Elliptic integrals II

Series involving $K(p)$ and the Catalan constant

Today we show the proof of this result posted by @seriesbot_q on Twitter: $$\beta(2) = \frac{\pi}{4}\sum_{n=0}^{\infty}\frac{ \binom{2n}{n}^2}{2^{4n}(2n+1)}$$ This series turned out being a slight variation of the series expansion of the complete elliptic integral of the first kind $K(p)$ .

Proof

Recall the definition of the complete elliptic integral of the first kind: $$K(p) = \int_{0 }^{\frac{\pi}{2}} \frac{1}{\sqrt{1-p^2 \sin^2 \theta}} d\theta \quad |p|<1$$

First, we are going to obtain its series expansion.

In a previous post we proved the following result: $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}x^n \quad 0\leq x <1$$ We know that $$-1\leq \sin \theta \leq 1$$ $$|p|<1$$ $$\Longrightarrow p^2\sin^2 \theta < 1$$ Hence $$\frac{1}{\sqrt{1-p^2\sin^2 \theta}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}p^{2n}\sin^{2n}(\theta)$$ Integrating from $0$ to $\displaystyle \frac{\pi}{2}$ $$\begin{align*} K(p)= \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-p^2\sin^2 \theta}}d\theta = & \int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}p^{2n} \frac{(2n-1)!!}{(2n)!!}\sin^{2n}(\theta)d\theta\\ =& \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}p^{2n}\int_{0}^{\frac{\pi}{2}}\sin^{2n}(\theta)d\theta\\ =& \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}p^{2n}\int_{0}^{1}w^{2n}(1-w^2)^{-\frac{1}{2}} dw \quad (w \mapsto \sin\theta )\\ =& \frac{1}{2}\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}p^{2n}\int_{0}^{1}s^{n+\frac{1}{2}-1}(1-s)^{\frac{1}{2}-1} ds \quad (s \mapsto w^2 )\\ =& \frac{1}{2}\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}p^{2n}B\left(n+\frac{1}{2},\frac{1}{2}\right)\\ =& \frac{1}{2}\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} p^{2n}\frac{\Gamma\left(n+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(n+1)} \end{align*}$$ From the properties of the double factorial we have: $$(2n-1)!! = 2^{n} \frac{\Gamma\left(n+\frac{1}{2}\right) }{\sqrt{\pi} }$$ $$(2n)!! = 2^{n}n! = 2^{n} \Gamma(n+1)$$ Hence $$\begin{align*} K(p)= \frac{1}{2}\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} \frac{\Gamma\left(n+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(n+1)} p^{2n}= & \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n-1)!!^2}{(2n)!!^2} p^{2n} \end{align*}$$ Recall the we can also write $$\frac{(2n-1)!!}{(2n)!!} = \frac{ \binom{2n}{n}}{2^{2n}}$$ Therefore, this is the expansion series of the complete elliptic integral of the first kind: $$K(p) = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n-1)!!^2}{(2n)!!^2} p^{2n} = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{ \binom{2n}{n}}{2^{4n}}^2 p^{2n}$$ Now, integrating from $0$ to $1$ $$\int_{0}^{1}K(p)dp = \frac{\pi}{2}\sum_{n=0}^{\infty}\frac{ \binom{2n}{n}^2}{2^{4n}(2n+1)}$$ The right hand side is the series that we are looking for. Then, we just have to find the solution to the integral in the left hand side: $$\begin{align*}\int_{0}^{1}K(p)dp =& \int_{0}^{1}\int_{0 }^{\frac{\pi}{2}} \frac{1}{\sqrt{1-p^2 \sin^2 \theta}}d\theta dp\\ =& \int_{0 }^{\frac{\pi}{2}} \int_{0}^{1} \frac{1}{\sqrt{1-p^2 \sin^2 \theta}}dpd\theta\\ =& \int_{0 }^{\frac{\pi}{2}} \int_{0}^{\sin \theta} \frac{1}{\sqrt{1-w^2}\sin \theta }dwd\theta \quad (w \mapsto p\sin \theta )\\ =& \int_{0 }^{\frac{\pi}{2}} \frac{1}{\sin \theta} \int_{0}^{\sin \theta} \frac{1}{\sqrt{1-w^2} }dwd\theta \quad\\ =& \int_{0 }^{\frac{\pi}{2}} \frac{1}{\sin \theta} \arcsin(\sin \theta) d\theta \\ =& \int_{0 }^{\frac{\pi}{2}} \frac{\theta}{\sin \theta} d\theta \\ \end{align*}$$ We solved this last integral in another post and the solution turned out being $2 \beta(2)$. Hence $$\int_{0}^{1}K(p)dp = 2\beta(2)$$ Therefore $$\boxed{ \beta(2) = \frac{\pi}{4}\sum_{n=0}^{\infty}\frac{ \binom{2n}{n}^2}{2^{4n}(2n+1)} }$$

Taylor's theorem II

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

Today we show the proof of the following series found on Twitter: $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}x^n = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)$$ $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n^2} = \sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!n^2} = \frac{\pi^2}{6} -2\ln^2(2)$$ In reality these series are just a transformation of a very simple series as we will show

Proof

We previously showed the proof of the following series expansion with the help of Pochhammer polynomials: $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^n = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}}x^n ={}_{2}F_{1}\left(\frac{1}{2},1;1;x\right)$$ This series, as simple as it is, will allow us to calculate other integrals and series much easier and this is one of those cases.
We will start subtracting $1$ from each side of the equation: $$\frac{1}{\sqrt{1-x}}-1 = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^n$$ Now divide by $x$ $$\frac{1}{x\sqrt{1-x}}-\frac{1}{x} = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^{n-1}$$ Integrating from $0$ to $x$: $$\int_{0}^{x}\frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!} \int_{0}^{x} x^{n-1}dx = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx$$ The integral in the left hand side has the following antiderivative: $$\int \frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = -2\ln\left(\sqrt{1-x}+1\right)+C$$ Hence $$\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = -2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2) = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)$$ This is our first result: $$\boxed{\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)}$$ If we divide by $x$ again we have: $$\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n-1}dx = \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x}$$ Integrating from $0$ to $1$: $$\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n^2(2n)!!} = \int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx$$ We can also write the left hand side with this form: $$\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n^2(2n)!!} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{n^22^{2n}}$$ This follow from the fact that: $$(2n-1)!! = \frac{(2n)!}{2^n n!}$$ $$(2n)!! = 2^n n!$$ $$\Longrightarrow \frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{2^{2n}n!^2} = \frac{1}{2^{2n}}\binom{2n}{n}$$ Therefore: $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{n^22^{2n}} = \int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx \tag{1}$$ Integrating by parts: Let $$du = \frac{1}{x} \quad v = -2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)$$ then $$u = \ln(x) \quad dv = \frac{1}{(\sqrt{1-x}+1)\sqrt{1-x}}$$ $$\begin{align*} \int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx =& \ln(x)\left(-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)\right)\Big|_{0}^{1} - \int_{0}^{1} \frac{\ln(x)}{(\sqrt{1-x}+1)\sqrt{1-x}}\\ =& -\int_{0}^{1} \frac{\ln(x)}{(\sqrt{1-x}+1)\sqrt{1-x}}\\ =& -2\int_{0}^{1} \frac{\ln(1-w^2)}{w+1} dw\\ =& -2\int_{0}^{1} \frac{\ln\left((1+w)(1-w)\right)}{w+1} dw\\ =& -2\underbrace{\int_{0}^{1} \frac{\ln(1+w)}{w+1} dw}_{I_{1}} - 2\underbrace{\int_{0}^{1} \frac{\ln(1-w)}{w+1} dw}_{I_{2}} \tag{1} \end{align*}$$ The integral $I_{1}$ is easily solvable given that $$\int \frac{\ln(1+w)}{w+1} dw = \frac{1}{2} \ln^2(1+w) + C$$ Hence $$\displaystyle I_{1} = \frac{1}{2} \ln^2(2) \tag{3}$$ For $I_{2}$: $$\begin{align*} I_{2}= \int_{0}^{1} \frac{\ln(1-w)}{w+1} dw =& \int_{0}^{1} \sum_{n=0}^{\infty}(-w)^n\ln(1-w) dw\\ =& \sum_{n=0}^{\infty}(-1)^n\int_{0}^{1} w^n\ln(1-w) dw\\ =& \sum_{n=0}^{\infty}(-1)^n\int_{0}^{1} w^n\left(\frac{d}{dt}\Big|_{t=0+}(1-w)^t\right) dw\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^n\int_{0}^{1} w^n(1-w)^t dw\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^nB(n+1,t+1)\\ =& \lim_{t \to 0+} \sum_{n=0}^{\infty}(-1)^n \left(\psi^{(0)}(t+1)-\psi^{(0)}(n+t+2)\right)B(n+1,t+)\\ =& \sum_{n=0}^{\infty}(-1)^n \left(\psi^{(0)}(1)-\psi^{(0)}(n+2)\right)B(n+1,1)\\ =& \sum_{n=0}^{\infty}(-1)^{n+1} \frac{H_{n+1}}{n+1}\\ =& \sum_{j=1}^{\infty}(-1)^{j} \frac{H_{j}}{j} \quad (j=n+1) \end{align*}$$ Now recall the generating function for the harmonic numbers: $$\sum_{n=1}^{\infty}H_{n}z^n = -\frac{\ln(1-z)}{1-z}$$ Dividing by $z$ $$\sum_{n=1}^{\infty}H_{n}z^{n-1} = -\frac{\ln(1-z)}{z(1-z)} = -\frac{\ln(1-z)}{z} -\frac{\ln(1-z)}{1-z}$$ Integrating from $0$ to $-1$ $$\sum_{n=1}^{\infty}\frac{H_{n}(-1)^{n}}{n} = -\int_{0}^{-1}\frac{\ln(1-z)}{z}dz -\int_{0}^{-1}\frac{\ln(1-z)}{1-z} = -\int_{0}^{1}\frac{\ln(1+z)}{z}dz +\int_{0}^{1}\frac{\ln(1+z)}{1+z}$$ Hence $$\sum_{n=1}^{\infty}\frac{H_{n}(-1)^{n}}{n} = -\int_{0}^{1}\frac{\ln(1+z)}{z}dz +I_{1} \tag{4}$$ We just have to calculate $\displaystyle \int_{0}^{1}\frac{\ln(1+z)}{z}dz$ $$\begin{align*} \int_{0}^{1}\frac{\ln(1+z)}{z}dz =& \displaystyle \int_{0}^{1}\sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{k-1}}{k}dz\\ =& \sum_{k=1}^{\infty} (-1)^{k+1}\int_{0}^{1}\frac{z^{k-1}}{k}dz\\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2} \quad \textrm{(Dirichlet eta function)}\\ =& \eta(2)\\ =& \frac{\pi^2}{12} \end{align*}$$ Therefore $$I_{2}= \sum_{n=1}^{\infty}\frac{H_{n}(-1)^{n}}{n} = -\frac{\pi^2}{12}+ \frac{1}{2} \ln^2(2) \tag{5}$$ Hence, from (1),(2),(3),(4),(5) we can conclude: $$\int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx = -2I_{1} - 2I_{2} =\frac{\pi^2}{6} -2\ln^2(2)$$ Therefore
$$\boxed{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n^2} = \sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!n^2} = \frac{\pi^2}{6} -2\ln^2(2)}$$ $$\boxed{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}x^n = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)}$$

