a(a+1)\cdots(a+n)}\\ =& \sum_{n=0}^{\infty} \frac{z^n}{a\left[(a+1)\cdots(a+n)\right]}\\ =& \sum_{n=0}^{\infty} \frac{z^n}{a(a+1)_{n}}\\ =& \frac{1}{a} \sum_{n=0}^{\infty} \frac{(1)_{n}}{(a+1)_{n}} \frac{z^n}{(1)_{n}}\\ =& \frac{1}{a} \sum_{n=0}^{\infty} \frac{(1)_{n}}{(a+1)_{n}} \frac{z^n}{n!}\\ =& \frac{1}{a} {}_{1}F_{1}\left(1,a+1,z\right)\\ \end{align*}$$ ${}_{1}F_{1}(a,b,z)$ is the confluent hypergeometric function, also denoted $M(a,b,z)$. This function satisfies the Kummer’s Transformation: $${}_{1}F_{1}(a,b,z) = e^{z} {}_{1}F_{1}(b-a,b,-z)$$ Therefore $$\begin{align*} \frac{1}{a}{}_{1}F_{1}\left(1,a+1,z\right) =& \frac{e^{z}}{a} {}_{1}F_{1}(a,a+1,-z)\\ =& \frac{e^{z}}{a} \sum_{n=0}^{\infty} \frac{(a)_{n}}{(a+1)_{n}} (-z)^n\\ =& \frac{e^{z}}{a} \sum_{n=0}^{\infty} \frac{a(a+1)\cdots(a+n-1)}{(a+1)(a+2)\cdots(a+n-1)(a+n)} (-z)^n\\ =& \frac{e^{z}}{a}\sum_{n=0}^{\infty} \frac{a(a+1)\cdots(a+n-1)}{(a+1)(a+2)\cdots(a+n-1)(a+n)} (-z)^n\\ =& \frac{e^{z}}{a} \sum_{n=0}^{\infty} \frac{a}{(a+n)} (-z)^n\\ =& e^{z}\sum_{n=0}^{\infty} \frac{(-z)^n}{(a+n)} \end{align*}$$ Hence $$\boxed{ \sum_{n=0}^{\infty} \frac{z^n}{(a)_{n+1}} = e^{z}\sum_{n=0}^{\infty} \frac{(-z)^n}{(a+n)}}$$

Integral of the day XXII

Nnice integral representation of $\pi$

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = -\pi$$ For the proof we will use the Lobachevsky integral formula.

Proof

Recall the Lobachevsky integral formula:

Proposition. Let $f(x)$ a complex-valued, $\pi$-periodic function that is absolutely integrable over a single period. Therefore $$\int_{0}^{\infty} \frac{\sin^2 x}{x^2}f(x)dx = \int_{0}^{\infty} \frac{\sin x}{x} f(x) dx = \int_{0}^{\frac{\pi}{2}} f(x) dx$$ Note that $$\int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = \int_{0}^{\infty} \frac{\sin^2 x}{x^2}\frac{\ln \cos^2 x}{\sin^2 x} dx$$ Since the function $$f(x) = \frac{\ln \cos^2 x}{\sin^2 x} dx$$ is $\pi$-periodic it follows that $$I = \int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = \int_{0}^{\infty} \frac{\sin^2 x}{x^2}\frac{\ln \cos^2 x}{\sin^2 x} dx= \int_{0}^{\frac{\pi}{2}} \frac{\ln \cos^2 x}{\sin^2 x} dx$$ Hence $$\begin{align*} I = \int_{0}^{\frac{\pi}{2}} \frac{\ln \cos^2 x}{\sin^2 x} dx =& 2\int_{0}^{\frac{\pi}{2}} \frac{\ln \cos x}{\sin^2 x} dx\\ =& 2\int_{0}^{1} \frac{\ln(w)}{(1-w^2)^{\frac{3}{2}}} dw \quad \left(w\mapsto \cos x \right)\\ \stackrel{IBP}{=}& \underbrace{2\frac{\ln(w)w}{\sqrt{1-w^2}}\Big|_{0}^{1}}_{=0}-2\int_{0}^{1} \frac{1}{\sqrt{1-w^2}}dw \\ =& -2\arcsin(1)\\ =& -\pi \end{align*}$$ Therefore $$\boxed{\int_{0}^{\infty} \frac{\ln \cos^2 x}{x^2} dx = -\pi }$$

Integral of the day XXI

Integral involving binomial series

Today we show the proof of this integral proposed by @BenriBot_ccbs. $$\int_{0}^{1} \frac{\arcsin\left(\sqrt{x}\right)}{1+x} dx = \frac{\pi}{2}\ln(8) - \pi\ln(1+\sqrt{2})$$ The proof relies on the use of binomial series.

