Up to the challenge?

First, let's derive an expression for the $n^{th}$ root. We have -

$\int_{0}^{\frac {\pi}{2}} \sqrt [n] {\tan \eta} d\eta = \int_{0}^{\frac {\pi}{2}} \sin^{2(\frac {1}{2n} + \frac {1}{2}) - 1} \eta \cdot \cos^{2(\frac {1}{2} -\frac {1}{2n}) -1} \eta d\eta = \frac {1}{2}\Beta\left(\frac {1}{2n} + \frac {1}{2}, \frac {1}{2} - \frac {1}{2n}\right) = \frac {\Gamma(\frac {1}{2n} + \frac {1}{2}) \Gamma(\frac {1}{2} -\frac {1}{2n})}{2} = \frac {\pi}{2} \csc \left(\frac {\pi}{2n} + \frac {\pi}{2}\right) = \frac {\pi}{2} \sec \left(\frac {\pi}{2n}\right)$

by Euler's reflection formula.

In our case, $n = -i.$ Therefore,

$\int_{0}^{\frac {\pi}{2}} \frac {d\eta}{\sqrt[i] {\tan \eta}} =\frac {\pi}{2\cos (\frac {i\pi}{2})} = \frac {\pi}{2\cosh\left(\frac {\pi}{2}\right)} = \frac {\pi}{e^{\frac {\pi}{2}} + e^{\frac {-\pi}{2}}}$

$\color{#0C6AC7} {\boxed {\Large{\therefore \alpha + \beta + \gamma + \pi - 3 = \pi -2}}}$

NOTE - To extend this to any complex number, $\cos(a + bi)$ can be evaluated as follows

We know from Euler's Formula that -

$\cos x = \frac {e^{ix} +e^{-ix} } {2} \therefore \cos (a + bi) = \frac {e^{ai - b} + e^{-ai +b}}{2}$

as $\sin x$ is an odd function.

Separating real and imaginary parts by gives us -

$\cos (a + bi) = \frac {(e^{-b} + e^{b}) \cos a }{2} + \frac {i(e^{-b} - e^{b})\sin a}{2}$

Similar solution with @N. Aadhaar Murty's

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {dx}{\sqrt[i]{\tan x}} \\ & = \int_0^\frac \pi 2 \sin^i x \cos^{-i} x \ dx & \small \blue{\text{Beta function } B(m,n) = 2\int_0^\frac \pi 2 \sin^{2m-1} x \cos^{2n-1} x\ dx} \\ & = \frac 12 B \left(\frac {1+i}2, \frac {1-i}2 \right) & \small \blue{B(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\text{, where }\Gamma(\cdot) \text{ is gamma function.}} \\ & = \frac {\blue{\Gamma \left(\frac {1+i}2\right) \Gamma \left(\frac {1-i}2\right)}}{2\red{\Gamma (1)}} & \small \blue{\text{Euler's reflection: } \Gamma(z) \Gamma(1-z) = \frac \pi{\sin (\pi z)}} \\ & = \frac \blue \pi{2\red{(0!)}\blue{\sin \left(\frac {1+i}2\pi\right)}} & \small \red{\Gamma(n) = (n-1)!} \\ & = \frac {\pi i}{e^{\frac {1+i}2\pi}-e^{-\frac {1+i}2\pi}} & \small \blue{\text{Euler's formula: }e^{\theta i}=\cos \theta + i \sin \theta} \\ & = \frac {\pi i}{e^\frac \pi 2 (\cos \frac \pi 2 + i \sin \frac \pi 2) - e^{-\frac \pi 2}(\cos \frac \pi 2 - i\sin \frac \pi 2)} \\ & = \frac \pi{e^\frac \pi 2 + e^{-\frac \pi 2}} \end{aligned}$

Therefore $\alpha+\beta+\gamma+\pi - 3 = \pi - 2 \approx \boxed{1.14}$ .

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References:

• Beta function
• Gamma function
• Euler's formula