Integral of the day II

Integral related to the Sinc function

We show the proof of this mysterious integral:
$$\int_{0}^{\infty}\frac{\sin x}{x} \ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) = \pi \ln\left( \frac{\sqrt{a^2+1}+a}{\sqrt{b^2+1}+b}\right) \quad |a|,|b|> 1$$
The integral resisted various techniques (residues, laplace transform) but following a hint suggested by @mvs_rpi we managed to obtain the proof: It turned out that this integral is a Lobachevsky integral.

The Lobachevsky integral formula states that:

If $f(x)$ is a continous function satisfying the $\pi$-periodic assumption $f(x+\pi) = f(x)$, and $f(\pi-x)=f(x)$, for $0\leq x <\infty$

then
$$\int_{0}^{\infty} \frac{\sin^2 x}{x^2}f(x) dx = \int_{0}^{\infty} \frac{\sin x}{x} f(x)dx = \int_{0}^{\frac{\pi}{2}} f(x) dx$$
Clearly
$$f(x) = \ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right)$$
is a $\pi-$ periodic function.

Therefore we just have to calculate
$$\int_{0}^{\frac{\pi}{2}}\ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) dx$$
If we split the integral in two:
$$\int_{0}^{\frac{\pi}{2}}\ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) dx = \underbrace{\int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx}_{I_{1}} - \underbrace{\int_{0}^{\frac{\pi}{2}} \ln(b^2+\cos^2 x)dx}_{I_{2}}$$
if we find $I_{1}$ automatically we also have $I_{2}$


Proof:

First, we will show that $$\int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx = \pi \ln(\sqrt{a^2+1} +a)-\pi\ln(2) \quad |a|>1$$ $$\begin{align*} I_{1} = \int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx =& \int_{0}^{\frac{\pi}{2}} \ln\left(a^2\left(1+\frac{1}{a^2}\cos^2 x\right)\right)dx\\ =& \int_{0}^{\frac{\pi}{2}}\ln(a^2)dx + \int_{0}^{\frac{\pi}{2}}\ln\left(1+\frac{1}{a^2}\cos^2 x\right)dx\\ =& \pi\ln(a) + \int_{0}^{\frac{\pi}{2}}\sum_{k=1}^{\infty} \left[\frac{(-1)^{k+1}}{k} \frac{\cos^{2k}x}{a^{2k}} \right] dx \quad (|a|>1)\\ =& \pi\ln(a) + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{ka^{2k}} \int_{0}^{\frac{\pi}{2}}\cos^{2k} x dx\\ =& \pi\ln(a) - \sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \int_{0}^{\frac{\pi}{2}} w^{2k}(1-w^2)^{-\frac{1}{2}} dw \quad (w \mapsto \cos x)\\ =& \pi\ln(a) - \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \int_{0}^{\frac{\pi}{2}} s^{k+\frac{1}{2}-1}(1-s)^{\frac{1}{2}-1} ds \quad (s \mapsto w^2)\\ =& \pi\ln(a) - \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k B\left(k+\frac{1}{2}, \frac{1}{2}\right)\\ =& \pi\ln(a) - \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(k+1)}\\ \end{align*}$$ Recall that $$\Gamma\left(k+\frac{1}{2}\right) = \frac{(2k-1)!!\sqrt{\pi}}{2^k}$$ $$\Gamma(k+1)= \frac{(2k)!!}{2^k}$$ Hence $$\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma(k+1)} = \frac{(2k-1)!!\sqrt{\pi}}{(2k)!!}$$ Therefore $$I_{1} = \pi\ln(a) - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \frac{(2k-1)!!}{(2k)!!}$$ In another post we proved that $$\frac{1}{\sqrt{1-x}} = \sum_{k=0}^{\infty} \frac{(2k-1)!!}{(2k)!!} x^k$$ Then, substracting $-1$ and dividing by $\displaystyle \frac{1}{x}$ afterwards (the same trick we used in this post): $$\frac{1}{\sqrt{1-x}}-1 = \sum_{k=1}^{\infty} \frac{(2k-1)!!}{(2k)!!} x^k$$ $$\Longrightarrow \frac{1}{x\sqrt{1-x}}-\frac{1}{x} = \sum_{k=1}^{\infty} \frac{(2k-1)!!}{(2k)!!} x^{k-1}$$ Integrating: $$\int_{0}^{x}\frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = \sum_{k=1}^{\infty} \frac{(2k-1)!!}{(2k)!!} \int_{0}^{x}x^{k-1}dx = \sum_{k=1}^{\infty}\frac{1}{k} \frac{(2k-1)!!}{(2k)!!} x^k$$ But note that $$\int\frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = -2\ln\left(\sqrt{1-x}+1\right)+C$$ Therefore $$\sum_{k=1}^{\infty}\frac{1}{k} \frac{(2k-1)!!}{(2k)!!} x^k = -2 \ln\left(\sqrt{1-x}+1\right)+2\ln(2)$$ If we put $x = -\frac{1}{a^2}$ $$\sum_{k=1}^{\infty}\frac{1}{k} \frac{(2k-1)!!}{(2k)!!} \left(-\frac{1}{a^2}\right)^k = -2 \ln\left(\sqrt{1+\frac{1}{a^2}}+1\right)+2\ln(2)$$ Therefore $$\begin{align*} I_{1} =& \pi\ln(a) - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{1}{k} \left(-\frac{1}{a^2}\right)^k \frac{(2k-1)!!}{(2k)!!} \\ =& \pi\ln(a)+ \pi \ln\left(\sqrt{1+\frac{1}{a^2}}+1\right)-\pi \ln(2)\\ =& \pi\ln(a)+ \pi \ln\left(\sqrt{\frac{a^2+1}{a^2}}+\frac{a}{a}\right)-\pi \ln(2)\\ =& \pi\ln(a)+ \pi \ln\left(\frac{1}{a} (\sqrt{a^2+1}+a)\right)-\pi \ln(2)\\ =& \pi\ln(a) -\pi \ln(a) + \pi \ln\left( (\sqrt{a^2+1}+a)\right)-\pi \ln(2)\\ =& \pi \ln\left( \sqrt{a^2+1}+a\right)-\pi \ln(2) \end{align*}$$ Therefore $$I_{1}= \int_{0}^{\frac{\pi}{2}} \ln(a^2+\cos^2 x)dx = \pi \ln\left( \sqrt{a^2+1}+a\right)-\pi \ln(2)$$ $$I_{2}= \int_{0}^{\frac{\pi}{2}} \ln(b^2+\cos^2 x)dx = \pi \ln\left( \sqrt{b^2+1}+b\right)-\pi \ln(2)$$ Hence, we have the desired result $$\int_{0}^{\frac{\pi}{2}}\ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) dx = \pi \ln\left( \frac{\sqrt{a^2+1}+a}{\sqrt{b^2+1}+b}\right)$$ By the Lobachevsky integral formula: $$\boxed{\int_{0}^{\infty}\frac{\sin x}{x} \ln\left(\frac{a^2+\cos^2 x}{b^2+\cos^2 x}\right) = \pi \ln\left( \frac{\sqrt{a^2+1}+a}{\sqrt{b^2+1}+b}\right) \quad |a|,|b|>1}$$

