inverse trigonometric complex integral?

$\begin{aligned} I & = \int_0^{e^\pi} \sin^{-1} \left(\frac i2 \left(\sqrt x - \frac 1{\sqrt x}\right)\right) dx \\ & = \int_0^{e^\pi} \sin^{-1} \left(\frac i2 \left(e^{\frac {\ln x}2} - e^{-\frac {\ln x}2}\right)\right) dx \\ & = \int_0^{e^\pi} \sin^{-1} \left(\frac i2 \left(e^{-i \left(i\frac {\ln x}2\right)} - e^{i\left(i\frac {\ln x}2\right)}\right)\right) dx \\ & = \int_0^{e^\pi} \sin^{-1} \left(\sin \left(i\frac {\ln x}2\right)\right) dx \\ & = \int_0^{e^\pi} i\frac {\ln x}2 dx \\ & = \frac {ix(\ln x -1)}2 \bigg|_0^{e^\pi} \\ & = \frac {ie^\pi (\pi - 1)}2 \end{aligned}$

Therefore, $\Re(A) + \Im(A) = 0 + \dfrac {e^\pi (\pi - 1)}2 \approx \boxed{24.779}$ .

We will first consider the relation $\large{\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}}$ , as to why the sin relation is not used here as a more natural means is attributed to the latter fact that some minus signs will not work out, the $\cos \theta$ relation works out better

Now upon substitution of $\theta = -i\ln (i\sqrt {x})$ ,

$\cos (-i\ln (i\sqrt {x}))$ $=$ $\frac{e^{i \times i\ln (i\sqrt {x})} + e^{-i \times i\ln (i\sqrt {x})}}{2}$

$\frac{e^{i \times i\ln (i\sqrt {x})} + e^{-i \times i\ln (i\sqrt {x})}}{2}$ $=$ $\frac{i \sqrt {x} + \frac{1}{i\sqrt {x}}}{2}$ $=$ $\frac{i(x-1)}{2\sqrt {x}}$

$\therefore \cos (-i\ln (i\sqrt {x})) = \frac{i(x-1)}{2\sqrt {x}}*$

$-i\ln (i\sqrt {x}) = \arccos (\frac{i(x-1)}{2\sqrt {x}})$

Note: if the $\sin \theta$ relation were used, the result would have been $\frac{i(x+1)}{2\sqrt {x}}$

Now, $\arcsin \theta = -\arccos \theta + \frac{\pi}{2}$

$\therefore \arcsin \frac{i(x-1)}{2\sqrt {x}} = -\arccos \frac{i(x-1)}{2\sqrt {x}}+ \frac{\pi}{2}$ $=$ $-(-i\ln (i\sqrt {x})) + \frac{\pi}{2}$ $=$ $i\ln (i\sqrt {x}) + \frac{\pi}{2}$

Now, $i\ln (i\sqrt {x}) + \frac{\pi}{2} = i\ln (i) + \frac{\pi}{2} + i\ln (\sqrt {x})$

Where $i\ln (i) = \ln i^{i} = \ln (e^{-\frac{\pi}{2}}) = -\frac{\pi}{2}$ , the expression simplifies down to just $i\ln (\sqrt {x})$

The question at hand now becomes $\int_{0}^{e^{\pi}} \frac{1}{2} i \ln (x) dx$ and by means of contour integration,

$\int_{0}^{e^{\pi}} \frac{1}{2} i \ln (x) dx = i \int_{0}^{e^{\pi}} \frac{1}{2} \ln (x) dx = i[x\ln (x) - x]_{0}^{e^{\pi}} = \frac{1}{2}ie^{\pi}(\pi - 1)$

$\therefore ℜ(A)+ℑ(A) = 0 + \frac{1}{2}e^{\pi}(\pi - 1) = \boxed {24.77896867}$

*As to why $\cos (i\ln (i\sqrt {x})) = \frac{i(x-1)}{2\sqrt {x}}$ can't be used instead feels intuitively clear, but i don't have a rigorous justification.

First we look at how we arrive at the form $\displaystyle\arcsin \left(\frac{i \left(x-1\right)}{2\sqrt{x}}\right)$ from another form, from which we can equate this to a much simpler representation.

