Integral of the day IV

Integral related to the Mellin transform of $e^{ax}$

Today we show the proof of this integral posted by @mvs_rpi $$\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}} e^{-x}\sin x dx = \frac{\sqrt{\pi}}{2i}\left[ \frac{-\gamma -\ln(4-4i)}{\sqrt{1-i}} + \frac{( \gamma+ \ln(4+4i))}{\sqrt{1+i}} \right]$$

Proof

$$\begin{align*} \int_{0}^{\infty} \frac{\ln x}{\sqrt{x}} e^{-x}\sin x dx =& \int_{0}^{\infty} \frac{\left(\frac{d}{dt}\Big|_{t=0+} x^t\right) }{\sqrt{x}} e^{-x}\sin x dx\\ =&\frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t-\frac{1}{2}} e^{-x}\sin x dx\\ =&\frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t-\frac{1}{2}} e^{-x}\left[\frac{e^{ix}-e^{-ix}}{2i}\right] dx\\ =& \frac{d}{dt}\Big|_{t=0+} \underbrace{\frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}} e^{x(-1+i)} dx}_{I_{1}} + \frac{d}{dt}\Big|_{t=0+} \underbrace{\frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}} e^{x(-1-i)} dx}_{I_{2}} \end{align*}$$ Recall the Mellin transform: $$\int_{0}^{\infty} x^{s-1}e^{ax} dx \quad a\in \mathbb{C}$$ $$e^{ax} = \sum_{n=0}^{\infty} \frac{a^nx^n}{n!} = \sum_{n=0}^{\infty} \frac{(-a)^n(-x)^n}{n!} = \sum_{n=0}^{\infty} \frac{\phi(n)(-x)^n}{n!}$$ where $\displaystyle \phi(n) = (-a)^n$ By Ramanujan's master theorem $$\int_{0}^{\infty} x^{s-1}e^{ax} dx = \Gamma(s)(-a)^{-s} \quad \operatorname{Re}(s)>0, \; \operatorname{Re}(a) <0$$ If $\displaystyle s = t+\frac{1}{2}, \; \displaystyle a = -1+i$ $$\int_{0}^{\infty} x^{t-\frac{1}{2}}e^{x(-1+i)} dx = \Gamma\left(t+\frac{1}{2}\right)(1-i)^{-t-\frac{1}{2}}$$ $$\begin{align*} I_{1} = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}}e^{x(-1+i)} dx = & \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\Gamma\left(t+\frac{1}{2}\right)(1-i)^{-t-\frac{1}{2}}\\ =& \frac{\sqrt{\pi}( -\gamma -\ln(4-4i))}{2i\sqrt{1-i}} \end{align*}$$ If $\displaystyle s = t+\frac{1}{2}, \; \displaystyle a = -1-i$ $$\begin{align*} I_{2} = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\int_{0}^{\infty} x^{t-\frac{1}{2}}e^{x(-1-i)} dx =& I_{1} = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2i}\Gamma\left(t+\frac{1}{2}\right)(1+i)^{-t-\frac{1}{2}} \\ =& \frac{\sqrt{\pi}( -\gamma -\ln(4+4i))}{2i\sqrt{1+i}} \end{align*}$$ Therefore $$I = I_{1}-I_{2} =+ \frac{\sqrt{\pi}}{2i}\left[ \frac{-\gamma -\ln(4-4i)}{\sqrt{1-i}} + \frac{( \gamma+ \ln(4+4i))}{\sqrt{1+i}} \right] \approx -.0.23611$$
$$\boxed{ \int_{0}^{\infty} \frac{\ln x}{\sqrt{x}} e^{-x}\sin x dx = \frac{\sqrt{\pi}}{2i}\left[ \frac{-\gamma -\ln(4-4i)}{\sqrt{1-i}} + \frac{( \gamma+ \ln(4+4i))}{\sqrt{1+i}} \right] }$$

Inegral of the day III

Integral involving the binomial coefficient

Today we show the proof of the following integral involving the binomial coefficient posted by @infseriesbot. The bot has a typo but this is the correct formula: $$\int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx = \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2}$$
Proof

We need the following ingredients for the proof:


1. $\sin^{r}(x)$ can be expanded with the following formula:
$$sin^{r}(x) = \begin{cases} \displaystyle \frac{(-1)^n}{2^{r-1}} \sum_{j=0}^{n-1} (-1)^j \binom{r}{j} \cos\left[(r-2j)x\right] + \frac{1}{2^n} \binom{r}{n} &\quad \textrm{if} \quad r=2n \\ \displaystyle \frac{(-1)^n}{2^{r-1}}\sum_{j=0}^{n} (-1)^j \binom{r}{j} \sin\left[(r-2j)x\right] & \quad \textrm{if} \quad r=2n+1 \end{cases}$$