Laplace transform III

Laplace transform of the Sinhc and Sinc functions

We show the proof of the Laplace transforms of these slight variations of the Sinc and the Sinhc function from two problems posted by @BossGercek here and here. Both can be obtained from each other if we allow the parameters $\displaystyle b,c$ to have complex values. $$\int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = \ln\left(\sqrt{\frac{a+b}{a-b}}\right)$$ $$\int_{0}^{\infty} \frac{e^{-ax}\sin cx}{x} dx = -i\ln \left(\sqrt{\frac{a+ic}{a-ic}}\right)$$ Taking care where the inverse hyperbolic and trigonometric functions are defined, if we put $b=c=1$ we get the classic values for the Laplace transform of the Sinc and Sinhc functions: $$\int_{0}^{\infty} \frac{e^{-ax}\sinh x}{x} dx = \operatorname{arccoth} (a)$$ $$\int_{0}^{\infty} \frac{e^{-ax}\sin x}{x} dx = \frac{\pi}{2} - \arctan(a) = \operatorname{arccot}(a)$$ If we put $a=1$ and maintain $b,c$ as free parameters we have: $$\int_{0}^{\infty} \frac{e^{-x}\sinh bx}{x} dx = \operatorname{arctanh}(b)$$ $$\int_{0}^{\infty} \frac{e^{-x}\sin cx}{x} dx = \arctan(c)$$ From the above we have the particular value: $$\int_{0}^{\infty} \frac{e^{-x}\sin x}{x} dx = \frac{\pi}{4}$$

Proof

Differtiating under the integral sign:
$$\frac{d}{da} \int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = -\int_{0}^{ \infty} e^{-ax}\sinh bx dx = - \left\{\mathscr{L} \sinh bx \right\}\left(a\right)$$
Hence
$$\begin{align*} \left\{\mathscr{L} \sinh bx \right\}\left(a\right) =& \int_{0}^{\infty } e^{-ax} \left(\frac{e^{bx}-e^{-bx}}{2}\right)dx \\ = & \frac{1}{2} \int_{0}^{\infty } e^{-x(a-b)} dx - \frac{1}{2} \int_{0}^{\infty } e^{-x(a+b)} dx \\ =& \frac{1}{2}\left(\frac{1}{a-b} - \frac{1}{a+b}\right)\\ \end{align*}$$
therefore
$$\int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = -\int \frac{1}{2}\left(\frac{1}{a-b} - \frac{1}{a+b}\right)da + C = \ln\left(\sqrt{\frac{a+b}{a-b}}\right) +C$$
Taking the limit as $b \to 0 \Longrightarrow \sin bx \to 0 \Longrightarrow C = 0$
$$\boxed{ \int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = \ln\left(\sqrt{\frac{a+b}{a-b}}\right)}$$
For the Lapalce transform of the Sinc function:

Recall that
$$\sin x = -i \sinh ix$$
Therefore if we allow $\displaystyle b = ic$ :
$$\int_{0}^{\infty} \frac{e^{-ax}\sinh ic x}{x} dx = \ln \left(\sqrt{\frac{a+ic}{a-ic}}\right) \Longrightarrow -i\int_{0}^{\infty} \frac{e^{-ax}\sinh ic x}{x} dx = -i\ln\left(\sqrt{\frac{a+ic}{a-ic}}\right)$$
Hence $$\boxed{\int_{0}^{\infty} \frac{e^{-ax}\sin cx}{x} dx = -i\ln \left(\sqrt{\frac{a+ic}{a-ic}}\right)}$$

Now we will show a particular value:

Recall that in the complex plane $\displaystyle \arctan(z) = -\frac{i}{2}\ln\left(\frac{1+iz}{1-iz}\right)$ Hence if $\displaystyle a=1$ $$\int_{0}^{\infty} \frac{e^{-x}\sin cx}{x} dx = -i\ln \left(\sqrt{\frac{1+ic}{1-ic}}\right) = \arctan(c)$$ Moreover, if $c=1$ $$\int_{0}^{\infty} \frac{e^{-x}\sin x}{x} dx = -i\ln \left(\sqrt{\frac{1+ic}{1-ic}}\right) = \arctan(1) = \frac{\pi}{4}$$ Therefore $$\boxed{ \int_{0}^{\infty} \frac{e^{-x}\sin x}{x} dx = \frac{\pi}{4} }$$

Laplace transform II

Laplace transform related to an infinite product

We show the solution of this integral posted on Twitter by @diegorattaggi. The solution turned out to be the logarithm of the golden ratio. We also find a general formulation for this integral wich turned out to be a Laplace transform with a beutiful solution. $$\boxed{\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = \ln\left(\frac{1+\sqrt{5}}{2}\right)}$$ As a corollary we have the Laplace transform: $$\boxed{\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx = \ln\left[ \frac{\Gamma\left(\frac{1+s}{5}\right) \Gamma\left(\frac{4+s}{5}\right)}{\Gamma\left(\frac{2+s}{5}\right)\Gamma\left(\frac{3+s}{5}\right)} \right] }$$ Moreover, the integral and the Laplace transform can be further generalized:

If $a_{1}+a_{2}+...+a_{n} = b_{1}+b_{2}+...+b_{n}=c \;$ with $\; a_{i},b_{i} \in \mathbb{N}\setminus\left\{0\right\}$
Hence $$\boxed{ \int_{0}^{\infty} \frac{e^{-a_{1}x}+e^{-a_{2}x}+...+e^{-a_{n}x} -e^{-b_{1}x}-e^{-b_{2}x}-...-e^{-b_{n}x}}{x(1-e^{-cx})}dx = \ln\left(\frac{\Gamma\left(\frac{a_{1}}{c}\right)\Gamma\left(\frac{a_{2}}{c}\right)\cdots\Gamma\left(\frac{a_{n}}{c}\right)}{\Gamma\left(\frac{b_{1}}{c}\right)\Gamma\left(\frac{b_{2}}{c}\right)\cdots\Gamma\left(\frac{b_{n}}{c}\right)}\right)}$$ $$\boxed{ \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-a_{1}x}+e^{-a_{2}x}+...+e^{-a_{n}x} -e^{-b_{1}x}-e^{-b_{2}x}-...-e^{-b_{n}x}}{x(1-e^{-cx})}\right]dx = \ln\left(\frac{\Gamma\left(\frac{a_{1}+s}{c}\right)\Gamma\left(\frac{a_{2}+s}{c}\right)\cdots\Gamma\left(\frac{a_{n}+s}{c}\right)}{\Gamma\left(\frac{b_{1}+s}{c}\right)\Gamma\left(\frac{b_{2}+s}{c}\right)\cdots\Gamma\left(\frac{b_{n}+s}{c}\right)}\right)}$$

Proof
We show the proof of the particular case, the generalization can be shown in a similar manner.

