Elliptic integrals

Forgotten functions used to calculate $\pi$

Today we will show the proof of the following result:
$$1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2\cdot 4}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4 \cdot 6 }\right)^2 + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8 }\right)^2 + ... = \frac{4}{\pi}$$
We show the proof of this result using an elliptic integrals approach (the result can also be obtained trough the Vandermonde's identity). However, following the path of elliptic integrals turned out very interesting given that it needs a tour through some of the main results of the theory of this sepcial functions to prove it.

This is the outline of the proof:

1. Obtain the series expansion for $E(p), K(p)$ (the elliptic integrals of the first and second kind)
2. Prove a result know as the Landen transformation
3. Obtain the following series expansion as corollary of 1 and 2 (our result is a particular case of this expansion)

$$E(p) =\frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}$$ $$\quad p=\sqrt{1-q^2}$$
where $\displaystyle E(p)$ is the elliptic integral of the second kind, $\displaystyle p$ is known as the modulus and $q$ is known as de complementary modulus. If $E(p)=1$ when $p=1$ we get our result.


Proof


Consider the complete elliptic integral of the first kind:
$$K(p) = \int_{0}^{\frac{\pi}{2}} \frac{d \theta}{\sqrt{1-p^2\sin^2 \theta}}$$
where $\displaystyle p$ is known as the modulus.

We will obtain its series expansion:

Proposition.
$$K(p) = \frac{\pi}{2} \sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} p^{2j}$$ Proof

First, recall that
$$\arcsin (x) = \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} \frac{x^{2j+1}}{2j+1} \quad |x|\leq 1$$ Hence $$\frac{1}{\sqrt{1-x^2}} = \arcsin' (x)= \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} x^{2j} \quad |x|\leq 1$$ Now, given that $\displaystyle -1<\sin \theta < 1$ then $0< p^2 \sin^2 \theta < p^2$

Therefore
If $\displaystyle p^2 \leq 1$ $$\frac{1}{\sqrt{1-p^2\sin^2 \theta}} = \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j} \sin^{2j} \theta$$ Integrating $$\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{d\theta }{\sqrt{1-p^2\sin^2 \theta}} =& \int_{0}^{\frac{\pi}{2}} \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j} \sin^{2j} \theta d\theta \\ =& \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j} \int_{0}^{\frac{\pi}{2}} \sin^{2j} \theta d\theta \end{align*}$$ It is easy to show that (proof will be added in the Appendix) $$\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2j} \theta d\theta = \frac{(2j-1)!!}{(2j)!!} \frac{\pi}{2}$$ Hence $$K(p)= \int_{0}^{\frac{\pi}{2}}\frac{d\theta }{\sqrt{1-p^2\sin^2 \theta}} = \frac{\pi}{2} \sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} p^{2j} \tag{1}$$
Now we are going to prove the expansion series for the complete elliptic integral of the second kind:

Proposition
$$E(p)=\int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta = \frac{\pi}{2} \sum_{j=0}^{\infty } \frac{1}{1-2j} \left[\frac{(2j-1)!!}{(2j)!!} p^j \right]^2$$
Proof

Lets apply the Feynman's trick: $$\frac{d}{dp} E(p) = -\int_{0}^{\frac{\pi}{2}} \frac{p\sin^2 \theta}{\sqrt{1-p^2\sin^2 \theta }}d\theta$$ Hence $$-\frac{p \sin^2 \theta }{\sqrt{1-p^2\sin^2 \theta}} = -\sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \sin^{2j+2} (\theta)$$ Therefore $$\begin{align*} -\int_{0}^{\frac{\pi}{2}}\frac{p \sin^2 \theta }{\sqrt{1-p^2\sin^2 \theta}}d\theta = & -\int_{0}^{\frac{\pi}{2}}\sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \sin^{2j+2} (\theta) d\theta \\ =& -\sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \int_{0}^{\frac{\pi}{2}}\sin^{2j+2} (\theta) d\theta \\ =& -\sum_{j=0}^{\infty} \frac{\pi}{2} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \frac{(2j+1)!!}{(2j+2)!!}\\ =& -\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j+2)!!^2} \frac{(2j+1)}{(2j+2)} p^{2j+1} \end{align*}$$
where we make use of the fact that (the proof will be added to the appendix)
$$\int_{0}^{\frac{\pi}{2}}\sin^{2j+2} (\theta) d\theta = \frac{(2j+1)!!}{(2j+2)!!}$$ Hence, integrating with respect to $p$ $$\begin{align*} \int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta =& -\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} \frac{(2j+1)}{(2j+2)^2} p^{2j+2} + C \\ =& -\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j+1)!!^2}{(2j+2)!!^2} \frac{1}{(2j+1)} p^{2j+2} + C \end{align*}$$ Now if $\displaystyle p=0$ then $\displaystyle C=\frac{\pi}{2}$ therefore $$\int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta =\frac{\pi}{2}-\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j+1)!!^2}{(2j+2)!!^2} \frac{1}{(2j+1)} p^{2j+2}$$ If $m=j+1$ $$\int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta =\frac{\pi}{2}-\frac{\pi}{2}\sum_{m=1}^{\infty} \frac{(2m-1)!!^2}{(2m)!!^2} \frac{1}{(2m-1)} p^{2m} = \frac{\pi}{2}\sum_{m=0}^{\infty} \frac{(2m-1)!!^2}{(2m)!!^2} \frac{1}{(1-2m)} p^{2m}$$ Therefore $$\boxed{E(p)= \frac{\pi}{2}\sum_{m=0}^{\infty} \frac{(2m-1)!!^2}{(2m)!!^2} \frac{1}{(1-2m)} p^{2m}}\tag{2}$$
Now we have to make use of the Landen transformation (the proof is very extensive and will be added in the appendix later)

Proposition (Landen transformation) $$\boxed{K\left(\frac{1-q}{1+q}\right) = \frac{1+q}{2} K(p)} \tag{3}$$ $$\boxed{E\left(\frac{1-q}{1+q} \right) = \frac{E(p) +qK(p)}{1+q}} \tag{4}$$
Finally, we are going to show the proof for the following expansion of $E(p)$

Proposition $$E(p) =\frac{\pi(1+q)}{4} \sum_{n=0}^{\infty} \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}$$
Proof