Proof

$$\begin{align*} (2i\sin(x))^{r} = (e^{ix}-e^{-ix})^{r} = & \sum_{j=0}^{r} \binom{r}{j} (-1)^{r-j}e^{ixj}e^{-ix(r-j)} =\sum_{j=0}^{r} \binom{r}{j} (-1)^{r-j}e^{-ix(r-2j)} \\ =& \sum_{j=0}^{r} \binom{r}{j}(-1)^{r-j}\left[\cos(x(r-2j))-i\sin(x(r-2j))\right] \end{align*}$$ Now suppose that $r=2n+1$, then $$\begin{align*} 2^{2n}(-1)^{n}(2i)\sin^{2n+1}(x) = &\sum_{j=0}^{2n+1} \binom{2n+1}{j}(-1)^{2n+1-j} \left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\ =& \sum_{j=0}^{n} \binom{2n+1}{j}(-1)^{2n+1-j} \left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\ +& \sum_{j=n+1}^{2n+1} \binom{2n+1}{j}(-1)^{2n+1-j}\left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\ =& \sum_{j=0}^{n} \binom{2n+1}{j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]\\ +&\sum_{j=n+1}^{2n+1} \binom{2n+1}{2n+1-j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right] \end{align*}$$ Lets $k=2n+1-j$ then $$\begin{align*} &\sum_{j=n+1}^{2n+1} \binom{2n+1}{2n+1-j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]\\ =& \sum_{k=0}^{n} \binom{2n+1}{k}(-1)^{k}\left[\cos(-x(2n+1-2k))-i\sin(-x(2n+1-2k))\right] \end{align*}$$ Given that for $k=0,...,n$ $$\begin{eqnarray*} \cos(-x(2n+1-2k))-\cos(x(2n+1-2k))&=&0\\ i\sin(x(2n+1-2k))-i\sin(-x(2n+1-2k))& =& 2i\sin(x(2n+1-2k)) \end{eqnarray*}$$ $$\begin{equation} \therefore sin^{2n+1}(x) =\frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \sin\left[(2n+1-2j)x\right] \end{equation}$$ A similar argument shows the result for $r=2n$


2. From complex analysis we know that (proof in the appendix):


$$\begin{equation} \int_{0}^{\infty} \frac{\sin(x)}{x} = \frac{\pi}{2} \end{equation}$$


3. From combinatorics we kwnow that that (proof in the appendix):


$$\begin{equation} \sum _{j=0}^{k} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{k} \end{equation}$$ We just have tu put all the ingredients together $$\begin{align*} \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx =& \int_{0}^{\infty} \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx \quad \textrm{from (1)}\\ =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx \end{align*}$$ If $u= (2n+1-2j)x$ then $\displaystyle dx = \frac{1}{2n+1-2j} du$ $$\begin{align*} \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx\\ =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin(u)}{u} du\\ \overset{(2)}{=}& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \frac{\pi}{2} \quad \textrm{from (2)}\\ =& \frac{(-1)^n}{2^{2n}} (-1)^{n} \binom{2n}{n}\frac{\pi}{2} \quad \textrm{from (3)}\\ =& \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2} \end{align*}$$ Therefore $$\boxed{ \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx = \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2}}$$

Appendix: Here are the proofs of (2) and (3):

Proof of (2)

$$\int_{0}^{\infty} \frac{\sin(x)}{x} = \frac{\pi}{2}$$ The integral $$I = \operatorname{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x}dx$$ exists by the \emph{Cauchy principal value theroem}. Therefore $$\operatorname{Im}I = \operatorname{p.v.} \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx$$ also exists. Then $$\operatorname{p.v.} \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx = \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx$$ by the continuity of $\displaystyle \frac{\sin(x)}{x}$ at $x=0$. The existence of $\displaystyle \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx$ also implies the existence of $\displaystyle \int_{0}^{\infty} \frac{\sin(x)}{x}dx$ and $$\operatorname{Im} I = 2\int_{0}^{\infty} \frac{\sin(x)}{x}dx$$ Now applying the residue formula for the principal value: $$I = \pi i \sum \underset{z=0}{\textrm{Res}}\left(\frac{e^{ix}}{x}\right)=\pi i \Longrightarrow \int_{0}^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}$$

Proof of (3)

If $k,m \in \mathbb{N}$ $$\sum _{j=0}^{k} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{k}$$ Base case Lets $k=1$ then $$(-1)^{0}\binom{m}{0} +(-1)^{1}\binom{m}{1} = 1-m$$ On the other hand $$(-1)^{1}\binom{m-1}{1} = 1-m$$ Inductive hypothesis Assume the proposition is true for $k=n$, then $$\sum _{j=0}^{n} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{n}$$ Inductive step Lets $k=n+1$ then $$\begin{align*} \sum _{j=0}^{n+1} (-1)^j\binom{m}{j} = &\sum _{j=0}^{n} (-1)^j\binom{m}{j} + (-1)^{n+1}\binom{m}{m+1}\\ =& (-1)^{n} \binom{m-1}{n} +(-1)^{n+1} \left[\binom{m-1}{n} +\binom{m-1}{n+1}\right]\\ =& (-1)^{n+1}\binom{m-1}{n+1} \end{align*}$$