Consider the Laplace transform:
$$\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx$$ Then
$$\frac{d}{ds} \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx = -\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{(1-e^{-5x})}\right] dx$$ Hence
$$\begin{align*} \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{(1-e^{-5x})}\right] dx = & \int_{0}^{\infty} e^{-sx} \sum_{n=0}^{\infty} e^{-5nx} (e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}) dx\\ =& \sum_{n=0}^{\infty}\int_{0}^{\infty} e^{-x(s+5n)} (e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}) dx\\ =& \sum_{n=0}^{\infty}\int_{0}^{\infty} \left[ e^{-x(s+5n+1)} -e^{-x(s+5n+2)} - e^{-x(s+5n+3)} + e^{-x(s+5n+4)} \right]dx\\ =& \sum_{n=0}^{\infty}\left[\int_{0}^{\infty} e^{-x(s+5n+1)} dx -\int_{0}^{\infty} e^{-x(s+5n+2)}dx - \int_{0}^{\infty} e^{-x(s+5n+3)}dx + \int_{0}^{\infty} e^{-x(s+5n+4)}dx \right]\\ =& \sum_{n=0}^{\infty}\int_{0}^{\infty} \left[ e^{-x(s+5n-1)} -e^{-x(s+5n-2)} - e^{-x(s+5n-3)} + e^{-x(s+5n-4)} \right]dx\\ =& \sum_{n=0}^{\infty}\left[ \frac{1}{s+5n+1} - \frac{1}{s+5n+2} - \frac{1}{s+5n+3} +\frac{1}{s+5n+4} \right] \end{align*}$$ Then
$$\begin{align*} \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx= &-\int \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{(1-e^{-5x})}\right] dxds+C\\ =& - \int \sum_{n=0}^{\infty}\left[ \frac{1}{s+5n+1} - \frac{1}{s+5n+2} - \frac{1}{s+5n+3} +\frac{1}{s+5n+4} \right] ds + C\\ =& - \sum_{n=0}^{\infty}\left[ \int \frac{1}{s+5n+1}ds - \int\frac{1}{s+5n+2}ds - \int\frac{1}{s+5n+3}ds +\int\frac{1}{s+5n+4}ds \right] + C\\ =& - \sum_{n=0}^{\infty}\left[ \ln(s+5n+1) - \ln(s+5n+2) -\ln(s+5n+3) +\ln(s+5n+4)\right] + C\\ =& - \sum_{n=0}^{\infty} \ln\left[\frac{(s+5n+1)(s+5n+4)}{(s+5n+2)(s+5n+3)}\right]+C \end{align*}$$ Therefore
If $\displaystyle s \to 0$ $$\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = - \sum_{n=0}^{\infty} \ln\left[\frac{(5n+1)(5n+4)}{(5n+2)(5n+3)}\right]+C$$ If $\displaystyle s \to \infty$ $$\begin{align*} 0 =& - \lim_{s \to \infty} \sum_{n=0}^{\infty} \ln\left[\frac{(s+5n+1)(s+5n+4)}{(s+5n+2)(s+5n+3)}\right]+C \\ = & - \sum_{n=0}^{\infty} \ln\left[ \lim_{s \to \infty} \frac{(s+5n+1)(s+5n+4)}{(s+5n+2)(s+5n+3)}\right]+C \\ = & - \sum_{n=0}^{\infty} \ln(1)+C \\ \Longrightarrow C=& 0 \end{align*}$$ Then
$$\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = - \sum_{n=0}^{\infty} \ln\left[\frac{(5n+1)(5n+4)}{(5n+2)(5n+4)}\right] = - \ln\left[ \prod_{n=0}^{\infty} \frac{(5n+1)(5n+4)}{(5n+2)(5n+3)} \right]$$ Recall the following infinite product representation for a quotient of products of Gamma functions (proof in the Appendix):

If $\displaystyle \sum_{j=1}^{m} a_{j} = \sum_{j=1}^{m} b_{j}$ then: $$\prod_{k=0}^{\infty}\frac{(a_{1}+k)(a_{2}+k)\cdots(a_{m}+k)}{(b_{1}+k)(b_{2}+k% )\cdots(b_{m}+k)}=\frac{\Gamma\left(b_{1}\right)\Gamma\left(b_{2}\right)\cdots% \Gamma\left(b_{m}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right)% \cdots\Gamma\left(a_{m}\right)}$$ Hence, we can write:
$$\prod_{n=0}^{\infty} \frac{(5n+1)(5n+4)}{(5n+2)(5n+3)} = \prod_{n=0}^{\infty} \frac{\left(n+\frac{1}{5}\right)\left(n+\frac{4}{5}\right)}{\left(n+\frac{2}{5}\right)\left(n+\frac{3}{5}\right)} = \frac{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}{\Gamma\left(\frac{1}{5}\right) \Gamma\left(\frac{4}{5}\right)} = \frac{2}{1+\sqrt{5}}$$ Therefore
$$\boxed{\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = \ln\left(\frac{1+\sqrt{5}}{2}\right)}$$ As a corollary we have the Laplace transform:
$$\boxed{\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx = \ln\left[ \frac{\Gamma\left(\frac{1+s}{5}\right) \Gamma\left(\frac{4+s}{5}\right)}{\Gamma\left(\frac{2+s}{5}\right)\Gamma\left(\frac{3+s}{5}\right)} \right] }$$

Appendix: Euler's infinite product representation for a quotient of Gamma functions

Recall the Euler's definition of the Gamma function $$\Gamma(z) = \lim_{n \to \infty} \frac{(n+1)^z n!}{z(z+1)(z+2)\cdots (z+n)}$$ Supposse that $\displaystyle \sum_{j=1}^{m} a_{j} = \sum_{j=1}^{m} b_{j}$ Hence $$\frac{\Gamma\left(b_{1}\right)\Gamma\left(b_{2}\right)\cdots \Gamma\left(b_{m}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right) \cdots\Gamma\left(a_{m}\right)} = \lim_{n \to \infty} \prod_{j=1}^{m} \left[(n+1)^{b_{j}-a_{j}} \prod_{k=0}^{n} \frac{a_{j}+k}{b_{j}+k}\right]=\lim_{n \to \infty} (n+1)^{\sum_{j=1}^{m} a_{j} - \sum_{j=1}^{m} b_{j}}\prod_{j=1}^{m} \prod_{k=0}^{n} \frac{a_{j}+k}{b_{j}+k} = \lim_{n \to \infty} \prod_{k=0}^{n} \prod_{j=1}^{m} \frac{a_{j}+k}{b_{j}+k}$$ Therefore: $$\boxed{\prod_{k=0}^{\infty}\frac{(a_{1}+k)(a_{2}+k)\cdots(a_{m}+k)}{(b_{1}+k)(b_{2}+k% )\cdots(b_{m}+k)}=\frac{\Gamma\left(b_{1}\right)\Gamma\left(b_{2}\right)\cdots% \Gamma\left(b_{m}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right)% \cdots\Gamma\left(a_{m}\right)} \; \textrm{ if }\; \sum_{j=1}^{m} a_{j} = \sum_{j=1}^{m} b_{j}}$$

Laplace transform I

What is the Laplace transform of $e^{-x}\ln(1-e^{-x})$

We repost here the solution of this problem posted by @SrinivasR1729 on Twitter.
$$\int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = -\frac{H_{a+1}}{a+1}$$

Proof $$\begin{align*} \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = & \int_{0}^{1} (1-w)^a \ln(w) dw \quad (w \mapsto 1-e^{-x})\\ =& \int_{0}^{1} (1-w)^a \left(\frac{d}{dt}\Big|_{t=0+} w^t\right) dw\\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} w^t(1-w)^a dw\\ =& \frac{d}{dt}\Big|_{t=0+} B\left(t+1,a+1\right)\\ \end{align*}$$ Recall that the derivative of the complete beta function: $$\frac{d}{da} B(a,b) = B(a,b)\left(\psi^{(0)}(a) - \psi^{(0)}(a+b)\right)$$ Therefore: $$\begin{align*} \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = &\frac{d}{dt}\Big|_{t=0+} B\left(t+1,a+1\right)\\ =&\lim_{t \to 0+} \left(\psi^{0}(t+1)-\psi^{0}(a+t+2)\right)B(t+1,a+1) \\ =& \left(\psi^{0}(1)-\psi^{0}(a+2)\right)B(1,a+1)\\ =& \left(-\gamma-\psi^{0}(a+2)\right)B(1,a+1)\\ \end{align*}$$ where we used the fact that $\psi^{(0)}(1) = -\gamma$, the Euler constant.