Proof

$$\begin{align*} I = \int_{0}^{1} \frac{\arcsin\left(\sqrt{x}\right)}{1+x} dx =& 2\int_{0}^{1} \frac{\arcsin(w)w}{1+w^2} dw \quad \left( w \mapsto \sqrt{x} \right)\\ \stackrel{IBP}{=}& \ln(w^2+1)\arcsin(w)\Big|_{0}^{1} - \int_{0}^{1} \frac{\ln(w^2+1)}{ \sqrt{1-w^2}}dw \\ =& \frac{\pi}{2}\ln(2) - \int_{0}^{1} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\frac{w^{2n}}{\sqrt{1-w^2}}dw \\ =& \frac{\pi}{2}\ln(2) - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1} w^{2n}(1-w^2)^{-\frac{1}{2}}dw \\ =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1} s^{n-\frac{1}{2}}(1-s)^{-\frac{1}{2}}ds \quad (s \mapsto w^2) \\ =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}B\left(n+\frac{1}{2},\frac{1}{2}\right) \\ =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\frac{\Gamma\left(n+\frac{1}{2}\right) \sqrt{\pi}}{n!} \end{align*}$$ From the formulas $$\Gamma\left(n+\frac{1}{2}\right) =\frac{(2n-1)!! \sqrt{\pi}}{2^n} \tag{1}$$ $$(2n)!! = 2^nn! \tag{2}$$ $$\binom{2n}{n} = \frac{2^{2n}(2n-1)!!}{(2n)!!} \tag{3}$$ we have $$\begin{align*} I =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\frac{(2n-1)!! \pi}{2^nn!} \quad \textrm{ from } (1)\\ =& \frac{\pi}{2}\ln(2) + \frac{\pi}{2}\sum_{n=1}^{\infty} (-1)^n\frac{(2n-1)!!}{n(2n)!!} \quad \textrm{ from } (2)\\ =& \frac{\pi}{2}\ln(2) + \frac{\pi}{2}\sum_{n=1}^{\infty} (-1)^n\frac{\binom{2n}{n}}{2^{2n}n} \quad \textrm{ from } (3) \\ \end{align*}$$ We previously proved that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}x^n = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)$$ If we put $x = -1$ $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}(-1)^n = 2\ln\left(\frac{2}{\sqrt{2}+1}\right)$$ Hence, we can conclude $$\boxed{\int_{0}^{1} \frac{\arcsin\left(\sqrt{x}\right)}{1+x} dx = \frac{\pi}{2}\ln(8) - \pi\ln(1+\sqrt{2})}$$

Integral of the day XX

Nice integral representation of the Apéry's constant $\zeta(3)$

Today we show the proof of the following integral representation of the Aperys constant posted by @sounansya_29 $$\frac{32}{7}\int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx = \zeta(3)$$ The proof relies on some properties of the $\operatorname{arctanh}$ and $arctan$ functions and Fourier series.