From (3) we can write: $$E(p) =E\left(\frac{1-q}{1+q} \right)-qK(p)$$ Substituting (4) in (3) we can write $$E(p) =E\left(\frac{1-q}{1+q} \right)-qK(p) =E\left(\frac{1-q}{1+q} \right)-\frac{2q}{1+q} K\left(\frac{1-q}{1+q}\right)$$ Then, using the power series for $E,K$ in (1) and (2): $$K\left(\frac{1-q}{1+q} \right) = \frac{\pi}{2} \sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} \left(\frac{1-q}{1+q} \right)^{2j}$$ $$E\left(\frac{1-q}{1+q} \right)= \frac{\pi}{2}\sum_{m=0}^{\infty} \frac{(1-2m)!!^2}{(2m)!!^2} \frac{1}{(1-2m)} \left(\frac{1-q}{1+q} \right)^{2m}$$ Hence $$\begin{align*} E(p) = & \frac{\pi(1+q)}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-q}{1+q}\right)^{2n} -\frac{2q}{1+q} \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-q}{1+q}\right)^{2n}\\ =& \frac{\pi(1+q)}{2} \sum_{n=0}^{\infty}\left\{ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-q}{1+q}\right)^{2n} \left(\frac{1}{1-2n} - \frac{2q}{(1+q)^2}\right)\right\}\\ =& \frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\left[\frac{2}{1-2n} - 1 + \left(\frac{1-q}{1+q}\right)^2\right] \\ =&\frac{\pi(1+q)}{4} \left\{ 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2(n+1)}\right\} \\ =&\frac{\pi(1+q)}{4} \left\{ 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n-2)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \right\} \\ =& \frac{\pi(1+q)}{4} \left\{1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \left[(2n)^2 - \frac{2n+1}{2n-1} (2n-1)^2\right] \right\}\\ =& \frac{\pi(1+q)}{4} \left\{ 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \right\}\\ =& \frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \end{align*}$$ Hence $$\boxed{ E(p) =\frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} }$$ Note that if $\displaystyle p=1$ then $q=0$. Moreover, $$E(1) = \int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin^2 }\theta d\theta = \int_{0}^{\frac{\pi}{2}} cos \theta d\theta = 1$$ Therefore, we can conclude that $$\boxed{ 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2\cdot 4}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4 \cdot 6 }\right)^2 + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8 }\right)^2 + ... = \frac{4}{\pi} }$$
Appendix I will be adding the omitted results the following days

Euler's work II

Corollary to Euler's expansion for $arctan(x)$

Today we show the proof of this result:
$$\frac{\arcsin(w)}{\sqrt{1-w^2} } = \sum_{n=0}^{\infty} \frac{2^{2n} n!^2}{(2n+1)!}w^{2n+1}$$
We have proved it earlier but this time we also prove the Euler's expansion for $\arctan(x)$ on which this proof relies on.

Proof

Recall the famous Euler's formula for $arctan(x)$ (Proof below) which states that:
$$\arctan (x) = \frac{x}{1+x^2} \sum_{n=0}^{\infty}\left[\frac{(2n)!!}{(2n+1)!!} \left(\frac{x^2}{1+x^2}\right)^{n}\right]$$
Now consider the following identity for inverse trigonometric functions:
$$\arcsin(w) = \arctan\left(\frac{w}{\sqrt{1-w^2}}\right)$$ Then if $x= \displaystyle \frac{w}{\sqrt{1-w^2}}$ we have $$\begin{align*} \arcsin(w) =& \arctan\left(\frac{w}{\sqrt{1-w^2}}\right) = w\sqrt{1-w^2}\left[1+\frac{2}{3}w^2 +\frac{2}{3}\frac{4}{5}w^4 + \frac{2}{3}\frac{4}{5}\frac{6}{7}w^6...\right]\\ \Longrightarrow \frac{\arcsin(w)}{\sqrt{1-w^2} }=& w+\frac{2}{3}w^3+\frac{2}{3}\frac{4}{5}w^5+\frac{2}{3}\frac{4}{5}\frac{6}{7}w^7+...\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} \end{align*}$$ Now from the definition of double factorial for even integers: $$(2n)!! = 2^{n} n!$$ And for odd integers: $$(2n+1)!! = \frac{(2n+1)!}{2^n n!}$$ Hence $$\frac{\arcsin(w)}{\sqrt{1-w^2} } = \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} = \frac{2^{2n} n!^2}{(2n+1)!}w^{2n+1}$$ Therefore $$\boxed{\frac{\arcsin(w)}{\sqrt{1-w^2} } = \sum_{n=0}^{\infty}\frac{2^{2n} n!^2}{(2n+1)!}w^{2n+1}}$$

Appendix.

But now it is time to prove Euler's formula for $\arctan(x)$. We know that there are fancy proofs for this formula but we want to show a proof more in line with the spirit of Euler. We need this lemma:

Proposition If $\displaystyle (1+x^2)(1-y^2) =1$ then
$$\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1} = \sqrt{1-y^2} \sum_{n=0}^{\infty} \Delta^n a_{0} y^{2n+1}$$
where $$\Delta^n a_{0} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} a_{2k+1}$$
Proof
If $\displaystyle (1+x^2)(1-y^2)=1$ then $\displaystyle x^2=\frac{y^2}{1-y^2}$
Hence
$$\begin{align*} \sum_{k=0}^{\infty} (-1)^k a_{2k+1} x^{2k+1} = &\sum_{n=0}^{\infty} (-1)^k a_{2k+1} x^{2k}\cdot x\\ =& \sum_{k=0}^{\infty} (-1)^k a_{2k+1} \left(\frac{y^2}{1-y^2}\right)^{k}\frac{y}{\sqrt{1-y^2}}\\ =& \sum_{k=0}^{\infty} (-1)^ka_{2k+1} \frac{y^{2k}}{(1-y^2)^{k+1}}\frac{y(1-y^2)}{\sqrt{1-y^2}}\\ =& \sum_{k=0}^{\infty} (-1)^ka_{2k+1} \frac{y^{2k}}{(1-y^2)^{k+1}} y\sqrt{1-y^2}\\ =& \sum_{k=0}^{\infty} (-1)^ka_{2k+1} \left[\sum_{n=0}^{\infty}\binom{n}{k} y^{2n}\right] y\sqrt{1-y^2}\\ =& \sum_{n=0}^{\infty}\left[\sum_{k=0}^{\infty} \binom{n}{k} (-1)^ka_{2k+1} \right] y^{2n+1} \sqrt{1-y^2}\\ =& \sum_{n=0}^{\infty}\Delta^n a_{0} y^{2n+1} \sqrt{1-y^2}\\ \end{align*}$$ Hence $$\boxed{\sum_{k=0}^{\infty} a_{2k+1} x^{2k+1} = \sqrt{1-y^2} \sum_{n=0}^{\infty} \Delta^n a_{0} y^{2n+1} \quad \textrm{ if } (1+x^2)(1-y^2)=1 }$$