Integral of the day III

Corollary of the integral representation of the $sec(x)$ function

Today we show the proof of this beautiful integral posted by @gureponnn on Twitter. $$\int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x + \sin^{2n} x}dx = \frac{\pi^2}{8n}\sum_{j=1}^{n}\binom{n-1}{j-1} \csc \left(\frac{\pi(2j-1)}{2n}\right)$$ The proof of this result rely on the well-known integral representation of the $sec(x)$ function

Proof

$$\begin{align*} I=\int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x + \sin^{2n} x}dx =& \int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x (1+\tan^{2n} x)}dx \\ =& \int_{0}^{\infty} \frac{\arctan(w)(w^2+1)^{n-1}}{1+w^{2n}}dw \quad (w \mapsto \tan x)\\ =& \int_{0}^{\infty} \frac{\arctan\left(\frac{1}{s}\right)(s^2+1)^{n-1}}{s^{2n}+1}ds \quad \left(s \mapsto \frac{1}{w}\right)\\ =& \int_{0}^{\infty} \frac{\left(-\arctan(s)+\frac{\pi}{2}\right)(s^2+1)^{n-1}}{s^{2n}+1}ds \quad \left(\arctan \left(\frac{1}{a}\right) = -\arctan(a)+\frac{\pi}{2}\right)\\ =& -\int_{0}^{\infty}\frac{\arctan(s)(s^2+1)^{n-1}}{s^{2n}+1}ds + \frac{\pi}{2} \int_{0}^{\infty}\frac{(s^2+1)^{n-1}}{s^{2n}+1}ds \end{align*}$$ Hence $$2I = \frac{\pi}{2} \int_{0}^{\infty}\frac{(s^2+1)^{n-1}}{s^{2n}+1}ds$$ Therefore $$\begin{align*} I = \frac{\pi}{4} \int_{0}^{\infty}\frac{(s^2+1)^{n-1}}{s^{2n}+1}ds = &\frac{\pi}{4n} \int_{0}^{\infty}\frac{(r^{\frac{2}{n}}+1)^{n-1}}{(r^2+1)r^{\frac{n-1}{n}}}dr \quad (r^2 \mapsto s^{2n})\\ =& \frac{\pi}{4n} \int_{0}^{\infty}\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{r^{\frac{2k}{n}}}{(r^2+1)r^{\frac{n-1}{n}}}dr \quad (\textrm{binomial theorem})\\ =& \frac{\pi}{4n}\sum_{k=0}^{n-1}\binom{n-1}{k} \int_{0}^{\infty}\frac{r^{\frac{2k+1-n}{n}}}{r^2+1}dr \end{align*}$$ Recall the integral representation of the secant function: $$\boxed{\sec z = \frac{2}{\pi} \int_{0}^{\infty} \frac{t^{\frac{2z}{\pi}}}{t^2+1}dt \quad |\operatorname{Re}(z)|<\frac{\pi}{2}}$$ If $\displaystyle z = \frac{\pi(2k+1-n)}{2n}$ $$\int_{0}^{\infty} \frac{t^{\frac{2k+1-n}{n}}}{t^2+1}dt = \frac{\pi}{2} \sec \left(\frac{\pi(2k+1)}{2n} - \frac{\pi}{2} \right) = \frac{\pi}{2} \csc \left(\frac{\pi(2k+1)}{2n} \right)$$

Hence
$$I = \frac{\pi}{4n}\sum_{k=0}^{n-1}\binom{n-1}{k} \int_{0}^{\infty}\frac{r^{\frac{k-1+n}{n}}}{r^2+1}dr = \frac{\pi^2}{8n}\sum_{k=0}^{n-1}\binom{n-1}{k} \csc \left(\frac{\pi(2k+1)}{2n} \right)$$ If $\displaystyle k = j-1$ $$I = \frac{\pi^2}{8n}\sum_{j=1}^{n}\binom{n-1}{j-1} \csc \left(\frac{\pi(2j-1)}{2n}\right)$$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{x}{\cos^{2n} x + \sin^{2n} x}dx = \frac{\pi^2}{8n}\sum_{j=1}^{n}\binom{n-1}{j-1} \csc \left(\frac{\pi(2j-1)}{2n}\right)}$$

Euler's work V

Consequences of the series expansion of $\displaystyle \frac{1}{\sqrt{1-x}}$ vol II

