Proving the transform of the Q-function

Question

I have the Gaussian Q-function, given by:

and I want to prove that it can be also expressed as:

Can somebody help explaining how to obtain the second integral from the first?

Answer 1

I will give one possible approach here. Four other recently given methods to show this result can be found here.

We begin by noting that $$Q(x) = \frac{1}{2} \text{erfc} \left (\frac{x}{\sqrt{2}} \right ), \qquad (\dagger)$$ where $\text{erfc} (x)$ is the complementary error function define by $$\text{erfc} (x) = \frac{2}{\sqrt{\pi}} \int^\infty_x e^{-u^2} \, du.$$ The result ($\dagger$) is obtained from your definition for the $Q$-function using a substitution of $t \sqrt{2}$ for $y$.

We start by considering the function $$I(x) = \int^\infty_0 \frac{e^{-x^2 t^2}}{1 + t^2} \, dt, \quad x > 0.$$ On differentiating under the integral sign (using Feynman's trick) we have $$I'(x) = -2x \int^\infty_0 \frac{t^2}{1 + t^2} e^{-x^2 t^2} \, dt.$$ Noting that $$\frac{t^2}{1 + t^2} = 1 - \frac{1}{1 + t^2},$$ our expression for $I'(x)$ can be rewritten as $$I'(x) = -2x \int^\infty_0 e^{-x^2 t^2} \, dt + 2x \int^\infty_0 \frac{e^{-x^2 t^2}}{1 + t^2} \, dt = -\sqrt{\pi} + 2x I(x), \qquad (*)$$ were the well-known result for the value of the Gaussian (probability) integral for the first integral has been used after a substitution of $u = xt$ has been made.

Eq. ($*$) is a first-order linear differential equation. An integrating factor for this equation is $R(x) = e^{-x^2}$. Thus one can rewrite the linear differential equation in the following separable form $$\frac{d}{dx} [e^{-x^2} I(x)] = - \sqrt{\pi} e^{-x^2}.$$ On separating and integrating we have $$e^{-x^2} I(x) = -\frac{\pi}{2} \cdot \frac{2}{\sqrt{\pi}} \int^x_0 e^{-u^2} \, du + C. \qquad (**)$$ To find the value for the arbitrary constant $C$, let $x \to 0^+$. From the definition for our function $I(x)$, when this is done we find $$I(x) \to \int^\infty_0 \frac{dt}{1 + t^2} = \frac{\pi}{2}.$$ Now letting $x \to 0^+$ in Eq. ($**$) we find $C = \pi/2$. So Eq. ($**$), can be written as $$e^{-x^2} I(x) = \frac{\pi}{2} \left [1 - \frac{2}{\sqrt{\pi}} \int^x_0 e^{-u^2} \, du \right ]. \qquad (\ddagger)$$ Recalling the error function $\text{erf} (x)$, which is defined as $$\text{erf} (x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-u^2} \, du,$$ is related to its complement by $\text{erf} (x) + \text{erfc} (x) = 1$, Eq. ($\ddagger$), becomes $$e^{-x^2} I(x) = \frac{\pi}{2} \text{erfc} (x),$$ which after rearranging may be written as $$\text{erfc} (x) = \frac{2}{\pi} \int^\infty_0 \frac{e^{-x^2(1 + t^2)}}{1 + t^2} \, dt.$$ Making a substitution of $t = \cot \theta$ it follows that $$Q(x) = \frac{1}{\pi} \int^{\frac{\pi}{2}}_0 \exp \left (-\frac{x^2}{2 \sin^2 \theta} \right ) \, d\theta, \quad x > 0,$$ while if a substitution of $t = \tan \theta$ is made, as a bonus, we get $$Q(x) = \frac{1}{\pi} \int^{\frac{\pi}{2}}_0 \exp \left (-\frac{x^2}{2 \cos^2 \theta} \right ) \, d\theta, \quad x > 0$$ almost for free!

Answer 2

Both expressions tend to zero as $x\to +\infty$, so it is sufficient to show that their derivatives agree.

Assume that $x>0$ (I leave the other case to you). Let us differentiate the second integral, $$ Q'(x)=-\frac{1}{\pi}\int_0^{\pi/2}\exp(-x^2/(2\sin^2\theta))\frac{x}{\sin^2\theta}\,d\theta. $$ Now, $$ \frac{1}{\sin^2\theta}=1+\cot^2\theta, $$ so, with $y=x\cot\theta$ it holds that $dy=-x/\sin^2\theta\,d\theta$, and hence $$ Q'(x)=-\exp(-x^2/2)\frac{1}{\pi}\int_0^{+\infty}\exp(-y^2/2)\,dy=-\frac{1}{\sqrt{2\pi}}\exp(-x^2/2). $$ Using the fundamental theorem of calculus, we see that the derivative of the first expression agrees with what we just found.