Proof

Recall the definition in terms of logarithms of $\operatorname{arctanh}(z)$ $$\operatorname{arctanh}(z) = \frac{1}{2}\ln\left(\frac{1+z}{1-z}\right)$$ Therefore $$\begin{align*} I = \int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx =&\frac{1}{2}\int_{0}^{\frac{\pi}{4} } x\ln\left(\frac{1+\tan x}{1-\tan x}\right) dx \\ =& \frac{1}{2}\int_{1}^{\infty} \frac{\arctan\left(\frac{w-1}{w+1}\right)\ln(w)}{1+w^2} dw \quad \left( w \mapsto \frac{1+\tan x}{1-\tan x}\right) \end{align*}$$ Recall the following addition formula for $\arctan(x)$: $$\arctan(w) - \frac{\pi}{4} = \arctan\left(\frac{w-1}{w+1}\right)$$ Hence $$\begin{align*} I = \int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx =& \frac{1}{2}\int_{1}^{\infty} \frac{\arctan\left(\frac{w-1}{w+1}\right)\ln(w)}{1+w^2} dw\\ =& \frac{1}{2}\int_{1}^{\infty} \frac{\left(\arctan\left(w\right) - \frac{\pi}{4}\right)\ln(w)}{1+w^2} dw\\ =& \frac{1}{2}\underbrace{\int_{1}^{\infty} \frac{\arctan(w)\ln(w)}{1+w^2}dw}_{J} - \frac{\pi}{8}\underbrace{\int_{1}^{\infty} \frac{\ln(w)}{1+w^2} dw}_{K}\\ \end{align*}$$ $$\begin{align*} K = \int_{1}^{\infty} \frac{\ln(w)}{1+w^2} dw = & -\int_{0}^{1} \frac{\ln(s)}{1+s^2} ds \quad \left(s \mapsto \frac{1}{w} \right)\\ =& -\int_{0}^{1} \sum_{n=0}^{\infty} (-1)^ns^{2n} \ln(s) ds\\ =& -\sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} s^{2n} \left(\frac{d}{dt}\Big|_{t=0+} s^t \right) ds\\ =& \frac{d}{dt}\Big|_{t=0+} -\sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} s^{2n+t} ds\\ =& \frac{d}{dt}\Big|_{t=0+} -\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+t+1} ds\\ =& -\sum_{n=0}^{\infty} \left[ \frac{d}{dt}\Big|_{t=0+} \frac{(-1)^n}{2n+t+1} \right] ds\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} \\ =& \beta(2) \end{align*}$$ $$\begin{align*} J = \int_{1}^{\infty} \frac{\arctan(w)\ln(w)}{1+w^2}dw =& \int_{0}^{1} \frac{\arctan\left(\frac{1}{s}\right)\ln\left(\frac{1}{s}\right)}{1+s^2} ds \quad \left( w \mapsto \frac{1}{s}\right)\\ =& -\int_{0}^{1} \frac{\left(-\arctan(s) + \frac{\pi}{2}\right)\ln(s)}{1+s^2} \quad \left(\arctan\left(\frac{1}{b}\right) = -\arctan(b)-\frac{\operatorname{sign}(b)\pi}{2}\right)\\ =& \int_{0}^{1} \frac{\arctan(s)\ln(s)}{1+s^2} ds - \frac{\pi}{2} \int_{0}^{1} \frac{\ln(s)}{1+s^2} ds \\ =& \int_{0}^{1} \frac{\arctan(s)\ln(s)}{1+s^2} ds +\frac{\pi}{2}\beta(2)\\ =& \int_{0}^{\frac{\pi}{4}} t\ln(\tan(t)) dt +\frac{\pi}{2}\beta(2)\\ =& -\int_{0}^{\frac{\pi}{4}} t\ln(\cot(t)) dt +\frac{\pi}{2}\beta(2)\\ \end{align*}$$ Recall the following Fourier series $$\sum_{k=1}^{\infty} \frac{\cos\left((2k-1)x\right)}{2k-1} = \frac{1}{2}\ln\left(\cot \frac{x}{2}\right)$$ Hence $$\begin{align*} J = -\int_{0}^{\frac{\pi}{4}} t\ln(\cot(t)) dt +\frac{\pi}{2}\beta(2) =& -2\int_{0}^{\frac{\pi}{4}} t\sum_{k=1}^{\infty} \frac{\cos\left((2k-1)2t\right)}{2k-1} dt +\frac{\pi}{2}\beta(2) \\ =& -\sum_{k=1}^{\infty} \frac{2}{2k-1}\int_{0}^{\frac{\pi}{4}} t\cos\left((2k-1)2t\right) dt +\frac{\pi}{2}\beta(2)\\ =& -\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3}\int_{0}^{\frac{\pi(2k-1)}{2}} y \cos\left(y\right) dy +\frac{\pi}{2}\beta(2)\\ \stackrel{IBP}{=}& -\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \left[\sin(y)y\Big|_{0}^{\frac{\pi(2k-1)}{2}} - \int_{0}^{\frac{\pi}{8(2k-1)}}\sin(y)dy \right] +\frac{\pi}{2}\beta(2)\\ =& -\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \left[\sin\left(\frac{\pi(2k-1)}{2}\right)\frac{\pi(2k-1)}{2} + \cos\left(\frac{\pi(2k-1)}{2}\right) -1\right] +\frac{\pi}{2}\beta(2)\\ =& -\frac{\pi}{4}\sum_{k=1}^{\infty} \frac{\sin\left(\frac{\pi(2k-1)}{2}\right)}{(2k-1)^2} - \frac{1}{2}\sum_{k=1}^{\infty} \frac{\cos\left(\frac{\pi(2k-1)}{2}\right)}{(2k-1)^3}+\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3}+\frac{\pi}{2}\beta(2)\\ =& \frac{\pi}{4}\sum_{k=1}^{\infty} \frac{\cos(\pi k)}{(2k-1)^2} - \frac{1}{2}\sum_{k=1}^{\infty} \frac{\sin(\pi k)}{(2k-1)^3}+\frac{1}{2}\lambda(3)+\frac{\pi}{2}\beta(2) \end{align*}$$ Note that $$\left(\sin(\pi k)\right)_{k\in \mathbb{N}} = (0,0,0,0,...)$$ $$\left(\cos(\pi k)\right)_{k\in \mathbb{N}} = (-1,1,-1,1,-1,..)$$ Therefore $$J = \frac{\pi}{4}\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2k-1)^2} + \frac{1}{2}\lambda(3)+\frac{\pi}{2}\beta(2) = \frac{\pi}{4}\beta(2) + \frac{1}{2}\lambda(3)$$ From the identity $$\lambda(v) = (1-2^{-v}) \zeta(v)$$ we have $$J = \int_{1}^{\infty} \frac{\arctan(w)\ln(w)}{1+w^2}dw = \frac{7}{16}\zeta(3) + \frac{\pi}{4}\beta(2)$$ Hence $$I = \int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx = \frac{1}{2}J - \frac{\pi}{8}K = \frac{7}{32}\zeta(3) - \frac{\pi}{8}\beta(2) + \frac{\pi}{8}\beta(2)$$ Therefore, we can conclude $$\boxed{ \frac{32}{7}\int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx = \zeta(3)}$$