We continue developing some results of series involving the binomial coefficient. This time we show the proof of the following beautiful result posted by @seriesbot_q:
$$\sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = \frac{\sqrt{3}}{2}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}-...\right)$$

Proof:

From the basic series $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n} }x^n = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!}x^{n}$$ If we make the transformation $\displaystyle x \mapsto x^2$ $$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n} }x^{2n}$$ Integrating from $0$ to $t$: $$\int_{0}^{t} \frac{1}{\sqrt{1-x^2}} dx = \arcsin(t) = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n+1) }t^{2n+1}$$ Dividing by $t$: $$\frac{\arcsin(t)}{t} = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n+1) }t^{2n}$$ Integrating again from $0$ to $\displaystyle \frac{1}{2}$: $$\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}(2n+1) }\int_{0}^{\frac{1}{2}}t^{2n}dt = \frac{1}{2}\sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2 }$$ $$\Longrightarrow \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = 2\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt \tag{1}$$
The left hand side is the series that we are looking for. Despite the fact that the integral $\displaystyle \int \frac{\arcsin x}{x}dx$ cannot be expressed as a finite combination of elementary functions, we can express it in the form of a new infinite series.

First we have to make a change of variable: $$\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \int_{0}^{\frac{\pi}{6}} \frac{w\cos(w)}{\sin(w)} dw = \int_{0}^{\frac{\pi}{6}} w\cot(w) dw\quad (w\mapsto \arcsin(t))$$ Now, recall the following series: $$\cot(w) = \lim_{t \to 0}2\sum_{k=1}^{\infty} e^{-tk}\sin(2kw)$$ This proof of this result is easy: Consider the "discrete" Laplace transform: $$\sum_{k=1}^{\infty} e^{-kt}\sin(kw) = \frac{1}{2} \frac{\sin w}{\cosh t - \cos w} \quad t>0$$ This follows from: $$\begin{align*} \sum_{k=1}^{\infty} e^{-kt}\sin(kw) =& \sum_{k=0}^{\infty} e^{-kt}\left(\frac{e^{kiw}-e^{-kiw}}{2i}\right)\\ =& \frac{1}{2i} \sum_{k=0}^{\infty} e^{-kt}\left(e^{kiw}-e^{-kiw}\right)\\ =& \frac{1}{2i} \sum_{k=0}^{\infty} e^{k(iw-t)} -\frac{1}{2} \sum_{k=0}^{\infty} e^{k(-t-iw)}\\ =& \frac{1}{2i} \frac{1}{1-e^{iw-t}}- \frac{1}{2i} \frac{1}{1-e^{-t-iw}} \\ =& \frac{1}{2i} \left(\frac{e^t}{e^t-e^{iw}}- \frac{e^t}{e^t-e^{-iw}} \right)\\ =& \frac{1}{2i} \left(\frac{e^{t+iw}-e^{t-iw}}{e^{2t}-e^{t-iw}-e^{t+iw}+1}\right)\\ =& \frac{1}{2i} \left(\frac{e^{iw}-e^{-iw}}{e^{t}+e^{-t}-(e^{-iw}+e^{iw})}\right)\\ =& \frac{1}{2} \frac{\frac{e^{iw}-e^{-iw}}{2i}}{\frac{e^{t}+e^{-t}-(e^{-iw}+e^{iw})}{2}}\\ =& \frac{1}{2} \frac{\sin w}{\cosh t - \cos w}\\ \end{align*}$$ If we take the limit as $\displaystyle t \to 0+$ $$\lim_{t \to 0+ }\sum_{k=1}^{\infty}e^{tk} \sin(kw) = \frac{1}{2} \frac{\sin w}{1 - \cos w}$$ and using the following identities $$\sin\left(\theta\right) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$$ $$1-\cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)$$ we have $$\lim_{t \to 0+} \sum_{k=1}^{\infty} e^{-tk}\sin(kw) = \frac{1}{2} \frac{\sin w}{1 - \cos w} = \frac{1}{2}\cot\left(\frac{w}{2}\right)$$ therefore $$\cot\left(w\right) = \lim_{t \to 0+} 2\sum_{k=1}^{\infty} e^{-tk}\sin(2kw) \tag{2}$$ This seems to imply that $$\cot\left(w\right) \stackrel{?}{=} 2\sum_{k=1}^{\infty} \sin(2kw)$$
However, this is not true... at least formally: note that the series $\displaystyle \sum_{k=1}^{ \infty}\sin(2kw)$ does not converge.