Now, we will apply the Lemma to $arctan(x)$

Recall the series expansion for $arctan(x)$
$$\arctan(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1}$$ Therefore $$\displaystyle a_{2k+1} = \frac{1}{2k+1}, \quad \Delta^n a_{0} = \sum_{k=0}^{n} (-1)^k \binom{n}{k}\frac{1}{2k+1}$$ Hence $$\begin{align*} \arctan(x) = & \sum_{j=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1}\\ =& \sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\sum_{k=0}^{n} (-1)^k \binom{n}{k}\frac{1}{2k+1}\right] y^{2n+1} \\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\sum_{k=0}^{n} (-1)^k \binom{n}{k} \int_{0}^{1} t^{2k} dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[ \int_{0}^{1}\sum_{k=0}^{n} (-1)^k \binom{n}{k} t^{2k} dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[ \int_{0}^{1} (1-t^2)^n dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\frac{1}{2} \int_{0}^{1} (1-w)^n w^{-\frac{1}{2}} dt \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\frac{1}{2} B\left(n+1,\frac{1}{2}\right) \right] y^{2n+1}\\ =&\sqrt{1-y^2} \sum_{n=0}^{\infty} \left[\frac{1}{2} \frac{\sqrt{\pi}{n!}}{\Gamma \left(n+\frac{3}{2}\right)} \right] y^{2n+1}\\ \end{align*}$$ From Legendre duplication formula we can write: $$\Gamma\left(n+\frac{3}{2}\right) = \frac{\Gamma(2n+2)\sqrt{\pi}}{2^{2n+1} \Gamma(n+1) } = \frac{(2n+1)! \sqrt{\pi}}{ 2^{2n+1} n!}$$ Then $$\arctan(x) = \sqrt{1-y^2} \sum_{n=0}^{\infty} \left[ \frac{2^{2n} n!^2}{(2n+1)!} \right] y^{2n+1} = \sqrt{1-y^2} \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} y^{2n+1}$$ Given that $\displaystyle y=\frac{x}{\sqrt{1+x^2}}$ Therefore $$\boxed{\arctan (x) = \frac{x}{1+x^2} \left[\frac{(2n)!!}{(2n+1)!!} \left(\frac{x^2}{1+x^2}\right)^n\right]}$$

This type of proof is called Euler's transform

Hardy's work

Hardy and Mellin transform

In a previous post (link) we used the following identity. We repost here the proof based on the work of Hardy: a mix of the Mellin transform and theta functions properties. $$\mathop{\sum_{n=-\infty }^{\infty}\sum_{m=-\infty}^{\infty}}_{n\neq 0 \wedge m\neq 0}\frac{1}{(m^2+n^2)^s} = 4\zeta(s)\beta(s)$$
Proof
Let $n,m \neq 0$ then, $$\int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt = \frac{1}{(n^2+m^2)} \int_{0}^{\infty} w^{s-1}e^{-w}dw \quad \left(w \mapsto (n^2+m^2)t\right)$$ Hence $$\int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt = \frac{1}{(n^2+m^2)} \Gamma(s) \Longrightarrow \frac{1}{(n^2+m^2)}= \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt$$ Therefore $$\begin{align*} \mathop{\sum_{n=-\infty}^{\infty}\sum_{m=-\infty }^{\infty}}_{n\neq 0 \wedge m\neq 0} \frac{1}{(m^2+n^2)^s} =& \mathop{\sum_{n=-\infty }^{\infty}\sum_{\substack{m=-\infty}}^{\infty}}_{n\neq 0 \wedge m\neq 0}\frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1} e^{-(n^2+m^2)t}dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\mathop{\sum_{n=-\infty }^{\infty}\sum_{m=-\infty }^{\infty}}_{n\neq 0 \wedge m\neq 0} e^{-(n^2+m^2)t}dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\left[\sum_{\substack{n=-\infty }}^{\infty}\sum_{\substack{m=-\infty }}^{\infty} e^{-(n^2+m^2)t} -1\right]dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\left[\sum_{\substack{n=-\infty}}^{\infty} e^{-n^2t}\sum_{\substack{m=-\infty}}^{\infty}e^{-m^2t}-1\right]dt \\ =& \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}\left[\left(\sum_{\substack{n=-\infty}}^{\infty} e^{-n^2t}\right)^2-1\right]dt \\ \end{align*}$$ Now recall the definition of the theta function $\displaystyle \theta_{3}(q,z)=\sum_{n=-\infty}^{\infty} q^{n^2}e^{2inz}$
If $\displaystyle q=e^{-t}$ and $z=0$ then:
$$\displaystyle \theta_{3}(e^{-t},0)=\sum_{n=-\infty}^{\infty} e^{-n^2t}\Longrightarrow \theta^2_{3}(e^{-t},0)=\left(\sum_{n=-\infty}^{\infty} e^{-n^2t} \right)^2$$ Therefore $$\left(\sum_{\substack{n=-\infty}}^{\infty} e^{-n^2t}\right)^2 -1 = \theta^2_{3}(e^{-t},0)-1$$ Now recall the following identity for $\displaystyle \theta_{3}(q,0)$: $$\theta^2_{3}(q,0) =1+4\sum_{n=1}^{\infty} \frac{q^n}{(1+q^{2n})}$$
In this case with $q=e^{-t}$ $$\theta^2_{3}(e^{-t},0) -1=4\sum_{n=1}^{\infty} \frac{e^{-tn}}{(1+e^{-2tn})} = 4 \sum_{n=1}^{\infty}\sum_{m=0}^{\infty}e^{-tn}(-1)^m e^{-2tnm}$$ Therefore $$\begin{align*}\left\{\mathscr{M} \left[\theta^2_{3}(e^{-t},0) -1\right] \right\}(s) =& \int_{0}^{\infty} t^{s-1} \left[\theta^2_{3}(e^{-t},0) -1\right] dt\\ =& \int_{0}^{\infty} t^{s-1}4 \sum_{n=1}^{\infty}\sum_{m=0}^{\infty}(-1)^m e^{-n(2m+1)t} dt\\ =& \Gamma(s) 4\sum_{n=1}^{\infty}\frac{1}{n^s}\sum_{n=0}^{\infty}(-1)^m\frac{1}{(2m+1)^s}\\ =& 4\Gamma(s)\zeta(s)\beta(s) \end{align*}$$ Therefore $$\boxed{\mathop{\sum_{n=-\infty }^{\infty}\sum_{m=-\infty}^{\infty}}_{n\neq 0 \wedge m\neq 0}\frac{1}{(m^2+n^2)^s} = 4\zeta(s)\beta(s)}$$