Euler's work XIX

Another Euler sum involving the Catalan's constant $\beta(2)$

Today we show the proof of this nice Euler sum proposed by @Ali39342137 $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} = \frac{3\pi\ln(2)}{2} + 2 - \pi - 2\beta(2)$$ For the proof we will use some properties of Harmonic numbers and Fourier series.

Proof

We need the following integral representation (proof in Appendix 1): $$H_{n} - H_{2n} + \ln(2) = \int_{0}^{1} \frac{x^{2n}}{1+x} dx \tag{1}$$ and the following series representation (proof in Appendix 2): $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}x^{2n} = \sqrt{1-x^2} + x\arcsin(x) -1 \quad |x|\leq 1 \tag{2}$$ Therefore $$\begin{align*} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} =& \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2} \left( \ln(2) - \int_{0}^{1} \frac{x^{2n}}{1+x} dx \right) \quad \textrm{ from (1)} \\ =& \ln(2)\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}- \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}\int_{0}^{1} \frac{x^{2n}}{1+x} dx \end{align*}$$ From (2) if we put $x=1$ $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2} = \arcsin(1) -1 = \frac{\pi}{2} - 1$$ Hence $$\begin{align*} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} =& \frac{\pi\ln(2)}{2} - \ln(2)-\int_{0}^{1} \frac{1}{1+x}\left(\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}x^{2n}\right) dx \\ =& \frac{\pi\ln(2)}{2} - \ln(2)- \int_{0}^{1} \frac{\sqrt{1-x^2} + x\arcsin(x) -1}{1+x} dx \quad \textrm{ from (2) }\\ =& \frac{\pi\ln(2)}{2} - \ln(2)- \underbrace{\int_{0}^{1} \frac{\sqrt{1-x^2}}{1+x} dx}_{I} - \underbrace{\int_{0}^{1}\frac{x\arcsin(x)}{1+x}dx}_{J} +\underbrace{\int_{0}^{1}\frac {1}{1+x} dx}_{K} \end{align*}$$ $$\begin{align*} I = \int_{0}^{1} \frac{\sqrt{1-x^2}}{1+x} dx =& 4\int_{0}^{1} \frac{w^2}{(1+w^2)^2} dw \quad \left( w \mapsto \frac{\sqrt{1-x^2}}{1+x} \right) \\ =& 4\int_{1}^{\infty} \frac{1}{(1+s^2)^2} ds \quad \left( s\mapsto \frac{1}{w} \right)\\ =& 4\int_{\frac{\pi}{2} }^{\frac{\pi}{2}} \frac{\sec^2(w)}{(1+\tan^2{w})^2} dw \quad \left(w \mapsto \arctan(s) \right) \\ =& 4\int_{\frac{\pi}{2} }^{\frac{\pi}{2}} \cos^2(w) dw \quad \left( \tan^2(\theta) +1 = \sec^2(\theta)\right) \\ =& 2\int_{\frac{\pi}{2} }^{\frac{\pi}{2}} \cos(2w)dw + \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dw \quad \left( 2\cos^2(\theta)-1 = \cos(2\theta) \right)\\ =& 2\int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \cos(2w)dw + \frac{\pi}{8} \\ =& \int_{\frac{\pi}{2}}^{\pi} \cos(r)dr + \frac{\pi}{2} \quad (r \mapsto 2w)\\ =& -1 + \frac{\pi}{2}\\ \end{align*}$$ $$\begin{align*} J = \int_{0}^{1}\frac{x\arcsin(x)}{1+x}dx \stackrel{IBP}{=} & \left[x\arctan(x)-\ln(1+x)\arctan(x)\right]_{0}^{1} -\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}dx +\int_{0}^{1} \frac{\ln(x+1)}{\sqrt{1-x^2}} dx \\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -\left[-\sqrt{1-x^2}\right]_{0}^{1} +\int_{0}^{1} \frac{\ln(x+1)}{\sqrt{1-x^2}}dx\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\int_{0}^{1} \frac{\ln(x+1)}{\sqrt{1-x^2}}dx\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\int_{0}^{\frac{\pi}{2}} \ln(1+\cos(w))dw \quad (w \mapsto \operatorname{arccos}(x))\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\int_{0}^{\frac{\pi}{2}} \ln\left(2\cos^2\left(\frac{w}{2}\right)\right)dw \quad \left(\cos(2\theta) = 2\cos^2(\theta)-1 \right)\\ =& \frac{\pi}{2} -\frac{\pi\ln(2) }{2} -1 +\ln(2) \int_{0}^{\frac{\pi}{2}} dw +2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos\left(\frac{w}{2}\right)\right)dw \\ =& \frac{\pi}{2} -1 -2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos\left(\frac{w}{2}\right)\right)dw \end{align*}$$ Recall the Fourier series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kx)}{k} = \ln\left(2\cos\left(\frac{x}{2}\right)\right)$$ Therefore $$\begin{align*} J=& \frac{\pi}{2} -1 -2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos\left(\frac{w}{2}\right)\right)dw \\ =& \frac{\pi}{2} -1 -2\int_{0}^{\frac{\pi}{2}} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kw)}{k}dw -2\ln(2)\int_{0}^{\frac{\pi}{2}} dw\\ =& \frac{\pi}{2} -\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}\int_{0}^{\frac{\pi}{2}} \cos(kw)dw \\ =& \frac{\pi}{2}-\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2}\int_{0}^{\frac{k\pi}{2}} \cos(r)dr \\ =& \frac{\pi}{2}-\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2}\left[\sin(r)\right]_{0}^{\frac{k\pi}{2}} \\ =& \frac{\pi}{2} -\pi\ln(2) -1 -2\sum_{k=1}^{\infty} \frac{(-1)^{k-1} \sin\left(\frac{k \pi}{2} \right)}{k^2} \\ \end{align*}$$ Note that $$\left(\sin\left(\frac{k \pi}{2}\right)\right)_{k\in \mathbb{N}} = (1,0,-1,0,1,0,-1,0,..)$$ Hence $$J = \frac{\pi}{2} -\pi\ln(2) -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^2} = \frac{\pi}{2} -\pi\ln(2) -1 + 2\beta(2)$$ $$K = \int_{0}^{1} \frac{1}{1+x} dx = \left[\ln(1+x)\right]_{0}^{1} = \ln(2)$$ Therefore $$\begin{align*} \sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} =& \frac{\pi\ln(2)}{2} - \ln(2)- I - J + K \\ =& \frac{3\pi\ln(2)}{2} + 2 - \pi - 2\beta(2) \end{align*}$$ Hence, we can conclude $$\boxed{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}\left(H_{2n}- H_{n} \right)}{2^{2n}(2n-1)^2} = \frac{3\pi\ln(2)}{2} + 2 - \pi - 2\beta(2) }$$

