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 \[\boxed{ \int_{0}^{1} \left(\frac{1-x}{x}\right)^\frac{x}{\pi}\sin(x)dx =\frac{1}{2 e^\frac{1}{\pi}}}\]