Euler's work

Euler and Catalan constant

We show the proof of this interesting series related to the work of Euler and the Catalan constant. $$\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} = 2\beta(2)$$
Proof
From a previous post (link) we found that $$\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2\beta(2)$$ Therefore $$2\beta(2) = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} dx =\int_{0}^{1} \frac{\arcsin w}{w \sqrt{1-w^2}} dw \quad (w \mapsto \sin x) \tag{1}$$ Now we will show that $$\frac{\arcsin w}{\sqrt{1-w^2}} = \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1}$$ Euler found the following formula for $\arctan(x)$: $$\arctan(x) = \frac{x}{1+x^2}\left[1+\frac{2}{3}\frac{x^2}{1+x^2}+\frac{2}{3}\frac{4}{5}\left(\frac{x^2}{1+x^2}\right)^2+\frac{2}{3}\frac{4}{5}\frac{6}{7}\left(\frac{x^2}{1+x^2}\right)^3+...\right]=\frac{x}{1+x^2}\sum_{n=0}^{\infty}\prod_{k=1}^{\infty} \frac{2kx^2}{(2k+1)(1+x^2)} \tag{2}$$ Now, consider the following identity for inverse trigonometric functions: $$\arcsin(w) = \arctan\left(\frac{w}{\sqrt{1-w^2}}\right)$$ Then if we substitute $x= \displaystyle \frac{w}{\sqrt{1-w^2}}$ in (2) we have $$\begin{align*} \arcsin(w) =& \arctan\left(\frac{w}{\sqrt{1-w^2}}\right) = w\sqrt{1-w^2}\left[1+\frac{2}{3}w^2 +\frac{2}{3}\frac{4}{5}w^4 + \frac{2}{3}\frac{4}{5}\frac{6}{7}w^6...\right]\\ \Longrightarrow \frac{\arcsin(w)}{\sqrt{1-w^2} }=& w+\frac{2}{3}w^3+\frac{2}{3}\frac{4}{5}w^5+\frac{2}{3}\frac{4}{5}\frac{6}{7}w^7+...\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w ^{2n+1} \tag{3} \end{align*}$$ Hence, from (1) and (3) $$\begin{align*} \int_{0}^{1} \frac{\arcsin w}{w \sqrt{1-w^2}} dw =& \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \frac{w ^{2n+1}}{w} dw \\ =& \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} w^{2n} dw\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \int_{0}^{1} w^{2n} dw\\ =& \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} \end{align*}$$ Therefore $$\boxed{\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!(2n+1)} = 2\beta(2)}$$

Feynman's trick VIII

Integral found on Instagram

We show the proof of this result found on Instagram using the Feynman's trick again $$\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \ln(\sin x)dx = -\frac{\pi}{32}- \frac{3\pi\ln(4)}{16}$$
Proof $$\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \ln(\sin x)dx = \underbrace{\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x\right) \ln(\sin x)dx}_{A} +\underbrace{\int_{0}^{\frac{\pi}{2}} \left(cos^4 x\right) \ln(\sin x)}_{B}$$ Hence $$\begin{align*} A = \int_{0}^{\frac{\pi}{2}} \left(\sin^4 x\right) \ln(\sin x)dx = & \int_{0}^{1} \frac{w^4 \ln(w)}{\sqrt{1-w^2}}dw \quad (w \mapsto \sin x)\\ =& \int_{0}^{1} \frac{w^4 \frac{d}{dt}\Big|_{t=0+}w^t}{\sqrt{1-w^2}}dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \frac{w^{4+t}}{\sqrt{1-w^2}}dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} w^{4+t} (1-w^2)^{-\frac{1}{2}}dw \\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2}\int_{0}^{1} r^{\frac{t+5}{2}-1}(1-r)^{\frac{1}{2}-1}dr \quad (r \mapsto w^2) \\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2}B\left(\frac{t+5}{2},\frac{1}{2}\right)\\ =& \lim_{t \to 0+}\frac{1}{2}\left[\psi^{0}\left(\frac{t+5}{2}\right)-\psi^{0}\left(\frac{t}{2}+3\right)\right]B\left(\frac{t+5}{2},\frac{1}{2}\right)\\ =& \frac{1}{2}\left[\psi^{0}\left(\frac{5}{2}\right)-\psi^{0}\left(3\right)\right]B\left(\frac{5}{2},\frac{1}{2}\right)\\ =&\frac{1}{2}\left[\frac{8}{3}-\gamma -\ln(4) -\frac{3}{2}+\gamma \right]\frac{3\pi}{8}\\ =& \frac{\pi}{64}(7-6\ln(4)) \end{align*}$$ $$\begin{align*} B = \int_{0}^{\frac{\pi}{2}} \left(\cos^4 x\right) \ln(\sin x) =& \int_{0}^{1} (1-w^2)^{\frac{3}{2}} \ln(w) dw \quad (w \mapsto \sin x)\\ =& \int_{0}^{1} (1-w^2)^{\frac{3}{2}} \frac{d}{dt}\Big|_{t=0+}w^t dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} (1-w^2)^{\frac{3}{2}} w^t dw \\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2} \int_{0}^{1} (1-r)^{\frac{3}{2}}r^{\frac{t}{2}-\frac{1}{2}} dr \quad (r \mapsto w^2)\\ =& \frac{d}{dt}\Big|_{t=0+} \frac{1}{2} B\left(\frac{t+1}{2},\frac{5}{2}\right)\\ =& \lim_{t \to 0+} \frac{1}{2} \left[ \psi^{0} \left(\frac{t+1}{2}\right) - \psi^{0}\left(\frac{t}{2}+3\right)\right]B\left(\frac{t+1}{2},\frac{5}{2}\right)\\ =& \frac{1}{2}\left[-\gamma-\ln(4) -\frac{3}{2}+\gamma \right]\frac{3\pi}{8}\\ =& -\frac{3\pi }{64}(3+2\ln(4)) \end{align*}$$ Therefore $$A+B = -\frac{\pi}{32}- \frac{3\pi\ln(4)}{16}$$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \ln(\sin x)dx = -\frac{\pi}{32}- \frac{3\pi\ln(4)}{16}}$$

Feynman's trick VII

Integral involving the eta Dirichlet function

We repost here the solution of this integral previously posted on Twitter:
$$\int_{0}^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx = \frac{\pi^2}{6}$$

Proof
$$\begin{align*} \int_{0}^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx =& 2\int_{1}^{\infty} \frac{\ln(w)}{w(1+w)}dw \quad (w \mapsto e^{\sqrt{x}})\\ =& -2 \int_{0}^{1} \frac{\ln(r)}{(1+r)}dr \quad (r \mapsto \frac{1}{w})\\ =& -2 \int_{0}^{1} \ln(r) \sum_{n=0}^{\infty} (-1)^n r^n dr \\ =& 2 \sum_{n=0}^{\infty}(-1)^{n+1} \int_{0}^{1} \ln(r) r^n dr \\ =& 2 \sum_{n=0}^{\infty}(-1)^{n+1} \int_{0}^{1} \frac{d}{dt}\Big|_{t=0+} r^t r^n dr \\ =& 2 \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^{n+1} \int_{0}^{1} r^{t+n} dr \\ =& 2 \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^{n+1} \frac{1}{(n+t+1)} \\ =& 2 \lim_{t \to 0+ }\sum_{n=0}^{\infty}(-1)^n \frac{1}{(n+t+1)^2} \\ =& 2 \sum_{n=0}^{\infty}(-1)^n \frac{1}{(n+1)^2} \\ =& 2 \eta(2) \quad (\textrm{ Dirichlet eta function})\\ =& 2 \left(\frac{\pi^2}{12}\right) = \frac{\pi^2}{6} \end{align*}$$ If we do not know the value of $\eta(2)$ we can obtain it through the zeta function with this relation: $$\eta(v) = (1-\frac{1}{2^{v-1}})\zeta(v)$$ Then, if $v=2$, $$\eta(2) = \frac{1}{2} \zeta(2) = \frac{\pi^2}{12}$$ Hence
$$\boxed{\int_{0}^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx = \frac{\pi^2}{6}}$$

