So I was bored yet again, and decided to evaluate some integrals. After a while, I came up with this:$$\int_{-1}^{+1}(-1)^{i\pi e^{-x\pi}}dx$$which is based off of $\color{red}{\text{this}}$ integral I did $4$ days ago, which I thought that I might be able to evaluate. Here is my attempt at doing so:$$\begin{align}\int_{-1}^{+1}(-1)^{i\pi e^{-x\pi}}dx\implies&\int_{-1}^{+1}\left(e^{i\pi}\right)^{i\pi (-1)^{ix}}dx\\\implies&\int_{[-1,+1]}\left(e^{i\pi}\right)^{(-1)^{ix}\ln(-1)}dx\\\implies&\int_{[-1,+1]}\left(e^{i\pi\ln(-1)}\right)^{(-1)^{ix}}dx\\\implies&\int_{[-1,+1]}(-1)^{i\pi(-1)^{ix}}dx\\\implies&\int_{[-1,+1]}e^{i\pi e^{i\pi\cdot ix}}dx\end{align}$$Now, this was getting a bit difficult to integrate, so now we use $x=\ln(y)$ to get$$\int_{[\frac1e,e]}e^{i\pi y^{-\pi}}dy\implies\int_{[\frac1e,e]}(-1)^{\frac1{y^\pi}}dy$$Then, letting $y=\dfrac1{\ln(z)}$, we get$$-\int_{\left[\frac1{\sqrt[e]e},\frac1{e^e}\right]}\require{cancel}\cancel e^{i\pi^2\cancel\ln(z)}dz\implies-\int_{\left[\frac1{\sqrt[e]e},\frac1{e^e}\right]}z^{i\pi^2}dz=-\left(\dfrac{(1-i\pi^2)\left(\left(\dfrac1{e^e}\right)^{1+i\pi^2}-\left(\dfrac1{\sqrt[e]e}\right)^{1+i\pi^2}\right)}{1+\pi^4}\right)$$which is the exact value of the integral.
My question
Did I evaluate the integral correctly, or what could I do to evaluate the integral correctly/more quickly?
You made some wrong steps such as your fifth step $$ (-1)^{\pi i (-1)^{i x}}=e^{(\pi i)^2 e^{i \pi . ix}}\ne e^{i\pi e^{i \pi . ix}} $$ and also, the next step when you used $x=\ln(y)$, you didn't take the derivative of $\ln(y)$ to get $dy$ correctly. You can get the indefinite integral of the function:$$\text{let}\quad y=e^{-\pi x}\implies x=-\frac{\ln y}\pi\to dx=\frac{-1}{\pi y}dy$$ $$\implies\int(-1)^{i\pi e^{-x\pi}}dx=\int(-1)^{i\pi y}\frac{-1}{\pi y}dy=\frac{-1}{\pi}\int\frac{e^{-\pi^2 y}}{y}dy=\frac{-1}{\pi}\operatorname{Ei}(-\pi^2y )+c=\frac{-1}{\pi}\operatorname{Ei}(-\pi^2e^{-\pi x})+C$$ $$\therefore\int_{-1}^1(-1)^{i \pi e^{-x\pi}}dx=\left(\frac{-1}{\pi} \operatorname{Ei}(-\pi^2e^{-\pi x})\right)^{+1}_{-1}=\frac{\operatorname{Ei}(-\pi^2e^{\pi})-\operatorname{Ei}(-\pi^2e^{-\pi} )}{\pi}\approx 0.210063$$