Appendix 1

$$\ln(2) + H_{n} - H_{2n} = \int_{0}^{1} \frac{x^{2n}}{1+x} dx$$

Proof

$$\begin{align*} \int_{0}^{1} \frac{x^{2n}}{1+x} dx\stackrel{IBP}{=}& \left[\ln(1+x)x^{2n}\right]_{0}^{1} - 2n\int_{0}^{1} \ln(1+x)x^{2n-1} dx \\ =& \ln(2) - 2n\int_{0}^{1}\sum_{j=1}^{\infty} \frac{(-1)^{j+1}}{j} x^{j} x^{2n-1} dx\\ =& \ln(2) - 2n\sum_{j=1}^{\infty} \frac{(-1)^{j+1}}{j} \int_{0}^{1} x^{j+2n-1} dx\\ =& \ln(2) + 2n\sum_{j=1}^{\infty} \frac{(-1)^{j}}{j(j+2n)} \\ =& \ln(2) + n\sum_{j=1}^{\infty} \frac{1}{j(j+n)} - 2n\sum_{j=1}^{\infty} \frac{1}{j(j+2n)}\\ \end{align*}$$ From the series expansion for Harmonic numbers: $$H_{x} = x \sum_{k=1}^{\infty} \frac{1}{k(x+k)}$$ we obtain $$\ln(2) + H_{n} - H_{2n} = \int_{0}^{1} \frac{x^{2n}}{1+x} dx$$

Appendix 2

$$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}x^{2n} = \sqrt{1-x^2} + x\arcsin(x) -1 \quad |x|\leq 1$$