Beta Dirichlet function

Integral involving the beta Dirichlet function

We repost here the proof of this result:
$$\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx = 2\beta(2)$$
Proof

Do the following change of variable $$w=\ln \cot(\frac{x}{2}) \Longrightarrow e^w = \cot(\frac{x}{2}) \Longrightarrow 2\cot^{-1}(e^w)=x$$ $$w\to \infty \textrm{ if } x \to 0, \quad w\to 0 \textrm{ if } x \to \frac{\pi}{2}$$ Then $$dw = -\csc(x)dx = -\frac{1}{\sin(x)}dx$$ Therefore $$\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx =-\int_{\infty}^{0} 2\cot^{-1}(e^w)dw = \int_{0}^{\infty} 2\cot^{-1}(e^w)dw$$ Recall the series expansion of $\cot^{-1}$ $$\cot^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)}x^{-2n-1} \quad x\geq 0$$ Therefore $$\int_{0}^{\infty} 2\cot^{-1}(e^w)dw = \int_{0}^{\infty} 2\sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)}(e^w)^{-2n-1} dw= 2\sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)}\int_{0}^{\infty}(e^w)^{-2n-1} dw$$ We have that $$\int (e^w)^{-2n-1} dw = \frac{(e^w)^{-2n-1}}{-2n-1}+C \Longrightarrow \int_{0}^{\infty}(e^w)^{-2n-1} dw = \left[\frac{(e^w)^{-2n-1}}{-2n-1} \right]_{0}^{\infty} = \frac{1}{2n+1}$$ $$\therefore \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx = \int_{0}^{\infty} 2\cot^{-1}(e^w)dw = 2\sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)^2} =2\beta(2)$$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx = 2\beta(2)}$$

Lattices

Triple sum and summation theorem

Today we are going to prove this triple sum posted by @infesieres on Twitter. $$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = \frac{ 7\pi^4}{720} + \frac{\pi^2\beta(2)}{6}$$
Proof.

First, recall the following super theorem of complex analysis which we will be using extensively:

Let $f$ be analytic in $\mathbb{C}$ except for finitely isolated singularities. If there is a constant $M>0$ such that $\displaystyle |f(z)|\leq \frac{M}{z^{k}}$ where $k>1$ then we have the summation formula $$\lim_{N \to \infty } \sum_{n=-N}^{n=N} \{f(n) | \; n \textrm{ is not a singularity of } f\} = - \sum \{\textrm{Residues of } \pi\cot \pi z \;\textrm{ at the singularities of } f \}$$ If none of the singularities of $f$ are at integers, then $\displaystyle \lim_{N \to \infty} \sum_{n=-N}^{n=N} f(n)$ exists, is finite and $$\lim_{N \to \infty } \sum_{n=-N}^{n=N} f(n) = - \sum\{\textrm{Residues of } \pi\cot \pi z \; \textrm{ at the singularities of } f \}$$

