Q function

[luca.barletta]

Compute the following integral
$int_0^infty Q(sqrt(2x))1/mue^(-x/mu)dx$

where $Q(x)=1/sqrt(2pi)int_x^infty e^(-z^2/2)dz$

[elgiovo]

Let's start with the value of $Q(sqrt(2x))$: $1/(sqrt(2pi))int_(sqrt(2x))^(oo) e^(- z^2/2) dz=1/2 [1-erf(sqrtx)]$.
So the integral becomes $int_0^(oo) 1/(2mu) [1-erf(sqrtx)] e^(- x/mu) dx=I$. This can be evaluated by parts:
$I=[1/2 [1-erf(sqrtx)](1-e^(-x/mu))]_0^(oo) + int_0^(oo) (e^(-x))/(2sqrt(pi x)) (1-e^(-x/mu)) dx=1/(2sqrt pi) int_0^oo (e^(-x)-e^(-x-x/mu))/(sqrtx)dx$
$=1/(2sqrtpi)* lim_(a to oo) [sqrtpi erf(a)-(sqrtpi erf(sqrt(a(mu+1))/mu)))/sqrt((mu+1)/mu) =1/(2sqrtpi) (sqrtpi-sqrtpi/(sqrt((mu+1)/(mu))))=1/2(1-1/sqrt((mu+1)/mu))$.

[luca.barletta]

Ok. Another way?

[elgiovo]

$Q(x)$ looks like a Fourier transform (not too much), while the whole integral looks like a Laplace transform. Bad idea?

[luca.barletta]

mmh, I think it can't be useful...
Hint: try to write $Q(.)$ in another way