Proof

Recall the following basic series expansion: $$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n} \quad |x|\lt 1$$ Dividing by $x^2$ $$\frac{1}{x^2\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n-2} = \frac{1}{x^2} + \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n-2}$$ $$\Longrightarrow \frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} x^{2n-2}$$ Integrating from $0$ to $r$: $$\int_{0}^{r} \left[\frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} \right] dx = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}} \int_{0}^{r}x^{2n-2}dx = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}r^{2n-1}$$ - Note $$\begin{align*} \int \left[\frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} \right] dx =& \int \frac{1}{x^2\sqrt{1-x^2}}dx- \int \frac{1}{x^2} dx\\ =& \int \frac{1}{x^2\sqrt{1-x^2}}dx+ \frac{1}{x} + C\\ =& \int \csc^2(w) dw + \frac{1}{x} + C \quad \left(x \mapsto \sin(w)\right)\\ =& -\cot(w) + \frac{1}{x} + C\\ =& -\cot(\arcsin(x)) + \frac{1}{x} + C\\ =& -\frac{\sqrt{1-x^2}}{x} + \frac{1}{x} + C \quad \left( \arcsin(x) = \operatorname{arccot}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)\\ =& \frac{1-\sqrt{1-x^2}}{x} + C \end{align*}$$ Hence $$\begin{align*} \int_{0}^{r} \left[\frac{1}{x^2\sqrt{1-x^2}}- \frac{1}{x^2} \right] dx = & \left[ \frac{1-\sqrt{1-x^2}}{x}\right]_{0}^{r} \\ =& \frac{1-\sqrt{1-r^2}}{r} - \lim_{x \to 0 } \frac{1-\sqrt{1-x^2}}{x} \\ =& \frac{1-\sqrt{1-r^2}}{r} \end{align*}$$ Therefore $$\frac{1-\sqrt{1-r^2}}{r} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}r^{2n-1}$$ Dividing by $r$: $$\frac{1-\sqrt{1-r^2}}{r^2} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}r^{2n-2}$$ Integrating from $0$ to $t$: $$\int_{0}^{t} \frac{1-\sqrt{1-r^2}}{r^2} dt = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)}\int_{0}^{t} r^{2n-2} dt = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n-1}$$ Note that $$\begin{align*} \int \frac{1-\sqrt{1-r^2}}{r^2} dr =& \int \frac{1}{r^2} dr - \int \frac{\sqrt{1-r^2}}{r^2} dr \\ =& -\frac{1}{r} - \int \frac{\sqrt{1-r^2}}{r^2} dr + C\\ =& -\frac{1}{r} - \int \cot^2(w) dw + C \quad \left( r \mapsto \sin(w)\right)\\ =& -\frac{1}{r} - \int (\csc^2(w)-1) dw + C \quad \left( \csc^2(\theta) - \cot^2(\theta) = 1 \right)\\ =& -\frac{1}{r} + \cot(w) + w + C\\ =& -\frac{1}{r} + \cot(\arcsin(r)) + \arcsin(r) + C\\ =& -\frac{1}{r} + \frac{\sqrt{1-r^2}}{r} + \arcsin(r) + C \quad \left( \arcsin(x) = \operatorname{arccot}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)\\ \end{align*}$$ Therefore $$\int_{0}^{t} \frac{1-\sqrt{1-r^2}}{r^2} dt = \left[-\frac{1}{r} + \frac{\sqrt{1-r^2}}{r} + \arcsin(r)\right]_{0}^{t} = -\frac{1}{t} - \frac{\sqrt{1-t^2}}{t} + \arcsin(t)$$ Hence $$-\frac{1}{t} + \frac{\sqrt{1-t^2}}{t} + \arcsin(t) = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n-1}$$ $$\Longrightarrow \sqrt{1-t^2} + t\arcsin(t) -1 = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n-1)^2}t^{2n}$$

Integral of the day XIX

Integral involving a finite series

Today we show the proof of the following integral posted by @infseriesbot $$\int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx = \frac{1}{2^{n+1}}\sum_{k=1}^{n} \frac{2^k}{k}$$ The proof will rely on the Euler's formula and some properties of the binomial coefficient.