But we can calculate the integral taking the limit as $t \to 0+$: $$\begin{align*} I = \int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt =& \int_{0}^{\frac{\pi}{6}} \frac{w\cos(w)}{\sin(w)} dw\\ =& \int_{0}^{\frac{\pi}{6}} w\cot(w) dw\\ =& 2\int_{0}^{\frac{\pi}{6}} w\left[\lim_{t \to 0+} \sum_{k=1}^{\infty}e^{-tk}\sin(2kw)\right] dw \quad (\textrm{from (2)})\\ =& \lim_{t \to 0+} 2\sum_{k=1}^{\infty} e^{-tk}\int_{0}^{\frac{\pi}{6}} w\sin(2kw) dw\\ =& \lim_{t \to 0+} \frac{1}{2}\sum_{k=1}^{\infty}e^{-tk} \frac{1}{k^2}\int_{0}^{\frac{\pi k}{3}} s\sin(s) dw \quad (s \mapsto 2kw)\\ \stackrel{IBP}{=}& \lim_{t \to 0+} \sum_{k=1}^{\infty}e^{-tk}\frac{1}{2k^2} \left[-\frac{\pi k}{3}\cos\left(\frac{k\pi}{3}\right) + \int_{0}^{\frac{\pi k}{3}} \cos sds \right]\\ =& \lim_{t \to 0+} \sum_{k=1}^{\infty}e^{-tk}\frac{1}{2k^2} \left[\sin\left(\frac{\pi k}{3}\right)-\frac{\pi k}{3}\cos\left(\frac{k\pi}{3}\right) \right]\\ =& \lim_{t \to 0+} \frac{1}{2} \sum_{k=1}^{\infty} e^{-tk}\frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}-\lim_{t \to 0} \frac{\pi}{6}\sum_{k=1}^{\infty}e^{-tk}\frac{\cos\left(\frac{k\pi}{3}\right)}{k}\\ =& \frac{1}{2} \sum_{k=1}^{\infty}\frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}- \frac{\pi}{6}\sum_{k=1}^{\infty}\frac{\cos\left(\frac{k\pi}{3}\right)}{k}\\ \end{align*}$$ where the last equality follows from the fact that the series $$\sum_{k=1}^{\infty}\frac{\sin\left(\frac{\pi k}{3}\right)}{k^2} \quad \sum_{k=1}^{\infty}\frac{\cos\left(\frac{k\pi}{3}\right)}{k}$$ are well known Fourier series that converge.

Now we will prove that the second series is equal to zero:

From Euler's formula we have $$\sum_{k=1}^{\infty} \frac{\cos\left(\frac{k\pi}{3}\right)}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{\frac{i\pi k}{3}}}{k}+ \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{-\frac{i\pi k}{3}}}{k}$$ We can split the sum into two because the series $$\sum_{k=1}^{\infty} \frac{z^k}{k}$$ converges when $\displaystyle |z|<1$. But this series is just the Laurent series for the principal branch of the $-\ln(1-z)$ function: $$\ln(1-z) = -\sum_{k=1}^{\infty} \frac{z^k}{k}$$ Hence we can write: $$\sum_{k=1}^{\infty} \frac{\cos\left(\frac{k\pi}{3}\right)}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{\frac{i\pi k}{3}}}{k}+ \frac{1}{2}\sum_{k=1}^{\infty} \frac{e^{-\frac{i\pi k}{3}}}{k} = -\frac{1}{2}\ln\left(1-e^{\frac{i\pi}{3}}\right)-\frac{1}{2}\ln\left(1-e^{-\frac{i\pi}{3}}\right) = -i\frac{\pi}{6}+ i\frac{\pi}{6} =0$$ Therefore $$I =\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \frac{1}{2} \sum_{k=1}^{\infty} \frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}$$ If we tabulate the sequence $\displaystyle \sin\left(\frac{\pi k}{3}\right) \textrm{ for } k=1,2,3,...$ we find: $$\left(\sin\left(\frac{\pi k}{3}\right)\right)_{k} = \left(\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 0 ,-\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2},0, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2},... \right) = \frac{\sqrt{3}}{2}(1,1,0,-1-1,0,1,1,...)$$ Therefore $$I =\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt = \frac{1}{2} \sum_{k=1}^{\infty} \frac{\sin\left(\frac{\pi k}{3}\right)}{k^2}= \frac{\sqrt{3}}{4}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}+...\right) \tag{3}$$ Putting all the ingredients together: $$\begin{align*} \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = & 2\int_{0}^{\frac{1}{2}}\frac{\arcsin(t)}{t}dt \quad (\textrm{from (1)}) \\ =& \frac{\sqrt{3}}{2}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}+...\right) \quad (\textrm{from (3)}) \end{align*}$$ Therefore $$\boxed{\sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{4n}(2n+1)^2} = \frac{\sqrt{3}}{2}\left(1+\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2} + \frac{1}{7^2}+\frac{1}{8^2}-...\right) }$$

Euler's work IV

The strong connection between the $\operatorname{Li}_{2}$ function and $\phi$

Today we show the proof of this result posted by @seriesbot_q $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2\binom{2n}{n}} =\frac{\pi^2}{6}-3\ln^2(\phi)$$
Not surprisingly, this series is related to the work of Euler, specially his work on the properties of the $\arctan$ and his work on the beautiful properties that relates the dilogarithm function with the golden ratio.