Lets go back to the left hand side of the triple sum $$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)} \sum_{c=1}^{\infty} \frac{1}{(a^2+c^2)}$$ Consider the function $\displaystyle f(z) = \frac{1}{a^2+z^2}$. This function has two singularities at $z =\pm ia$. Moreover, $\displaystyle \lim_{z\to \infty} |z^2f(z)|=0$ so the conditions of the summation theorem hold. Then $$\begin{align*} \sum_{n=-\infty}^{\infty}\frac{1}{a^2+n^2} =& -\operatorname{Re}\left(\frac{\pi\cot \pi z}{a^2+z^2}, ia\right)- \operatorname{Re}\left(\frac{\pi\cot \pi z}{a^2+z^2}, -ia\right)\\ =& -\lim_{z\to ia} (z-ia)\frac{\pi\cot \pi z}{a^2+z^2} -\lim_{z\to -ia} (z+ia)\frac{\pi\cot \pi z}{a^2+z^2}\\ =& \frac{\pi\coth(a\pi)}{2a} +\frac{\pi\coth(a\pi)}{2a}\\ =&\frac{\pi \coth(a\pi)}{a} \end{align*}$$ Now, we know that the function $f(n)$ is even, then: $$\begin{align*} \sum_{n=-\infty}^{\infty}\frac{1}{a^2+n^2} =&\sum_{n=-\infty}^{n=-1}\frac{1}{a^2+n^2} + \frac{1}{a^2}+ \sum_{n=1}^{\infty}\frac{1}{a^2+n^2} \\ =& \frac{1}{a^2} + 2 \sum_{n=1}^{\infty}\frac{1}{a^2+n^2}\\ \Longrightarrow \sum_{n=1}^{\infty}\frac{1}{a^2+n^2} =& \frac{1}{2}\sum_{n=-\infty}^{n=\infty}\frac{1}{a^2+n^2} -\frac{1}{2a^2}\\ =& \frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \end{align*}$$ Then $$\begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} =& \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)} \sum_{c=1}^{\infty} \frac{1}{(a^2+c^2)}\\ =& \sum_{a=1}^{\infty}\left[\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)}\left(\frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \right)\right]\\ =& \sum_{a=1}^{\infty}\left[\left(\frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \right)\sum_{b=1}^{\infty}\frac{1}{(a^2+b^2)}\right]\\ =& \sum_{a=1}^{\infty}\left(\frac{\pi \coth(a\pi)}{2a} -\frac{1}{2a^2} \right)^2\\ =& \sum_{a=1}^{\infty}\left[\frac{1}{4a^4}-\frac{\pi \coth \pi a}{2a^{3}}+\frac{\pi^2\coth^2\pi a}{4a^2} \right]\\ =& \frac{1}{4} \sum_{a=1}^{\infty} \frac{1}{a^4} - \frac{1}{2}\sum_{a=1}^{\infty}\frac{\pi \coth \pi a}{a^{3}} + \frac{1}{4}\sum_{a=1}^{\infty}\frac{\pi^2\cot^2\pi a}{a^2}\\ =& \frac{\pi^4}{360} - \frac{1}{2}\underbrace{\sum_{a=1}^{\infty}\frac{\pi \coth \pi a}{a^{3}}}_{A} + \frac{1}{4}\underbrace{\sum_{a=1}^{\infty}\frac{\pi^2\coth^2\pi a}{a^2}}_{B} \end{align*}$$ We just have to find $A$ and $B$. Lets start with $A$ which is a relatively well known result of complex analysis. Consider the function $k(z)=\displaystyle \frac{\pi \coth \pi z}{z^{3}}$, it has a singularity at $z=0$ and countable set of singularities at $z=\pm in \quad n\in \mathbb{N}$. Additionally, $\displaystyle \lim_{z\to \infty} |z^2k(z)| =0$, so the assumptions of the summation theorem apply for the non-integer singularities. Then $$\begin{align*} \sum_{n=-\infty }^{\infty } \frac{\pi \coth \pi n}{n^{3}} = &-\sum_{n=1}^{\infty}\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},in\right)- \sum_{n=1}^{\infty}\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},-in\right)-\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right)\\ =& -\sum_{n=1}^{\infty}\left(\lim_{z\to in} (z-in)\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}}\right)-\sum_{n=1}^{\infty}\left(\lim_{z\to -in} (z+in)\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}}\right)-\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right)\\ =& -\sum_{n=0}^{\infty} \frac{\pi \coth \pi n}{n^3}-\sum_{n=0}^{\infty} \frac{\pi \coth \pi n}{n^3}-\operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right) \end{align*}$$ For $\displaystyle \operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right)$ we cannot apply the summation theorem directly (given that the singularity is an integer) but we can expand the function $k(z)$ with the Laurent series: $$\begin{align*} \frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}} =& \frac{\pi^2}{z^3} \left(\frac{1}{\pi z} -\frac{\pi z}{3} -\frac{\pi^3z^3}{45}+\right)\left(\frac{1}{\pi z} +\frac{\pi z}{3} -\frac{\pi^3z^3}{45}+\right)\\ =& \frac{1}{z^5}-\frac{7\pi^4}{45z} - \frac{19 \pi^9 z^3}{14175} -... \end{align*}$$ Therefore $\displaystyle \operatorname{Res}\left(\frac{\pi^2 \coth \pi z \cot \pi z}{z^{3}},0\right) =-\frac{7\pi^4}{45}$ Given that $k(z)$ is even $$4 \sum_{n=1}^{\infty } \frac{\pi \coth \pi n}{n^{3}} = \frac{7\pi^4}{45} \Longrightarrow \sum_{n=1}^{\infty } \frac{\pi \coth \pi n}{n^{3}} = \frac{7\pi^4}{180}=A$$ Now for B first recall the following identity: $\coth^2 x = \operatorname{csch}^2x+1$. Therefore $$B= \sum_{n=1}^{\infty} \frac{\pi^2\coth^2 \pi n}{n^2} = \underbrace{\sum_{n=1}^{\infty} \frac{\pi^2\operatorname{csch}^2 \pi n}{n^2}}_{C} + \frac{\pi^4}{6}$$ It turned out that we found C before in this post (actually the triple sum and that result seems to be deeply related): $$C = \sum_{n=1}^{\infty} \frac{\pi^2 \operatorname{csch}^2 \pi n}{n^2 } = \frac{2\pi^2}{3}\beta(2)-\frac{11}{180}\pi^4$$ Therefore $$B =\frac{2\pi^2}{3}\beta(2)-\frac{11}{180}\pi^4 + \frac{\pi^4}{6}$$ Then $$\begin{align*} \sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = & \frac{\pi^4}{360} - \frac{1}{2}\underbrace{\sum_{a=1}^{\infty}\frac{\pi \coth \pi a}{a^{3}}}_{A} + \frac{1}{4}\underbrace{\sum_{a=1}^{\infty}\frac{\pi^2\coth^2\pi a}{a^2}}_{B}\\ =& \frac{\pi^4}{360} - \frac{1}{2} \underbrace{\frac{7\pi^4}{180}}_{A} + \frac{1}{4} \underbrace{\left[\frac{2\pi^2}{3}\beta(2)-\frac{11}{180}\pi^4 +\frac{\pi^4}{6}\right]}_{B} \\ =& \frac{ 7\pi^4}{720} + \frac{\pi^2\beta(2)}{6} \end{align*}$$ Therefore $$\boxed{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\sum_{c=1}^{\infty} \frac{1}{(a^2+b^2)(a^2+c^2)} = \frac{ 7\pi^4}{720} + \frac{\pi^2\beta(2)}{6}}$$

Feynman's trick VI

Interesting integral involving the integral representation of $\sec(x)$

We prove the following result using Feynman's trick and the integral respresentation of $\sec(x)$ $$\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}^2(\sin(x)) dx = \frac{\pi^3}{8}$$
Proof
$$\begin{align*} S=\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}^2(\sin(x)) dx = & \int_{0}^{1} \frac{\operatorname{arctanh}^2 w}{\sqrt{1-w^2}} dw \quad (w \mapsto \sin(x))\\ =& \frac{1}{4} \int_{0}^{1} \frac{\ln^2\left(\frac{1+w}{1-w}\right)}{\sqrt{1-w^2}} dw \\ =& \frac{1}{4} \int_{1}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr \quad \left(r \mapsto \frac{1+w}{1-w}\right)\\ =& \frac{1}{4} \int_{0}^{1} \frac{\ln^2(s) }{\sqrt{s}(1+s)} ds \quad (s \mapsto \frac{1}{r})\\ \end{align*}$$ $$\therefore S = \frac{1}{4} \int_{0}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr - \underbrace{\frac{1}{4} \int_{0}^{1} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr}_{S} = \underbrace{\frac{1}{4} \int_{1}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr}_{S}$$ $$\begin{align*} \therefore 2S =& \frac{1}{4} \int_{0}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr\\ =& \frac{1}{4} \int_{0}^{\infty} \frac{ \frac{d^2}{dt^2}\Big|_{t=0+} r^t }{\sqrt{r}(1+r)} dr\\ =& \frac{1}{4} \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{\infty} \frac{ r^{t-\frac{1}{2}} }{(1+r)} dr\\ \end{align*}$$ Recall the integral representation of $\sec(x)$ (Proof in the appendix at the end): $$\sec(x) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{\frac{2x}{\pi}}}{y^2+1} dy \; \underset{v=y^2}{\Longrightarrow}\; \pi \sec(\pi x) = \int_{0}^{\infty} \frac{v^{x-\frac{1}{2}}}{v+1}dv$$ Therefore $$\begin{align*} S = \frac{1}{8} \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{\infty} \frac{ r^{t-\frac{1}{2}} }{(1+r)} dr =& \frac{1}{8} \frac{d^2}{dt^2}\Big|_{t=0+} \pi \sec(\pi t)\\ =& \lim_{t \to 0+} \frac{\pi^3}{8}\sec(\pi t)\left[\tan^2(\pi t) + \sec^2(\pi t)\right]\\ =& \frac{\pi^3}{8} \end{align*}$$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}^2(\sin(x)) dx = \frac{\pi^3}{8}}$$

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 $$\boxed{\frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy = \sec x}$$