Recall that a hamonic number for real and complex values can be expressed with the following relation: $$H_{x} = \psi^{(0)}(x+1) + \gamma$$ Also note that $$B(1,a+1) = \frac{\Gamma(1)\Gamma(a+1)}{\Gamma(a+2)} = \frac{\Gamma(a+1)}{\Gamma(a+1)(a+1)} = \frac{1}{a+1}$$ Hence $$\begin{align*} \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = & \left(\gamma-\psi^{0}(a+2)\right)B(1,a+1)\\ =& -\frac{H_{a+1}}{a+1} \end{align*}$$ $$\boxed{ \int_{0}^{\infty} e^{-ax}e^{-x}\ln(1-e^{-x}) dx = -\frac{H_{a+1}}{a+1}}$$

Residue theorem IV

Beautiful integral involving a product of cosine functions

Today we show the proof of this beautiful result posted by @SrinivasR1729 on Twitter:

$$\int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x)dx =\pi \cos \left(\frac{1}{a}\right)$$ $$\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) = \pi \cos\left(\frac{a+1}{a^2}\right)$$ $$\int_{0}^{8 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos\left(\frac{1}{a^3} + x \right)\cos(3x) = \pi \cos\left(\frac{a^2+a+1}{a^3}\right)$$ $$\int_{0}^{16 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos\left(\frac{1}{a^3} + x \right)\cos\left(\frac{1}{a^4} + x \right)\cos(3x) = \pi \cos\left(\frac{a^3+a^2+a+1}{a^4}\right)$$ $$\int_{0}^{2^m \pi } \left[\cos\left( \frac{1}{a}+x\right)\cos\left( \frac{1}{a^2}+x\right)\cdots\cos\left( \frac{1}{a^m}+x\right)\right]\cos(mx) = \pi \cos\left(\frac{a^{m-1}+a^{m-2}+a^{m-3}+...+1}{a^m}\right)$$
Proof

Recall that under the assumptions of the residue theorem with winding numbers, if $\gamma$ is a closed curve: $$\oint_{\gamma} fdz = 2\pi i \sum_{i=1}^{n} \left\{\operatorname{Res}\left(f,z_{i}\right)I(\gamma,z_{i})\right\}$$ where $I(\gamma,z_{i})$ is the winding number of $\gamma$ with respect to $z_{i}$.

Also recall that the circle $\displaystyle \gamma(t) = e^{it} \quad 0\leq t \leq 2\pi n$ has an index $\displaystyle I(\gamma,0) = n$ with respect to $0$

Case $n=1$ (the circle winds one time around 0)

The fact that $x$ varies from $0$ to $2\pi$ suggest we can consider $x$ as an argument of a point $z$ on the unit circle $C$ centered at the origin; hence we write $\displaystyle z = e^{ix} \quad 0\leq x \leq 2\pi \Longrightarrow dx = \frac{dz}{iz}$

Then
$$\begin{align*} I= \int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x) =& \int_{0}^{2 \pi} \left(\frac{e^{i\left(\frac{1}{a}+x\right)} + e^{-i\left(\frac{1}{a}+x\right)}}{2}\right)\left(\frac{e^{ix}+e^{-ix}}{2}\right)dx \\ =& \int_{0}^{2 \pi} \left(\frac{e^{\frac{i}{a}}e^{ix} + e^{-\frac{i}{a}}e^{-ix}}{2}\right)\left(\frac{e^{ix}+e^{-ix}}{2}\right)dx \\ =& \frac{1}{2^2}\oint_{C} \frac{\left(e^{\frac{i}{a}}z + \frac{e^{-\frac{i}{a}}}{z}\right)\left(z+\frac{1}{z}\right)}{iz}dz\\ =& \frac{1}{2^2i}\oint_{C} \left\{ e^{\frac{i}{a}}z + \frac{1}{z}\left(e^{\frac{i}{a}}+e^{\frac{-i}{a}}\right) +\frac{e^{\frac{-i}{a}}}{z^3}\right\}dz\\ \end{align*}$$ Hence, if $\displaystyle g(z) = e^{\frac{i}{a}}z + \frac{1}{z}\left(e^{\frac{i}{a}}+e^{\frac{-i}{a}}\right) +\frac{e^{\frac{-i}{a}}}{z^3}$
$$\Longrightarrow \operatorname{Res}(g,0) = e^{\frac{i}{a}}+e^{\frac{-i}{a}}$$ Therefore $z=0$ is the only singularity of $g$ and: $$\begin{align*} I=\int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x)dx=& 2 \pi i \left(\frac{1}{2^2i}\right) \operatorname{Res}\left( g, 0 \right)\underbrace{I(C,0)}_{=1}\\ =& \frac{\pi}{2} \left(e^{\frac{i}{a}}+e^{\frac{-i}{a}}\right)\\ =& \pi \cos \left(\frac{1}{a} \right) \end{align*}$$ $$\boxed{ \int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x)dx =\pi \cos \left(\frac{1}{a}\right) }$$

Case n = 2 (the circle winds 2 times around 0)
$$\begin{align*} I=\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) dx=& \int_{0}^{4 \pi} \left(\frac{e^{i\left(\frac{1}{a}+x\right)} + e^{-i\left(\frac{1}{a}+x\right)}}{2}\right)\left(\frac{e^{i\left(\frac{1}{a^2}+x\right)} + e^{-i\left(\frac{1}{a^2}+x\right)}}{2}\right)\left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx \\ =& \int_{0}^{4 \pi} \left(\frac{e^{\frac{i}{a}}e^{ix} + e^{-\frac{i}{a}}e^{-ix}}{2}\right)\left(\frac{e^{\frac{i}{a^2}}e^{ix} + e^{-\frac{i}{a^2}}e^{-ix}}{2}\right)\left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx \\ =& \frac{1}{2^3i}\oint_{C} \frac{\left(e^{\frac{i}{a}}z + \frac{e^{-\frac{i}{a}}}{z}\right)\left(e^{\frac{i}{a^2}}z + \frac{e^{-\frac{i}{a^2}}}{z}\right)\left(z^2+\frac{1}{z^2}\right)}{z}dz \quad (z = e^{ix} )\\ =& \frac{1}{2^3i}\oint_{C} \underbrace{\frac{\left(e^{\frac{i}{a}}z^2 + e^{-\frac{i}{a}}\right)\left(e^{\frac{i}{a^2}}z^2 + e^{-\frac{i}{a^2}}\right)\left(z^4+1\right)}{z^5}}_{g(z)}dz\\ \end{align*}$$ It is easy to see that $$\operatorname{Res}(g,0) = e^{\frac{i}{a}}e^{\frac{i}{a^2}} + e^{-\frac{i}{a}}e^{-\frac{i}{a^2}} = e^{i\left(\frac{1}{a} + \frac{1}{a^2}\right)} + e^{-i\left(\frac{1}{a}+\frac{1}{a^2}\right)}$$ Therefore, again $z=0$ is the only singularity of $g$ and $$\begin{align*} I =\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) dx=& 2\pi i \left(\frac{1}{2^3i } \right)\operatorname{Res}(g,0)\underbrace{I(C,0)}_{=2}\\ =& 2^2\pi i \left(\frac{1}{2^3i } \right)\left( e^{\frac{i}{a}}e^{\frac{i}{a^2}} + e^{-\frac{i}{a}}e^{-\frac{i}{a^2}} \right)\\ =& \frac{\pi}{2} \left[ e^{i\left(\frac{1}{a} + \frac{1}{a^2}\right)} + e^{-i\left(\frac{1}{a}+\frac{1}{a^2}\right)} \right]\\ =& \pi \cos\left(\frac{1}{a}+\frac{1}{a^2}\right)\\ =& \pi \cos\left(\frac{a+1}{a^2}\right) \end{align*}$$ $$\boxed{\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) = \pi \cos\left(\frac{a+1}{a^2}\right)}$$

Case n = m (the circle winds $2^{m-1}$ times around $0$)

$$\begin{align*} I= \int_{0}^{2^{m} \pi } \left(\prod_{j=1}^{m} \cos\left( \frac{1}{a^j}+x\right)\right)\cos(mx) = &\int_{0}^{2^m \pi } \left( \prod_{j=1}^{m} \frac{e^{i\left(\frac{1}{a^j}+x\right)} +e^{-i\left(\frac{1}{a^j}+x\right)}}{2}\right)\left(\frac{e^{imx}+e^{-imx}}{2}\right)dx\\ =& \int_{0}^{2^{m-1} 2 \pi } \left( \prod_{j=1}^{m} \frac{e^{\frac{i}{a^j}}e^{ix} +e^{-\frac{i}{a^j}}e^{-ix}}{2}\right)\left(\frac{e^{imx}+e^{-imx}}{2}\right)dx\\ =& \frac{1}{2^{m+1}}\oint_{C} \left[\prod_{j=1}^{m} \left(e^{\frac{i}{a^j}}z +\frac{e^{-\frac{i}{a^j}}}{z}\right)\right]\left(z^m + \frac{1}{z^m}\right)\frac{dz}{iz} \quad (z = e^{ix} )\\ =& \frac{1}{i2^{m+1}}\oint_{C} \frac{\left[\prod_{j=1}^{m} \left(e^{\frac{i}{a^j}}z^2 +e^{-\frac{i}{a^j}}\right)\right]\left(z^{2m} + 1\right)}{z^{2m+1}}dz \end{align*}$$ Therefore if $$g(z) = \frac{\left[\prod_{j=1}^{m} \left(e^{\frac{i}{a^j}}z^2 +e^{-\frac{i}{a^j}}\right)\right]\left(z^{2m} + 1\right)}{z^{2m+1}}$$ Then $$\operatorname{Res}(g,0) = \prod_{j=1}^{m} e^{\frac{i}{a^j}} + \prod_{j=1}^{m} e^{-\frac{i}{a^j}} = e^{i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})} + e^{-i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})}$$ Hence $$\begin{align*} I= \int_{0}^{2^m \pi } \left(\prod_{j=1}^{m} \cos\left( \frac{1}{a^j}+x\right)\right)\cos(mx) =& 2 \pi i \left( \frac{1}{i2^{m+1}}\right) \operatorname{Res}(g,0) \underbrace{I(C,0)}_{=2^{m-1}}\\ =& \frac{\pi}{2}\left(e^{i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})} + e^{-i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})}\right)\\ =& \pi \cos\left(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m}\right)\\ =& \pi \cos\left(\frac{a^{m-1}+a^{m-2}+a^{m-3}+...+1}{a^m}\right) \end{align*}$$ Hence $$\boxed{\int_{0}^{2^m \pi } \left[\cos\left( \frac{1}{a}+x\right)\cos\left( \frac{1}{a^2}+x\right)\cdots\cos\left( \frac{1}{a^m}+x\right)\right]\cos(mx) = \pi \cos\left(\frac{a^{m-1}+a^{m-2}+a^{m-3}+...+1}{a^m}\right)}$$