Proof

In a previous post we showed the proof of the following series expansion:
$$\frac{\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}x^{2n+1} = \sum_{n=0}^{\infty} \frac{2^{2n} n!^2}{(2n+1)!}x^{2n+1}$$

This series expansion is a consequence of the Euler's formula for $\arctan x$.

If we allow x to take complex values we can transform this series in another very useful:

Let $\displaystyle x= -iv$ and recall that $\displaystyle i\arcsin (-ia) = \operatorname{arcsinh}(a)$

Hence $$\frac{\arcsin(-iv)}{\sqrt{1+v^2}} = -i\sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n+1}$$ Then, multiplying by $i$ both sides $$\frac{i\arcsin(-iv)}{\sqrt{1+v^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n+1}$$ Therefore $$\frac{\operatorname{arcsinh}(v)}{\sqrt{1+v^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n+1}$$ If we multiply by $\displaystyle \frac{1}{v}$ $$\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}} = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)}(-1)^nv^{2n}$$ Integrating from $0$ to $x$ $$\int_{0}^{x}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = \sum_{n=0}^{\infty} \frac{2^{2n}}{\binom{2n}{n}(2n+1)^2}(-1)^nx^{2n+1}$$ If we let $\displaystyle x = \frac{1}{2}$ $$\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{\binom{2n}{n}(2n+1)^2}$$ $$\Longrightarrow 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = \sum_{n=0}^{\infty} \frac{(-1)^n}{\binom{2n}{n}(2n+1)^2}$$
The series in the right hand side is the series that we want in a closed-expression form. Therefore we have to find the value of the integral in the left hand side:

Recall that
$$\operatorname{arcsinh}(a) = \ln\left(a+\sqrt{1+a^2}\right)$$ Therefore, if we denote $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$ the golden ratio: $$\begin{align*} I = 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv =& 2\int_{0}^{\frac{1}{2}}\frac{\ln\left(v+\sqrt{1+v^2}\right)}{v\sqrt{1+v^2}}dv\\ =& 4\int_{1}^{\phi }\frac{\ln(w)}{(w+1)(w-1)}dw \quad \left(w \mapsto v+\sqrt{1+v^2} \right)\\ =& \underbrace{2\int_{1}^{\phi} \frac{\ln(w)}{w-1}dw}_{I_{1}} -\underbrace{2\int_{1}^{\phi} \frac{\ln(w)}{w+1}dw}_{I_{2}} \end{align*}$$ Recall the definition of the dilogarithm function: $$\operatorname{Li}_{2}(z) = \int_{z}^{0} \frac{\ln(1-t)}{t} dt$$ Hence $$\begin{align*} I_{1}= 2\int_{1}^{\phi} \frac{\ln(w)}{w-1}dw =& 2\int_{0}^{\phi-1} \frac{\ln(1+s)}{s}ds \quad (s \mapsto w-1)\\ =& -2\int_{1-\phi}^{0} \frac{\ln(1-r)}{r}ds \quad (s \mapsto -w)\\ =&-2\operatorname{Li}_{2}(1-\phi)\\ =&-2\operatorname{Li}_{2}(-\phi^{-1}) \quad (1-\phi = -\phi^{-1}) \end{align*}$$ For $I_{2}$: $$\begin{align*} I_{2}= 2\int_{1}^{\phi} \frac{\ln(w)}{w+1}dw \stackrel{IBP}{=} &2\ln(w)\ln(w+1)\Big|_{1}^{\phi} - 2\int_{1}^{\phi}\frac{\ln(w+1)}{w}dw \\ =& 2\ln(\phi)\ln(\phi+1) + 2\int_{-\phi}^{-1}\frac{\ln(1-s)}{s}ds \quad (s\mapsto -w)\\ =& 2\ln(\phi)\ln(\phi+1) + 2\int_{-\phi}^{0}\frac{\ln(1-s)}{s}ds - 2\int_{-1}^{0}\frac{\ln(1-s)}{s}ds\\ =& 2\ln(\phi)\ln(\phi+1) + 2\operatorname{Li}_{2}(-\phi) - 2\operatorname{Li}_{2}(-1)\\ \end{align*}$$ Therefore $$I= 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv = -2\operatorname{Li}_{2}(-\phi^{-1}) - 2\ln(\phi)\ln(\phi+1) - 2\operatorname{Li}_{2}(-\phi) + 2\operatorname{Li}_{2}(-1)$$
A famous result for the dilogarithm function states that $\displaystyle \operatorname{Li}_{2}(-1)=-\frac{1}{12}\pi^2$
Now, this is the interesting part: it turned out that the relationship between the dilogarithm function and the golden rate is very strong. Euler proved the following identities that relate both concepts (I will add the proof of this results in an appendix later): $$\operatorname{Li}_{2}(-\phi^{-1}) = -\frac{1}{15}\pi^2 + \frac{1}{2}\ln^2(\phi)$$ $$\operatorname{Li}_{2}(-\phi) -\frac{1}{10}\pi^2 -\ln^2(\phi)$$ Here you can find more of this identities. Additionally, we can make the following transformation: $$\begin{align*} \ln(\phi)\ln(1+\phi) =& \ln(\phi)\ln\left(\phi(1+\phi^{-1})\right)\\ =& \ln^2(\phi)+ \ln(\phi)\phi(1+\phi^{-1}) \end{align*}$$ where $$1+\phi^{-1} = 1+\frac{2}{1+\sqrt{5}}\frac{1-\sqrt{5}}{1-\sqrt{5}}=1 - \frac{1-\sqrt{5}}{2} = \frac{1+\sqrt{5}}{2} = \phi$$ Hence $$\ln(\phi)\ln(1+\phi) = \ln^2(\phi)+ \ln(\phi)\ln(1+\phi^{-1}) = 2\ln^2(\phi)$$ Putting all the ingredients togheter: $$\begin{align*} I= 2\int_{0}^{\frac{1}{2}}\frac{\operatorname{arcsinh}(v)}{v\sqrt{1+v^2}}dv =& \underbrace{-2\operatorname{Li}_{2}(-\phi^{-1})}_{\frac{2}{15}\pi^2-\ln^2(\phi)} \underbrace{-2\ln(\phi)\ln(\phi+1)}_{-4\ln^2(\phi)} \underbrace{-2\operatorname{Li}_{2}(-\phi)}_{\frac{2}{10}\pi^2+2\ln^2(\phi)} + \underbrace{2\operatorname{Li}_{2}(-1)}_{-\frac{\pi^2}{6}}\\ =& \frac{\pi^2}{6}-3\ln^2(\phi) \end{align*}$$ Then we can conclude: $$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2\binom{2n}{n}} =\frac{\pi^2}{6}-3\ln^2(\phi)}$$