Consider the integral $\displaystyle \int \frac{1}{1+x^{2}} dx =\arctan\left(x\right)$

$\displaystyle \frac{1}{1+x^{2}} =\displaystyle \frac{1}{2}\left(\frac{1}{1+ i x} +\frac{1}{1-ix} \right)$

$\displaystyle \int \frac{1}{2}\left(\frac{1}{1+ i x} +\frac{1}{1-ix} \right) dx =\arctan x$

$\displaystyle \frac{1}{2i}\left( \ln\left(1+ i x\right)-\ln\left(1-ix \right) \right) =\arctan x$

$\displaystyle \frac{1}{2i}\left( \ln\frac{\left(1+ i x\right)}{\left(1-ix \right)} \right) =\arctan x$

Let $\displaystyle m=\frac{\left(1+ i x\right)}{\left(1-ix \right)}$ , rearranging we have

$\displaystyle x= i\times \frac{1-m}{m+1}$

$\displaystyle \frac{1}{i} =\frac{-1\times \left(i^{2}\right)}{i}=-i$

$\displaystyle \frac{1}{2i}\left( \ln m \right) =\arctan \left( i\times \frac{1-m}{m+1}\right)$

we know,

$\arctan\left(q\right) =\arcsin \left(\frac{q}{\sqrt{1+q^{2}}}\right)$

Let $q= i\times \frac{1-m}{m+1}$

we have $\displaystyle \arctan \left( i\times \frac{1-m}{m+1}\right)=\displaystyle \arcsin\left(\frac{ i\times \frac{1-m}{m+1}}{\sqrt{1+\left( i\times \frac{1-m}{m+1}\right)^{2}}}\right)$

$\displaystyle \arcsin\left(\frac{ i\times \frac{1-m}{m+1}}{\sqrt{1+\left( i\times \frac{1-m}{m+1}\right)^{2}}}\right)=\displaystyle \arcsin\left(\frac{ i\times\left( 1-m\right)\times\left(m+1\right)}{\left(m+1\right)\sqrt{\left(m+1\right)^{2}+\left( i^{2}\times \left(1-m\right)^{2}\right)}}\right)$

$=\displaystyle \arcsin\left(\frac{ i\times\left( 1-m\right)}{\sqrt{\left(m+1\right)^{2}-\left(1-m\right)^{2}}}\right)=\displaystyle \arcsin\left(\frac{ i\times\left( 1-m\right)}{\sqrt{m^{2} +2m +1-\left(m^{2} -2m +1\right)}}\right) =\displaystyle \arcsin\left(\frac{ i\times\left( m-1\right)}{2\sqrt{m}}\right)$

We have

$\displaystyle \arcsin\left(\frac{ i\times\left( 1-m\right)}{2\sqrt{m}}\right) =\arctan \left( i\times \frac{1-m}{m+1}\right)=\displaystyle \frac{1}{2i}\left( \ln m \right)$

$\displaystyle \frac{1}{i} =\frac{-1\times \left(i^{2}\right)}{i}=-i$

$\displaystyle \arcsin\left(\frac{ i\times\left( 1-m\right)}{2\sqrt{m}}\right) =\displaystyle \frac{-i}{2}\left( \ln m \right)$

But we know $\arcsin\left(-x\right) =-\arcsin x$

So we rearrange again to obtain $-\displaystyle \arcsin\left(\frac{ i\times\left( m-1\right)}{2\sqrt{m}}\right) =\displaystyle \frac{-i}{2}\left( \ln m \right)$

$\displaystyle \arcsin\left(\frac{ i\times\left( m-1\right)}{2\sqrt{m}}\right) =\displaystyle \frac{i}{2}\left( \ln m \right)$

$\displaystyle \int_{0}^{e^{\pi}} \arcsin\left(\frac{ i\times\left( m-1\right)}{2\sqrt{m}}\right) dm=\int_{0}^{e^{\pi}} \displaystyle \frac{i}{2}\left( \ln m \right) dm$

$\int_{0}^{e^{\pi}} \displaystyle \frac{i}{2}\left( \ln m \right) dm =\displaystyle \frac{i\times m}{2}\left( \ln m -1\right)\Biggr|_{0}^{e^{\pi}}$

$I=\displaystyle \frac{i\times m}{2}\left( \ln m -1\right)\Biggr|_{0}^{e^{\pi}}=\frac{i\times e^{\pi} \left(\pi-1\right)}{2}=0+24.7789i$

$\Re\left(I\right)=0$

$\Im\left(I\right)=24.7789$

this can also be derived from hyperbolic sine function