Feynman's trick IX

Mellin transform of $\sin x$ and $\cos x$

We show the proof of the following result posted by @infseriesbot $$\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \frac{\pi^{\frac{3}{2}}}{\sqrt{2}}$$

Proof

First we can start differentiating under the integral sign: $$\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \int_{0}^{\infty} \frac{(\sin x - \cos x) \left(\frac{d}{dt}\Big|_{t=0+} x^t\right) }{\sqrt{x}} dx = \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}(\sin x - \cos x) dx$$ The right hand side is the derivative of the Mellin transform of the function $\displaystyle (\sin x - \cos x)$ at $\displaystyle t+\frac{1}{2}$: We can split the Mellin into two, the Mellin of $\sin x$ and $\cos x$: $$\left\{ \mathscr{M} \sin x \right\} \left(t+\frac{1}{2}\right) = \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\sin x dx$$ $$\left\{ \mathscr{M} \cos x \right\} \left(t+\frac{1}{2}\right) = \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\cos x dx$$ Now we can expand $\sin x$ and $\cos x$ using the Euler's formula and the de Moivre theorem in order to apply the Ramanujan's master theorem: $$\sin(w) = \frac{e^{ix}-e^{-ix}}{2i} = \sum_{n=0}^{\infty} \frac{(ix)^n}{2in!} -\sum_{n=0}^{\infty} \frac{(-ix)^n}{2in!}=\sum_{n=0}^{\infty} \frac{(-x)^n\left[(-i)^n-(i)^n\right]}{2in!}$$ By de Moivre's theorem $$(-i)^{n} = -(e^{-\frac{i\pi}{2}})^{n} = \cos\left(-\frac{n\pi}{2}\right)+i\sin\left(-\frac{n\pi}{2}\right) = \cos\left(\frac{n\pi}{2}\right)-i\sin\left(\frac{n\pi}{2}\right)$$ $$i^{n} = (e^{\frac{i\pi}{2}})^{n} = \cos\left(\frac{n\pi}{2}\right)+i\sin\left(\frac{n\pi}{2}\right)$$ Then $$\frac{\left[(-i)^n-(i)^n\right]}{2i} = \frac{-2i\sin\left(\frac{n\pi}{2}\right)}{2i} = -\sin\left(\frac{n\pi}{2}\right)$$ Therefore $$\sin x = \sum_{n=0}^{\infty}\frac{(-x)^n[-\sin\left(\frac{n\pi}{2}\right)]}{n!}dw$$ In a similar way we can show: $$\cos x = \sum_{n=0}^{\infty}\frac{(-x)^n[\cos\left(\frac{n\pi}{2}\right)]}{n!}dw$$ Applying the Ramanujan's master theorem and using the fact that $\sin$ is odd and $\cos$ is even $$\int_{0}^{\infty} x^{t+\frac{1}{2}-1}\sin x dx = \Gamma\left(t+\frac{1}{2}\right)\sin\left(\frac{t\pi}{2}+\frac{\pi}{4}\right)$$ $$\int_{0}^{\infty} x^{t+\frac{1}{2}-1}\cos x dx = \Gamma\left(t+\frac{1}{2}\right)\cos\left(\frac{t\pi}{2}+\frac{\pi}{4}\right)$$ Therefore: $$\begin{align*} \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\sin x dx = & \frac{d}{dt}\Big|_{t=0+} \Gamma\left(t+\frac{1}{2}\right)\sin\left(\frac{t\pi}{2}+\frac{\pi}{4}\right) \\ =& \lim_{t \to 0+} \frac{1}{2} -\Gamma\left(t+\frac{1}{2}\right)\left[ \pi \sin \left(\frac{\pi t}{2} +\frac{\pi}{4}\right) +2\cos\left(\frac{\pi t}{2} + \frac{\pi}{4}\right) \psi^{(0)}\left(t+\frac{1}{2}\right)\right]\\ =& \frac{\pi^{\frac{3}{2}}}{2\sqrt{2}} + \sqrt{\frac{\pi}{2}} \psi^{(0)}\left(\frac{1}{2}\right) \end{align*}$$ $$\begin{align*} \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}\cos x dx = & \frac{d}{dt}\Big|_{t=0+} \Gamma\left(t+\frac{1}{2}\right)\cos\left(\frac{t\pi}{2}-\frac{\pi}{4}\right) \\ =& \lim_{t \to 0+} \frac{1}{2} \Gamma\left(t+\frac{1}{2}\right)\left[2\cos\left(\frac{\pi t}{2} + \frac{\pi}{4}\right)\psi^{(0)}\left(t+\frac{1}{2}\right)- \pi \sin \left(\frac{\pi t}{2} +\frac{\pi}{4} \right)\right]\\ =& \sqrt{\frac{\pi}{2}} \psi^{(0)}\left(\frac{1}{2}\right) - \frac{\pi^{\frac{3}{2}}}{2\sqrt{2}} \end{align*}$$ Hence, we can conclude: $$\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} x^{t+\frac{1}{2}-1}(\sin x - \cos x) dx = \frac{\pi^{\frac{3}{2}}}{\sqrt{2}}$$ $$\boxed{\int_{0}^{\infty} \frac{(\sin x - \cos x) \ln x}{\sqrt{x}} dx = \frac{\pi^{\frac{3}{2}}}{\sqrt{2}}}$$

Pochhammer polynomials

Series expansion of $\displaystyle \frac{1}{\sqrt{1-x}}$

We prove the following result posted by @infseriesbot
$$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} x^n$$

Proof

$$\frac{1}{\sqrt{1-x}} = (1-x)^{-\frac{1}{2}}$$ Recall that Newton extended the binomial theorem to allow real and complex numbers as exponents: $$(x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k}x^{r-k}y^k \quad |x|>|y|$$ where $$\binom{r}{k} = \frac{(r)_{k}}{k!}$$ and $\displaystyle (r)_{k}$ is the Pochhammer symbol (falling factorial): $$(r)_{k} = r(r-1)\cdots(r-k+1)$$ Hence: $$\begin{align*} (1-x)^{-\frac{1}{2}} =& \sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n} (-1)^n x^{n} \\ =& \sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{n}}{n!} (-1)^n x^{n} \\ \end{align*}$$ One of the properties of the rising and falling factorial states that $$r^{(n)} = (-1)^n(r)_{n}$$ where $\displaystyle r^{(n)}$ is the rising factorial: $$r^{(n)} = r(r+1)(r+2)\cdots (x+n-1)$$ Hence $$\sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{n}}{n!} (-1)^n x^{n} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{n!} x^{n}$$ Note that we can express the factorial as a the pochhammer polynomial of $(1)^{(n)}$: $$(1)^{(n)} = 1\cdot 2\cdot 3\cdots(n-1)n = n!$$ Hence, $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{n!} x^{n} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}} x^{n}$$ On the other hand, double factorials can be expressed with the following form: $$(2n-1)!! = 2^n \frac{\Gamma\left(\frac{1}{2}+n\right)}{\sqrt{\pi}} = 2^n \frac{\Gamma\left(\frac{1}{2}+n\right)}{\Gamma\left(\frac{1}{2}\right)}$$ $$(2n)!! = 2^n n! = 2^n \Gamma(n+1) = 2^n \frac{\Gamma(n+1)}{\Gamma(1)}$$ while the pochhammer polynomial can be expressed as the quotient of two gamma functions: $$r^{(n)} = \frac{\Gamma(n+r)}{\Gamma(r)}$$ Hence, we can write: $$(2n-1)!!= 2^n \left(\frac{1}{2}\right)^{(n)}$$ $$(2n)!! = 2^n (1)^{(n)}$$ Therefore $$\frac{(2n-1)!!}{(2n)!!} = \frac{2^n \left(\frac{1}{2}\right)^{(n)}}{2^n (1)^{(n)}} = \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}}$$ Hence, we can conclude $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}} x^{n} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^{n}$$ This can also be expressed with the notation of the generalized hypergeometric function as $$\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}}{(1)^{(n)}} x^{n} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)^{(n)}(1)^{(n)}}{(1)^{(n)}} \frac{x^{n}}{n!} = {}_{2}F_{1}\left(\frac{1}{2},1;1;x\right)$$ However, in the hypergeometric function notation it is common to use $(x)_{n}$ to denote the rising factorial instead of the falling factorial, so it is very important to always check if we are referring to rising of falling factorials. $$\boxed{\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^{n}}$$

