An intriguing integral

The substitution $u = =e^{-xy}$ gives $A(y) \; = \; -y^{-2}\int_0^1 \ln u \,u^{y^{-1}-1}\sqrt{1-u}\,du \; = \; -y^{-2}F'(y^{-1})$ where $F$ is expressed in terms of the beta function: $F(t) \; = \; B(t,\tfrac32)$ Thus $\int_0^1 A(y)\,dy \; = \; -\int_1^\infty F'(t)\,dt \; = \; \lim_{T \to \infty}\big[F(1) - F(T)\big] \; = \; F(1) = B(1,\tfrac32) = \frac{\Gamma(\frac32)}{\Gamma(\frac52)} \; = \; \tfrac23$ making the answer $K = \boxed{2}$ .

$X=\frac { K }{ K+1 } =\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \infty }{ x{ e }^{ -x }\sqrt { 1-{ e }^{ -xy } } dxdy= } } \int _{ 0 }^{ \infty }{ x{ e }^{ -x }\int _{ 0 }^{ 1 }{ \sqrt { 1-{ e }^{ -xy } } dydx } }$

The substitution $u=arcos\left( { e }^{ -\frac { xy }{ 2 } } \right)$ gives:

$\int _{ 0 }^{ 1 }{ \sqrt { 1-{ e }^{ -xy } } dy } =\int _{ 0 }^{ arcos({ e }^{ -\frac { x }{ 2 } }) }{ sec(u)-cos(u)du} =\frac { 2 }{ x } \left( \ln { (1+\sqrt { 1-{ e }^{ -x } } ) } +\frac { x }{ 2 } -\sqrt { 1-{ e }^{ -x } } \right)$

Then:

$X=\int _{ 0 }^{ \infty }{ 2\left( \ln { (1+\sqrt { 1-{ e }^{ -x } } ) } +\frac { x }{ 2 } -\sqrt { 1-{ e }^{ -x } } \right) { e }^{ -x }dx } ={ 2\left[ -{ e }^{ -x }\ln { \left( 1+\sqrt { 1-{ e }^{ -x } } \right) } -\frac { 1 }{ 2 } \left( 1-{ e }^{ -x } \right) +\sqrt { 1-{ e }^{ -x } } -\frac { 1+x }{ 2 } { e }^{ -x }-\frac { 2 }{ 3 } { \left( 1-{ e }^{ -x } \right) }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ \infty }$

$X=\frac { K }{ K+1 }= { 2\left[ \left( -\frac { 1 }{ 2 } +1-\frac { 2 }{ 3 } \right) -\left( -\frac { 1 }{ 2 } \right) \right] =\frac { 2 }{ 3 } =\frac { 2 }{ 2+1 } }$

Thus

$K=2$