Feynman's trick V

Double sum and Feynman's trick

We prove the following double sum with the help of the Feynman's trick $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)} = 2\zeta(3)$$ We also found the following generalization $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(rn+m)} = \frac{1}{r}\sum_{k=1}^{\infty} \frac{H_{rk}}{k^2}$$ Proof Recall $$\int_{0}^{1} w^{n-1} dw = \frac{1}{n}$$ $$\int_{0}^{1} y^{n-1} dy = \frac{1}{m}$$ $$\int_{0}^{1} x^{n+m-1} dx = \frac{1}{n+m}$$ Therefore $$\frac{1}{nm(n+m)} = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} w^{n-1}y^{m-1}x^{n+m-1}dwdydx$$ Summing $$\begin{align*} \sum_{n=1}^{\infty}\frac{1}{nm(n+m)} =& \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \sum_{n=1}^{\infty} w^{n-1}y^{m-1}x^{n+m-1}dwdydx\\ & = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{y^{m-1} x^{m}}{(1-wx)}dwdydx \end{align*}$$ Then $$\begin{align*} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{nm(n+m)} =& \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sum_{m=1}^{\infty} \frac{y^{m-1} x^{m}}{(1-wx)}dwdydx\\ =& \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{x}{(1-wx)(1-yx)}dwdydx\\ =& \int_{0}^{1} \int_{0}^{1} \frac{x}{1-yx} \left[\int_{0}^{1} \frac{1}{(1-wx)}dw\right]dydx\\ =& \int_{0}^{1} \left[\int_{0}^{1} \frac{\ln(1-x)}{1-yx} dy\right]dx\\ =& \int_{0}^{1} \frac{\ln^2(1-x)}{x}dx\\ =& \int_{0}^{1} \frac{\frac{d^2}{dt^2}\Big|_{t=0+} (1-x)^t}{x}dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{1} \frac{(1-x)^t}{x}dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{1}\sum_{n=0}^{\infty} (1-x)^n(1-x)^tdx \\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \sum_{n=0}^{\infty}\int_{0}^{1} (1-x)^{n+t}dx \\ =& \lim_{t \to 0+ }\frac{d^2}{dt^2} \sum_{n=0}^{\infty} \frac{1}{n+t+1}\\ =& \lim_{t \to 0+ } \sum_{n=0}^{\infty} \frac{2}{(n+t+1)^3}\\ =& \sum_{n=0}^{\infty} \frac{2}{(n+1)^3}\\ =& 2\zeta(3) \end{align*}$$ $$\boxed{\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)} = 2\zeta(3)}$$ For the general case, $$\begin{align*} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{nm(rn+m)} =& \int_{0}^{1}\frac{ \ln(1-x)\ln(1-x^r)}{x} dx\\ = & \int_{0}^{1}\frac{ \frac{d}{dt}\Big|_{t=0+} (1-x)^t \ln(1-x^r)}{x} dx\\ = & \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1}\frac{ (1-x)^t \ln(1-x^r)}{x} dx\\ = & -\frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \sum_{k=1}^{\infty}\frac{1}{k} (1-x)^t x^{rk-1} dx \\ = & -\frac{d}{dt}\Big|_{t=0+} \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{1} (1-x)^t x^{rk-1} dx \\ = & -\frac{d}{dt}\Big|_{t=0+} \sum_{k=1}^{\infty}\frac{1}{k} B(rk,t+1) \\ =& \lim_{t \to 0+} -\sum_{k=1}^{\infty}\frac{\left(\psi^{0}(t+1)-\psi^{0} (kr+t+1)\right)B(kr,t+1)}{k}\\ =& \sum_{k=1}^{\infty}\frac{\left(\psi^{0}(kr+1) + \gamma \right)}{k^2r}\\ =& \frac{1}{r} \sum_{k=1}^{\infty}\frac{H_{kr}}{k^2} \end{align*}$$ $$\boxed{\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(rn+m)} = \frac{1}{r}\sum_{k=1}^{\infty} \frac{H_{rk}}{k^2}}$$

Feynman's trick IV

Diff under the integral sign

We show the proof for the following result using the Feynman's technique: $$\int_{0}^{\frac{\pi}{2}} \frac{\arctan \sin x}{\sin x} dx = \frac{\pi \ln(1+\sqrt{2})}{2}$$ Proof Consider the intgral $$\Phi(a)= \int_{0}^{\frac{\pi}{2}} \frac{\arctan a\sin x}{\sin x} dx$$ Differentiatinting the parameter $a$ under the integral sign: $$\begin{align*} \Phi'(a) =\frac{d}{da} \int_{0}^{\frac{\pi}{2}} \frac{\arctan a\sin x}{\sin x} dx =& \int_{0}^{\frac{\pi}{2}} \frac{d}{da} \frac{\arctan a\sin x}{\sin x} dx \\ =& \int_{0}^{\frac{\pi}{2}} \frac{1}{a^2\sin^2{x}+1} dx\\ =& \frac{\arctan \left(\tan x \sqrt{a^2+1} \right)}{\sqrt{a^2+1}}\Big|_{0}^{\frac{\pi}{2}} \\ =& \frac{\pi}{2\sqrt{a^2+1}} \end{align*}$$ Therefore $$\begin{align*} \Phi(a) =& \int \frac{\pi}{2\sqrt{a^2+1}} da \\ =& \frac{\pi\operatorname{arcsinh} a}{2} + C\\ =& \frac{\pi \ln(a+\sqrt(a^2+1))}{2} + C \end{align*}$$ Now if we evaluate $\displaystyle \Phi(1)$ Then $$\int_{0}^{\frac{\pi}{2}} \frac{\arctan \sin x}{\sin x} dx = \frac{\pi\ln(1+\sqrt(2))}{2} + C$$ Finally, if we evaluate $\displaystyle \Phi(0)$ then $\arctan a\sin(x)=0$ and $C=0$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{\arctan \sin x}{\sin x} dx = \frac{\pi\ln(1+\sqrt(2))}{2} }$$