Generalized hypergeometric function

Solution to integral involving ${}_{3}F_{2}$

We show the proof of the following result posted by @SrinivasR1729 $$\int_{0}^{\infty} \frac{\ln(1+e^{-x})}{\sqrt{1-e^{-x}}}e^{-\frac{x}{2}}dx = 2\pi \ln\left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right)$$

Proof

Recall that $${}_{p}F_{q}(a_{1},a_{2},...,a_{p};b_{1},b_{2},...,b_{q};x) = \sum_{j=0}^{\infty} \frac{(a_{1})(a_{2})\cdots(a_{p})}{(b_{1})(b_{2})\cdots(b_{q})} \frac{x^{j}}{j!}$$ where the symbol $\displaystyle (x)_{j}$ refers to the Pochhammer polynomial (also called rising factorial): $$(x)_{j} = x(x+1)(x+2)\cdots(x+j-1)$$ Hence, $$\begin{align*} \int_{0}^{\infty} \frac{\ln(1+e^{-x})}{\sqrt{1-e^{-x}}}e^{-\frac{x}{2}}dx =& \int_{0}^{1} \frac{\ln(1+w)}{\sqrt{1-w}\sqrt{w}} dw \quad (w \mapsto e^{-w})\\ =& \int_{0}^{1} \sum_{k=1}^{\infty} \frac{(-1)^{k+1} w^k}{k} \frac{1}{\sqrt{1-w}\sqrt{w}} dw\\ =& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \int_{0}^{1} w^{k-\frac{1}{2}}(1-w)^{-\frac{1}{2}} dw\\ =& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} B\left(k+\frac{1}{2},\frac{1}{2}\right)\\ =& \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} B\left(j+\frac{3}{2},\frac{1}{2}\right) \quad (k=j+1)\\ =& \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} \frac{\Gamma\left(j+\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(j+2)}\\ =&\sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)_{j}}{\Gamma(2) (2)_{j}} \quad \left(\Gamma(t+j) = \Gamma(t)(t)_{j}\right)\\ =& \frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} \frac{\left(\frac{3}{2}\right)_{j}}{(2)_{j}} \quad \left(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}, \; \Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{\pi}}{2}\right)\\ =&\frac{\pi}{2} \sum_{j=0}^{\infty} (-1)^{j} \frac{\left(\frac{3}{2}\right)_{j} (1)_{j} }{(2)_{j} (2)_{j}} \quad \left( \frac{(1)_{j}}{(2)_{j}}=\frac{1}{1+j}\right) \\ =&\frac{\pi}{2} \sum_{j=0}^{\infty} (-1)^{j} \frac{\left(\frac{3}{2}\right)_{j} (1)_{j} (1)_{j} }{(2)_{j} (2)_{j} (1)_{j}} \\ =&\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{\left(\frac{3}{2}\right)_{j} (1)_{j} (1)_{j} }{(2)_{j} (2)_{j} } \frac{(-1)^{j}}{j!} \quad (j! = (1)_{j}) \\ =&\frac{\pi}{2}{}_{3}F_{2}\left(\frac{3}{2},1,1;2,2;-1\right) \\ =&2\pi \ln\left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right) \end{align*}$$ Therefore $$\boxed{\int_{0}^{\infty} \frac{\ln(1+e^{-x})}{\sqrt{1-e^{-x}}}e^{-\frac{x}{2}}dx = 2\pi \ln\left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right)}$$

Hurwitz-zeta function

Integral representation of the Hurwitz-zeta function

We show the proof of the following result posted by @integralsbot: $$\int_{0}^{1} \frac{x^n \ln^{a-1} \frac{1}{x} }{1-x} dx = \Gamma(a) \left( \zeta (a)- H_{n}^{(a)} \right)$$ This result is the integral representation of the Hurwitz-zeta function. Recall that: $$\zeta(a,n+1) = \left( \zeta (a)- H_{n}^{(a)} \right)$$ The proof for the case $a \in \mathbb{R}$ is well known, so here we show a proof for $a \in \mathbb{N}$ using the Feynman's trick just for the sake to give a different proof.

Proof

For $\displaystyle a\in \mathbb{N}$
$$\begin{align*} \int_{0}^{1} \frac{x^n \ln^{a-1} \frac{1}{x} }{1-x} dx = & \int_{0}^{1} \frac{x^n \left(\frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \frac{1}{x^t}\right) }{1-x} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+}\int_{0}^{1} \frac{x^{n-t} }{1-x} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+}\int_{0}^{1} \sum_{k=0}^{\infty}x^kx^{n-t} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \sum_{k=0}^{\infty} \int_{0}^{1}x^{k+n-t} dx\\ =& \frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \sum_{k=0}^{\infty} \frac{1}{(k+n+1-t)} \\ =& \sum_{k=0}^{\infty} \left[\frac{d^{a-1}}{dt^{a-1}}\Big|_{t=0+} \frac{1}{(k+n+1-t)} \right] \\ =& \sum_{k=0}^{\infty} \left[\lim_{t \to 0+} (a-1)! \frac{1}{(k+n+1-t)^a} \right] \\ =& (a-1)! \sum_{k=0}^{\infty} \frac{1}{(k+n+1)^a} \\ =& (a-1)! \zeta(a,n+1) \\ =& \Gamma(a) \left( \zeta (a)- H_{n}^{(a)} \right) \\ \end{align*}$$ $$\boxed{ \int_{0}^{1} \frac{x^n \ln^{a-1} \frac{1}{x} }{1-x} dx =\Gamma(a) \left( \zeta (a)- H_{n}^{(a)} \right) }$$

Golden ratio

Integral that relates the secant function and the Golden ratio

Today we repost here the solution to this integral posted on Twitter by @integralsbot: $$\int_{0}^{\infty} \frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}} dx = \phi$$ In the proof we made use of the integral representation of the $\sec x$ function, a trick that we have used extensively in other occasions (link 1, link 2)

Proof


$$I= \int_{0}^{\infty} \frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}} dx = \frac{1}{\pi}\int_{0}^{\infty} \frac{y^{\frac{4}{5}}}{1+y^2} dw \quad (y \mapsto x^{\pi})$$ Recall the integral representation of $\displaystyle \sec(w)$ (proof below): $$\sec(w) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{\frac{2w}{\pi}}}{1+y^2} dy$$ Therefore if $\displaystyle w= \frac{2\pi}{5}$ $$I = \frac{1}{\pi}\int_{0}^{\infty} \frac{y^{\frac{4}{5}}}{1+y^2} dy = \frac{\sec\left(\frac{2\pi}{5}\right)}{2} = \frac{1+\sqrt{5}}{2} = \phi$$ Hence $$\boxed{\int_{0}^{\infty} \frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}} dx = \phi }$$

Appendix

We reproduce here the proof of the integral representation of the sec function again:
The proof can be obtained trough contour integration in a branch cut but here we will use a more straightforward approach with the help of the Ramanujan's master theorem

