I am trying to find the following definite integral:

\begin{equation} I = \int_{0}^{b} Q\left((b-x)\, a \right)\,\frac{x}{\sigma^2}\,\exp\left(-\frac{x^2}{2\sigma^2}\right)\,dx, \end{equation} where $a$, $b$, $\sigma^2$ are some positive constants, and $Q(u)\stackrel{\triangle}{=} \int_{u}^{+\infty}\frac{\exp(-t^2/2)}{\sqrt{2\pi}}\,dt$ is the Gaussian Q function.

I have tried to use integration by parts and use some table of integrals to solve it but in vain. Any help or hint would be highly appreciated!

  • $\begingroup$ Shouldn't you use another variable besides $u$ inside the integral that defines $Q$? $\endgroup$ – Lukas Rollier Dec 29 '19 at 21:53
  • $\begingroup$ Integration by parts is the way to go. I'm a bit too lazy to work it out in detail right now, but you should start by noting that your integral equals $-\int_0^b Q((b-x)a)d\exp(\frac{-x^2}{2\sigma^2}) $, and work from there. You get some constant plus another integral, in which two exponentials are multiplied with each other. This then equals an exponential of the sum of the two arguments in said previous exponentials. split that in a square of something plus a constant, and then make a change of varialbes. $\endgroup$ – Lukas Rollier Dec 29 '19 at 22:05
  • $\begingroup$ And no, I mean you use $u$ both as an edge of the domain over which you are integrating, and as the variable inside your integral. You should use something else inside the integral, because they're different concepts. $\endgroup$ – Lukas Rollier Dec 29 '19 at 22:07
  • $\begingroup$ @Interestedstudent You are absolutely right! I just noticed it and fixed it! Thank you for your the observation :) $\endgroup$ – Morteza Soltani Dec 29 '19 at 22:08
  • $\begingroup$ As it seems to be something related to the error-probability over a Rayleigh fading channel, a feasible way to find an approximate solution is to use the relation $Q(x) \approx \frac{1}{12} \exp(-x^2/2) + \frac{1}{4} \exp(-2x^2/3)$. This approximation is very accurate in the low-to-mid SNR range. The resulting integral is solvable in terms of Error functions / Q-functions (I have checked in Mathematica). Otherwise, you can use Chernoff bound for Q-functions. For a detailed discussion about the approximation, please see ieeexplore.ieee.org/document/1210748 (eqn. 14). $\endgroup$ – Nash J. Dec 29 '19 at 22:21

Using the integration by part, the solution to the above integral denoted by $I$ is as follows: \begin{align} I &= \int_{0}^{b}\underbrace{Q\left((b-x)\,a\right)}_{u}\,\underbrace{\left[-\frac{d}{dx}\left(\exp(-x^2/2\sigma^2)\right)\right]\,dx}_{dv}\\ &= Q(ab) - \frac{1}{2}\exp(-b^2/2\sigma^2) + \frac{a}{\sqrt{2\pi}}\exp(-a^2b^2/2 )\int_{0}^{b}\exp(-\beta^2 x^2 + a^2 b\, x)\,dx \\ &= Q(ab) - \frac{1}{2}\exp(-b^2/2\sigma^2) + \frac{\sqrt{\pi}}{2\beta}\exp(a^4y^2/4\beta^2)\left\{\text{erf}(a^2y/2\beta) + \text{erf}((\beta - a^2/2\beta)y)\right\}. \end{align} where $\beta \stackrel{\triangle}{=} a^2/2 + 1 / 2\sigma^2$, and $\text{erf}(x) \stackrel{\triangle}{=} \frac{2}{\sqrt{\pi}} \int_{0}^{x}\exp(-t^2)\,dt$.


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