2 subs to kill it!

$\begin{aligned} I & = \int_0^\infty \sin^{-1} (e^{-x}) \ dx & \small \color{#3D99F6} \text{Let }\sin \theta = e^{-x}; \ \cos \theta \ d \theta = - e^{-x} \ dx \\ & = \int_0^\frac \pi 2 \theta \cot \theta \ d \theta & \small \color{#3D99F6} \text{By integration by parts} \\ & = \theta \ln (\sin \theta) \bigg|_0 ^\frac \pi 2 - \int_0^\frac \pi 2 \ln (\sin \theta) \ d \theta \\ & = 0 - \int_0^\frac \pi 2 \ln \left(2 \sin \frac \theta 2 \cos \frac \theta 2 \right) d \theta & \small \color{#3D99F6} \text{Let }\phi = \frac \theta 2; \ d \phi = \frac 12 \ dx \\ & = - 2 \int_0^\frac \pi 4 \ln \left(2 \sin \phi \cos \phi \right) d \phi \\ & = - 2 \int_0^\frac \pi 4 (\ln (2 \sin \phi) + \ln (2\cos \phi) - \ln 2) \ d \phi \\ & = {\color{#3D99F6}- 2 \int_0^\frac \pi 4 (\ln (2 \sin \phi) \ d \phi} - {\color{#3D99F6}2 \int_0^\frac \pi 4 \ln (2\cos \phi) \ d \phi} + 2 \int_0^\frac \pi 4 \ln 2) \ d \phi & \small \color{#3D99F6} \text{See note.} \\ & = {\color{#3D99F6}G} - {\color{#3D99F6}G} + \frac {\pi \ln 2}2 & \small \color{#3D99F6} G \text{ is the Catalan's constant.} \end{aligned}$

$\implies \alpha^\alpha + \phi^\phi = 2^2 + 1^1 = \boxed{5}$

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Note:

$\begin{aligned} - 2 \int_0^\frac \pi 4 \ln (2 \sin \phi) \ d \phi = 2 \int_0^\frac \pi 4 \ln (2\cos \phi) \ d \phi = G \end{aligned}$

where, $G$ is the Catalan's constant .

Let substitute $e^{-x}=sinz \\ -e^{-x}dx=cosz dz$ .

The integration becomes

$\displaystyle \int_{0}^{\pi/2} zcotz dz$ .Integrating By-Parts:

$[zln(sinz)]_{0}^{\pi/2} - \int_{0}^{\pi/2} ln(sinz) dz \\ =-\int_{0}^{\pi/2} ln(sinz) dz$

Now consider $I=\int_{0}^{\pi/2} ln(sinz) dz$ .Using the formula

$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$ it becomes:

$I=\int_{0}^{\pi/2} ln(cosz) dz$ .Adding two integrals :

$2I=\int_{0}^{\pi/2} ln(sinzcosz) dz \\ =\int_{0}^{\pi/2} ln(\dfrac{sin2z}{2}) dz \\ =\int_{0}^{\pi/2} ln(sin2z) dz -\int_{0}^{\pi/2} ln2 dz$

Again $P=\int_{0}^{\pi/2} ln(sin2z) dz$ .

Put $2z=t \\ 2 dz=dt$

$\frac{1}{2} \times \int_{0}^{\pi} ln(sint) dt \\ =frac{1}{2} \times[\int_{0}^{\pi/2} ln(sint) dt + \int_{\pi/2}^{\pi} ln(sint) dt]$

Put $t=p+\pi/2$ in the second part we get $P=\frac{1}{2} \times 2 \int_{0}^{\pi/2} ln(sint) dt \\ = \int_{0}^{\pi/2} ln(sint) dt$

So, $2I=I-\int_{0}^{\pi/2} ln2 dz \\ \implies I=\int_{0}^{\pi/2} ln2 dz$

So, the first integral is $-I=\int_{0}^{\pi/2} ln2 dz=\dfrac{\pi}{2} ln2$