Let $|a|<1$

Recall $$\frac{1}{1+y^2} = \int_{0}^{1} t^{y^2} dt$$ Therefore $$\begin{align*} \int_{0}^{\infty}\frac{y^a}{1+y^2}dy =& \int_{0}^{\infty}\int_{0}^{1} y^a t^{y^2} dtdy\\ =&\int_{0}^{1} \int_{0}^{\infty} y^a t^{y^2} dydt\\ =& \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt \quad (w \mapsto y^2) \end{align*}$$ The integral $\displaystyle \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw$ is the Mellin transform of $\displaystyle t^w$ at $\displaystyle \frac{a+1}{2}$ Recall that $$t^w = e^{w\ln(t)} = \sum_{n=0}^{\infty} \frac{\ln^n(t)w^n}{n!} = \sum_{n=0}^{\infty} \frac{(-\ln(t))^n(-w^n)}{n!}$$ Therefore, by Ramanujan's master theorem $$\int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw = \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}}$$ Therefore $$\begin{align*} \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt =& \frac{1}{2}\int_{0}^{1} \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}v^{-\frac{a+1}{2}}e^{-v} dv \quad (v\mapsto -ln(t))\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) \end{align*}$$ Finally, recall the Euler's reflection formula $$\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin \pi z} = \pi \csc \pi z$$ Therefore $$\int_{0}^{\infty}\frac{y^a}{1+y^2}dy = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \frac{\pi}{2} \csc\left(\pi \frac{1+a}{2}\right)=\frac{\pi}{2} \sec\left(\pi \frac{a}{2}\right)$$ If we let $\displaystyle x = \pi \frac{a}{2}$ then $\displaystyle a=\frac{2x}{\pi}$. Then $$\boxed{\frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy = \sec x}$$

Catalan constant

Integral involving Catalan constant

Today we show the proof of the following result found on Twitter: $$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{\pi}}+1} dx dy= 2\beta(2)$$

Proof

$$I = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{\pi}}+1} dxdy = \int_{0}^{\frac{\pi}{2}} y\frac{2}{\pi}\int_{0}^{\infty} \frac{w^{\frac{2y}{\pi}-1}}{w^2+1} dwdy \quad (w \mapsto x^{\frac{\pi}{2y}})$$ Recall the integral representation fo $\sec(x)$ (proof in the Appendix): $$\sec(x) = \frac{2}{\pi}\int_{0}^{\infty} \frac{t^{\frac{2x}{\pi}}}{t^2+1}dt$$ Hence $$\frac{2}{\pi}\int_{0}^{\infty} \frac{w^{\frac{2y}{\pi}-1}}{w^2+1} dw = \sec\left(y-\frac{\pi}{2}\right) = \csc(y) = \frac{1}{\sin(y)}$$ Therefore $$I = \int_{0}^{\frac{\pi}{2}} \frac{y}{\sin(y)}dy = 2\int_{0}^{\infty} \operatorname{arccot}(e^w) dw \quad \left(w \mapsto \ln\left(\cot\left(\frac{y}{2}\right)\right)\right)$$ Recall the series expansion for $\operatorname{arccot}(x)$ $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)} x^{-2n-1}$$ Hence $$\begin{align*} I =& 2\int_{0}^{\infty} \operatorname{arccot}(e^w) dw \\ =& 2\int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)} e^{-w(2n+1)}dw\\ =& 2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)}\int_{0}^{\infty} e^{-w(2n+1)}dw\\ =& 2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}\\ =& 2\beta(2) \end{align*}$$ $$\boxed{\therefore \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{\pi}}+1} dxdy = 2\beta(2)}$$

Appendix: Integral representation of $\sec x$

How do we know that $\displaystyle \sec x= \frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy$?

The proof can be obtained trough contour integration in a branch cut but here we will use a more straigthforward approach with the help of the Ramanujan's master theorem

Let $|a|<1$
Recall $$\frac{1}{1+y^2} = \int_{0}^{1} t^{y^2} dt$$ Therefore $$\begin{align*} \int_{0}^{\infty}\frac{y^a}{1+y^2}dy =& \int_{0}^{\infty}\int_{0}^{1} y^a t^{y^2} dtdy\\ =&\int_{0}^{1} \int_{0}^{\infty} y^a t^{y^2} dydt\\ =& \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt \quad (w \mapsto y^2) \end{align*}$$ The integral $\displaystyle \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw$ is the Mellin transform of $\displaystyle t^w$ at $\displaystyle \frac{a+1}{2}$
Recall that $$t^w = e^{w\ln(t)} = \sum_{n=0}^{\infty} \frac{\ln^n(t)w^n}{n!} = \sum_{n=0}^{\infty} \frac{(-\ln(t))^n(-w^n)}{n!}$$ Therefore, by Ramanujan's master theorem $$\int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw = \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}}$$ Therefore $$\begin{align*} \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt =& \frac{1}{2}\int_{0}^{1} \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}v^{-\frac{a+1}{2}}e^{-v} dv \quad (v\mapsto -ln(t))\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) \end{align*}$$ Finally, recall the Euler's reflection formula $$\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin \pi z} = \pi \csc \pi z$$ Therefore $$\int_{0}^{\infty}\frac{y^a}{1+y^2}dy = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \frac{\pi}{2} \csc\left(\pi \frac{1+a}{2}\right)=\frac{\pi}{2} \sec\left(\pi \frac{a}{2}\right)$$ If we let $\displaystyle x = \pi \frac{a}{2}$ then $\displaystyle a=\frac{2x}{\pi}$. Then $$\boxed{\frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy = \sec x}$$

Taylor's theorem

Series related to the power series of $\arcsin(x)$

We show the proof of the following result. The proof relies on the series expansion of arcsin. $$\sum_{n=1}^{\infty} \frac{(2n-2)!}{2^{2n}(n!)^2 } = 1-\ln(2)$$ Proof

Lets start with the expansion series of $\arcsin(x)$: $$\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2 (2n+1)}x^{2n+1} = \arcsin(x) \quad \quad |x|\leq 1$$ Taking the derivative $$\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2 }x^{2n} = \frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$$ Multiplying by $\frac{1}{x^2}$ and subtracting $-\frac{1}{x^2}$ $$\sum_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2 }x^{2n-2} = \frac{1}{x^{2}\sqrt{1-x^2}} -\frac{1}{x^2}$$ Integrating $$\int_{0}^{x} \sum_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2 }x^{2n-2} dx = \sum_{n=1}^{\infty} \frac{(2n-2)!(2n)}{2^{2n}(n!)^2 }x^{2n-1}= \int_{0}^{x} \left(\frac{1}{x^{2}\sqrt{1-x^2}} -\frac{1}{x^2} \right)dx$$ Hence $$\int_{0}^{x} \left(\frac{1}{x^{2}\sqrt{1-x^2}} -\frac{1}{x^2} \right)dx = \left[ -\frac{\sqrt{1-x^2}}{x} +\frac{1}{x} \right]_{0}^{x} = -\frac{\sqrt{1-x^2}}{x} +\frac{1}{x}$$ Therefore $$\sum_{n=1}^{\infty} \frac{(2n-2)!(2n)}{2^{2n}(n!)^2 }x^{2n-1}= -\frac{\sqrt{1-x^2}}{x} +\frac{1}{x}$$ Integrating one more time from 0 to 1: $$\int_{0}^{1}\sum_{n=1}^{\infty} \frac{(2n-2)!(2n)}{2^{2n}(n!)^2 }x^{2n-1} dx=\sum_{n=1}^{\infty} \frac{(2n-2)!}{2^{2n}(n!)^2 }= \int_{0}^{1} \left(-\frac{\sqrt{1-x^2}}{x} +\frac{1}{x} \right)dx$$ If $x=\sin(w)$ and $dx =\cos(w)dw$ $$\int_{0}^{1} \left(-\frac{\sqrt{1-x^2}}{x} +\frac{1}{x}\right)dx = \int_{0}^{\frac{\pi}{2}} \left[ \cot(w) - \cot(w)\cos(w)\right]dw$$ Using the following identities $$\cos(w)\cot(w) = \csc(w)-\sin(w), \quad \cot(w)-\csc(x) = -\tan\left(\frac{w}{2}\right) \quad \textrm{and} \quad \sin(w)=2\sin\left(\frac{w}{2}\right)\cos\left(\frac{w}{2}\right)$$ $$\begin{align*} \sum_{n=1}^{\infty} \frac{(2n-2)!}{2^{2n}(n!)^2 } =& \int_{0}^{\frac{\pi}{2}} \left[ \cot(w) - \cot(w)\cos(w)\right]dw \\ = & \int_{0}^{\frac{\pi}{2}} \left[ \cot(w) - \csc(w) +\sin(w) \right]dw \\ = & \int_{0}^{\frac{\pi}{2}} \left[ \cot(w) - \csc(w) +\sin(w) \right]dw \\ = & \int_{0}^{\frac{\pi}{2}} \left[ -\tan\left(\frac{w}{2}\right) +2\sin\left(\frac{w}{2}\right)\cos\left(\frac{w}{2}\right) \right]dw \\ = & \left[ 2\ln(\cos\left(\frac{x}{2}\right)) -2\cos^{2}\left(\frac{x}{2}\right) \right]_{0}^{\frac{\pi}{2}} = 1-\ln(2) \end{align*}$$ Therefore $$\boxed{\sum_{n=1}^{\infty} \frac{(2n-2)!}{2^{2n}(n!)^2 } = 1-\ln(2) }$$