Euler's work III

Series involving $\zeta(2)$ and Fibonacci numbers

Today we show the proof of this results posted by @seriesbot_q and @infseriesbot $$3\sum_{k=1}^{\infty} \frac{1}{k^2\binom{2k}{k}}= \frac{\pi^2}{6} = \zeta(2)$$ $$\sum_{k=1}^{\infty} \frac{F_{2k}}{k^2\binom{2k}{k}} = \frac{4\pi^2}{25\sqrt{5}}$$ Both results are particular values of certain series expansion deeply related to the work of Euler. $$\sum_{k=1}^{\infty} \frac{r^k}{k^2\binom{2k}{k}} = 2\arcsin^2\left(\frac{\sqrt{r}}{2}\right)$$ From which we can also conclude $$\sum_{k=1}^{\infty} \frac{2^k}{k^2\binom{2k}{k}} = \frac{\pi^2}{8}$$ $$\sum_{k=1}^{\infty} \frac{3^k}{k^2\binom{2k}{k}} = \frac{3\pi^2}{9}$$

Proof

First, we will obtain the series expansion for $\displaystyle \arcsin^2(y)$:
$$\arcsin^2(y)= \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!}(2y)^{2k}=\frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}y^{2k}}{k^2\binom{2k}{k}}$$

This expansion will allow us to find closed-form expressions for the series.

To obtain the series expansion we will rely on the Euler's formula for $\arctan(x)$. We have been using the formula extensively in the past (you can find the proof in one of the propositions of this post) and we need to use it again.