Proof

$$\begin{align*} \int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx =& \Im\left( \int_{0}^{\frac{\pi}{2}} e^{inx}\cos^n(x) dx \right) \textrm{ (Euler's formula)} \\ =& \Im\left( \int_{0}^{\frac{\pi}{2}} e^{inx}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^ndx \right)\\ =& \Im\left( \frac{1}{2^n}\int_{0}^{\frac{\pi}{2}} e^{inx}\sum_{j=0}^{n}\binom{n}{j} e^{ix(n-2j)}dx \right) \quad \textrm{(Binomial expansion)}\\ =& \Im\left( \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j}\int_{0}^{\frac{\pi}{2}} e^{i2x(n-j)}dx \right)\\ =&\Im\left( \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \left[\cos\left(2x(n-j)\right)+i\sin\left(2x(n-j)\right) \right]dx\right) \textrm{ (Euler's formula)}\\ =&\Im\left( \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \left[\int_{0}^{\frac{\pi}{2}} \cos\left(2x(n-j)\right) dx +i \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx \right]\right)\\ =&\Im\left( \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \cos\left(2x(n-j)\right) dx +i \frac{1}{2^n}\sum_{j=0}^{n} \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx \right)\\ =& \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx \\ =& \frac{1}{2^n}\sum_{j=0}^{n-1}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx + \underbrace{\binom{n}{n} \int_{0}^{\frac{\pi}{2}} \sin\left(0\right) dx}_{=0} \\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1}{n-j} \int_{0}^{\pi(n-j)} \sin(w) dw \quad \left( w \mapsto 2x(n-j) \right)\\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1}{n-j} \int_{0}^{\pi(n-j)} \sin(w) dw \\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1}{n-j} \left[-\cos(w)\right]_{0}^{\pi(n-j)}\\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1-\cos\left(\pi(n-j)\right)}{n-j} \\ =& \frac{1}{2^{n+1}}\sum_{k=1}^{n}\binom{n}{n-k}\frac{1-\cos\left(\pi k \right)}{k} \\ =& \frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{n-k}\frac{\sin^2\left(\frac{\pi k}{2} \right)}{k} \quad \left(\sin^2 \theta =\frac{1-\cos(2\theta)}{2}\right)\\ =& \frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{k}\frac{\sin^2\left(\frac{\pi k}{2} \right)}{k} \;\;\quad \binom{n}{n-m} = \binom{n}{m} \\ \end{align*}$$ Note that $$\left(\sin^2\left(\frac{\pi k}{2} \right)\right)_{k\in \mathbb{N}} = (1,0,1,0,1,0,...)$$ Hence $$\begin{align*} I = \int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx =& \frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{k}\frac{\sin^2\left(\frac{\pi k}{2} \right)}{k}\\ =\frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{2k-1}\frac{1}{2k-1}\\ \end{align*}$$ Finally, we are going to prove that $$2\sum_{k=1}^{n}\binom{n}{2k-1}\frac{1}{2k-1} = \sum_{k=1}^{n} \frac{2^k}{k} \quad \forall n\in \mathbb{N}$$ using mathematical induction:

Base case $n=1$ $$2\binom{1}{2-1}\frac{1}{2-1} = 2$$ $$\frac{2^1}{1} = 2$$ Inductive step, suppose this is valid for $n=m$ $$2\sum_{k=1}^{m}\binom{m}{2k-1}\frac{1}{2k-1} = \sum_{k=1}^{m} \frac{2^k}{k}$$ We need to prove this for $\underline{n= m+1}$

From the recursion formula $$\binom{v+1}{m} =\binom{v}{m} + \binom{v}{m-1} \tag{1}$$ and the expression $$\sum_{j=0}^{J} \binom{n}{j} = 2^n \quad J\geq n \tag{2}$$ we have $$\begin{align*} 2\sum_{k=1}^{m+1}\binom{m+1}{2k-1}\frac{1}{2k-1} = & 2\sum_{k=1}^{m}\binom{m+1}{2k-1}\frac{1}{2k-1} + \underbrace{2\binom{m+1}{2m+1}\frac{1}{2m+1}}_{=0}\\ =& 2\sum_{k=1}^{m}\left[\binom{m}{2k-1}+\binom{m}{2k-2}\right]\frac{1}{2k-1} \quad \textrm{ from (1)}\\ =& 2\sum_{k=1}^{m}\binom{m}{2k-1}\frac{1}{2k-1}+2\sum_{k=1}^{m}\binom{m}{2k-2}\frac{1}{2k-1}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+2\sum_{k=1}^{m}\binom{m}{2k-2}\frac{1}{2k-1} \quad \textrm{(inductive hypothesis)}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+2\sum_{j=0}^{m-1}\binom{m}{2j}\frac{1}{2j+1} \quad (j = k-1)\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+2\sum_{j=0}^{m-1}\frac{m!}{(m-2j)!(2j)!(2j+1)} \\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{m-1}\frac{(m+1)!}{(m-2j)!(2j+1)!} \\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{m-1}\binom{m+1}{2j+1} \\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{m-1}\left[\binom{m}{2j+1}+ \binom{m}{2j}\right] \quad \textrm{ from (1)}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{2m-1}\binom{m}{j}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2^{m+1}}{m+1}\quad \textrm{ from (2)}\\ =& \sum_{k=1}^{m+1} \frac{2^k}{k} \end{align*}$$ Hence, we can conclude $$\boxed{\int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx = \frac{1}{2^{n+1}}\sum_{k=1}^{n} \frac{2^k}{k}}$$