Second tetration integrals

Integral involving tetration, pi and e

We prove the following result using contour integration $$\int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}\sin(x)dx =\frac{1}{2 e^\frac{1}{\pi}}$$
We are going to use the Euler's formula for sine: $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$ Then $$\int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}\sin(x)dx = \frac{1}{2i}\int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}e^{ix}dx - \frac{1}{2i} \int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}e^{-ix}dx$$ Note that: $$\int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}e^{ix}dx= \int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)+ix\right)}dx$$ $$\int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}e^{-ix}dx= \int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)-ix\right)}dx$$ Let's start with the first one. We will use the following change of variable: $\displaystyle w=\frac{\ln(1-x)}{\pi}-\frac{\ln(x)}{\pi}$ $$\int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)+ix\right)}dx = \int_{-\infty}^{\infty} e^{\left(\frac{w+i}{e^{\pi w}+1}\right)}\frac{\pi e^{\pi w}}{(1+e^{\pi w})^2}dw$$ where $\displaystyle x=\frac{1}{e^{\pi w}+1}$ and $\displaystyle dx = -\frac{\pi e^{\pi w}}{(1+e^{\pi w})^2}dw$. Now let $\displaystyle r=w+i$, then: $$\int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)+ix\right)}dx = -\int_{-\infty}^{\infty} e^{\left(\frac{w+i}{e^{\pi w}+1}\right)}\frac{\pi e^{\pi w}}{(1+e^{\pi w})^2}dw =-\int_{-\infty+i}^{\infty+i} e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr$$ A similar argument shows that $$\int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)-ix\right)}dx = -\int_{-\infty-i}^{\infty-i} e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr$$ Using contour integration Now consider the rectangle in the complex plane with vertices $t+i,-t+i,-t-i,t-i$ with $t\in \mathbb{R}^{+}$

And let $$f(z) = e^{\left(\frac{z}{1-e^{\pi z}}\right)}\frac{\pi e^{\pi z}}{(1-e^{\pi z})^2}$$ The function $f$ is analytic except at $z=0$. We can calculate a contour integral over the edges of the rectangle $C= C_{1}\cup C_{2}\cup C_{3} \cup C_{4}$.
Since the rectangle $C$ is a closed contour we can apply the Cauchy's residue theorem :
$$\oint_{C} f(z)dz= \oint_{C_{1}} f(z)dz + \oint_{C_{2}} f(z)dz + \oint_{C_{3}} f(z)dz + \oint_{C_{4}} f(z)dz = 2\pi i \underset{z=0}{\textrm{Res}}f(z)$$ where $z=0$ is the only pole of $f(z)$ inside $C$. Clearly $\displaystyle \oint_{C_{2}} f(z)dz, \oint_{C_{4}} f(z)dz$ vanish when $t \to \infty$. $$\lim_{t\to \infty}\int_{-t-i}^{-t +i } e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2} dr= \lim_{t\to \infty}\int_{t-i}^{t +i } e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr = 0$$ And $$\begin{align} \label{eu_eqn} \lim_{t\to \infty} \oint_{C_{1}} f(z)dz = &\int_{\infty+i}^{-\infty+i} e^{\left(\frac{z}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr = -\int_{-\infty+i}^{\infty+i} e^{\left(\frac{z}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr \\ \lim_{t\to \infty} \oint_{C_{3}} f(z)dz = &\int_{-\infty-i}^{\infty-i} e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr \end{align}$$ These are the integrals that we were looking for. We just have to find $\displaystyle \underset{z=0}{\textrm{Res}}f(z)$. We know that $z=0$ is a pole of order 2, then: $$\underset{z=0}{\textrm{Res}}f(z) = \lim_{z \to 0 } \frac{d}{dz} \left(z^2e^{\left(\frac{z}{1-e^{\pi z}}\right)}\frac{\pi e^{\pi z}}{(1-e^{\pi z})^2}\right) = \frac{1}{2\pi e^\frac{1}{\pi}}$$ This limit can be obtained with the l'Hôpital rule.
Therefore: $$\begin{align*} \int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}\sin(x)dx = & \frac{1}{2i}\int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)+ix\right)}dx - \frac{1}{2i}\int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)-ix\right)}dx\\ =& \frac{1}{2i} \left(\int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)+ix\right)}dx - \int_{0}^{1} e^{\left(\frac{x}{\pi}\ln(1-x)-\frac{x}{\pi}\ln(x)-ix\right)}dx \right)\\ =& \frac{1}{2i} \left(-\int_{-\infty+i}^{\infty+i} e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr + \int_{-\infty-i}^{\infty-i} e^{\left(\frac{r}{1-e^{\pi r}}\right)}\frac{\pi e^{\pi r}}{(1-e^{\pi r})^2}dr\right)\\ =& \frac{1}{2i} \left[ 2\pi i \underset{z=0}{\textrm{Res}}f(z)\right]\\ =& \pi \lim_{z \to 0 } \frac{d}{dz} \left(z^2e^{\left(\frac{z}{1-e^{\pi z}}\right)}\frac{\pi e^{\pi z}}{(1-e^{\pi z})^2}\right)\\ =& \frac{1}{2 e^\frac{1}{\pi}} \end{align*}$$ Therefore $$\boxed{ \int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}\sin(x)dx =\frac{1}{2 e^\frac{1}{\pi}}}$$

Feynman's trick III

Integral involging log-sine

We prove the following integral using the famous Feynman's trick $$\int_{0}^{\frac{\pi}{2}} \ln^2(\sin x) dx = \frac{\pi \ln^2 2}{2} + \frac{\pi^3}{24}$$
Proof
$$\begin{align*} \int_{0}^{\frac{\pi}{2}} \ln^2(\sin x) dx =& \int_{0}^{\frac{\pi}{2}} \frac{d^2}{dt^2}\Big|_{t=0+} \sin^t(x) dx \\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{\frac{\pi}{2}} \sin^t(x) dx \quad (w \mapsto \sin x)\\ =& \frac{d^2}{dt^2} \Big|_{t=0+}\int_{0}^{1} w^t(1-w^2)^{-\frac{1}{2}} dw \quad (y \mapsto w^2)\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \frac{1}{2}\int_{0}^{1} y^{\frac{t}{2}-\frac{1}{2}}(1-y)^{-\frac{1}{2}} dw \\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \frac{1}{2} B \left(\frac{t+1}{2},\frac{1}{2}\right)\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{1+t}{2}\right)}{\Gamma\left(1+ \frac{t}{2}\right)}\\ \end{align*}$$ Then $$\begin{align*} \frac{d}{dt}\Big|_{t=0+} \left[ \frac{d}{dt}\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{1+t}{2}\right)}{\Gamma\left(1+ \frac{t}{2}\right)} \right]=& \frac{d}{dt}\Big|_{t=0+} \frac{\sqrt{\pi} \Gamma(\frac{1+t}{2})\left(\psi^{0}(\frac{1+t}{2}) -\psi^{0}(\frac{t}{2}+1)\right)}{4\Gamma(\frac{t}{2}+1)}\\ =& \lim_{t \to 0+} \frac{\sqrt{\pi}\Gamma(\frac{1+t}{2})}{8\Gamma(\frac{t}{2}+1)} \left[\psi^0 (\frac{t}{2}+1)^2-2\psi^0 (\frac{t+1}{2})\psi^0 (\frac{t}{2}+1)+ \psi^0 (\frac{t+1}{2})^2-\psi^1 (\frac{t}{2}+1)+\psi^1 (\frac{t+1}{2})\right] \end{align*}$$ Taking the limit as $\displaystyle t \to 0+$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \ln^2(\sin x) dx = \frac{\pi \ln^2 2}{2} + \frac{\pi^3}{24}}$$