Recall the beautiful Euler's formula: $$\arctan (x) = \frac{x}{1+x^2} \sum_{n=0}^{\infty}\left[\frac{(2n)!!}{(2n+1)!!} \left(\frac{x^2}{1+x^2}\right)^{n}\right]$$ Now consider the following identity for inverse trigonometric functions: $$\arcsin(y) = \arctan\left(\frac{y}{\sqrt{1-y^2}}\right)$$ For $x= \displaystyle \frac{y}{\sqrt{1-y^2}}$ we have $$\begin{align*} \arcsin(y) =& \arctan\left(\frac{y}{\sqrt{1-y^2}}\right)\\ =& y\sqrt{1-y^2}\left[\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}y^{2n}\right]\\ =& y\sqrt{1-y^2}\left[1+\frac{2}{3}y^2 +\frac{2}{3}\frac{4}{5}y^4 + \frac{2}{3}\frac{4}{5}\frac{6}{7}y^6...\right]\\ \Longrightarrow \frac{\arcsin(y)}{\sqrt{1-y^2} }=& \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}y^{2n+1}\\ =& y+\frac{2}{3}y^3+\frac{2}{3}\frac{4}{5}y^5+\frac{2}{3}\frac{4}{5}\frac{6}{7}y^7+... \end{align*}$$ Now consider $$\begin{align*} \int_{0}^{y} \frac{\arcsin(t)}{\sqrt{1-t^2}dt} =& \int_{0}^{\arcsin(y)} udu \quad (u \mapsto \arcsin(t))\\ =& \left[\frac{u^2}{2}\right]_{0}^{\arcsin(y)} = \frac{\arcsin^2(y)}{2} \end{align*}$$ Therefore $$\begin{align*} \frac{\arcsin^2(y)}{2} =\int_{0}^{y}\frac{\arcsin(y)}{\sqrt{1-y^2} }dy=& \int_{0}^{y}\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}y^{2n+1}dy\\ & \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}\int_{0}^{y}y^{2n+1}dy\\ & \sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!(2n+2)}y^{2n+2}\\ =& \sum_{k=1}^{\infty} \frac{1}{2k}\frac{(2k-2)!!}{(2k-1)!!}y^{2k} \quad (n=k-1)\\ \end{align*}$$ $$\begin{align*} \Longrightarrow \arcsin^2(y) =& \sum_{k=1}^{\infty} \frac{1}{k}\frac{(2k-2)!!}{(2k-1)!!}y^{2k}\\ =& \sum_{k=1}^{\infty}\frac{1}{k} \frac{2^{k-1}(k-1)!2^k k!}{(2k)!}y^{2k}\\ =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!}(2y)^{2k} \\ =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k)!^2}{k^2(2k)!}2^{2k}y^{2k}\\ =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}y^{2k}}{k^2\binom{2k}{k}} \end{align*}$$ $$\boxed{\therefore\arcsin^2(y)= \frac{1}{2}\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!}(2y)^{2k}=\frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}y^{2k}}{k^2\binom{2k}{k}}}$$ Putting $\displaystyle y=\frac{1}{2}$ then $\displaystyle \arcsin^2\left(\frac{1}{2}\right)= \frac{\pi^2}{36}$
Then, we have our first result:
$$\boxed{3\sum_{k=1}^{\infty} \frac{(k-1)!^2}{(2k)!} = 3\sum_{k=1}^{\infty} \frac{1}{k^2\binom{2k}{k}}= \frac{\pi^2}{6} = \zeta(2)}$$
Now recall the closed-form expression for the Fibonacci number $\displaystyle F_{j}$:
$$F_{j} = \frac{\phi^{j}-(-\phi)^{-j}}{\sqrt{5}}$$
where $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio

Therefore if we put $\displaystyle y = \frac{\sqrt{5}+1}{4} \Longrightarrow \arcsin^2\left(\frac{\sqrt{5}+1}{4} \right) =\frac{9\pi^2}{100}$
Hence $$\frac{9\pi^2}{100}= \frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}\left(\frac{\sqrt{5}+1}{4}\right)^{2k}}{k^2\binom{2k}{k}} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{\phi^{2k}}{k^2\binom{2k}{k}}$$ Now if we put $\displaystyle y = -\frac{1}{\sqrt{5}+1} \Longrightarrow \arcsin^2\left( -\frac{1}{\sqrt{5}+1}\right) =\frac{\pi^2}{100}$
Hence $$\frac{\pi^2}{100}= \frac{1}{2}\sum_{k=1}^{\infty} \frac{2^{2k}\left( -\frac{1}{\sqrt{5}+1}\right)^{2k}}{k^2\binom{2k}{k}} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{(-\phi)^{-2k}}{k^2\binom{2k}{k}}$$
Therefore $$\sum_{k=1}^{\infty} \frac{\phi^{2k}}{k^2\binom{2k}{k}}- \sum_{k=1}^{\infty} \frac{(-\phi)^{-2k}}{k^2\binom{2k}{k}} = \sum_{k=1}^{\infty} \frac{\phi^{2k}- (-\phi)^{-2k}}{k^2\binom{2k}{k}} = \frac{8\pi^2}{50} = \frac{4\pi^2}{25}$$ Then $$\frac{1}{\sqrt{5}}\sum_{k=1}^{\infty} \frac{\phi^{2k}- (-\phi)^{-2k}}{k^2\binom{2k}{k}} = \frac{4\pi^2}{25\sqrt{5}}$$
Hence, we can conclude $$\boxed{ \sum_{k=1}^{\infty} \frac{F_{2k}}{k^2\binom{2k}{k}} = \frac{4\pi^2}{25\sqrt{5}}}$$