I am trying to compute a double integral. I will first define the functions that make up the integrand:

$$F(\gamma)= A \,\exp(-a \, \gamma^ {1/2})+B \, \gamma^{-1/2}\left(1-\exp(-b\gamma^ {1/4})\right)$$

$$ G(\gamma_1,\gamma_2) = \exp \left(\frac{T}{\gamma_1^{-1}+\gamma_2^{-1}}\right)$$

where $A, B,T, a, b $ are positive real numbers.

I am trying to analytically compute the double integral, I tried using MATLAB it doesn't work and I don't want to rely on numerical integration:

$$\int_0^{\infty}\int_{\gamma_1}^{\infty} F(\gamma_1)F(\gamma_2)\exp\left(-\int F(\gamma_2) d\gamma_2\right)G(\gamma_1,\gamma_2) d\gamma_2 d\gamma_1$$

Any tricks, ideas, or mathematical programming other than MATLAB you advise me to use is appreciated, or would you think its impossible to compute and numerical integration is the only way to solve this?

  • $\begingroup$ This integral makes me cringe $\endgroup$
    – AvZ
    Feb 15, 2015 at 6:29
  • $\begingroup$ It makes my eyes twitch. I've been staring at it for 5 minutes and still don't know where to begin. $\endgroup$ Feb 15, 2015 at 6:30
  • $\begingroup$ LOL @AvZ it made me cringe too, I am doing research in wireless communication and unfortunately I have arrived to such an integral.. $\endgroup$
    – Tyrone
    Feb 15, 2015 at 6:30
  • 1
    $\begingroup$ In your integral you have the product $F(\gamma_2)F(\gamma_2)$, but do you mean to have $F(\gamma_1)F(\gamma_2)$ instead? $\endgroup$
    – David H
    Feb 16, 2015 at 1:52
  • $\begingroup$ Did you mean to include bounds on your inner integral ($\int F(\gamma_2)\ \mathrm{d}\gamma_2$)? Otherwise, your integral is only defined up to a multiplicative constant. $\endgroup$ Feb 17, 2015 at 3:38

2 Answers 2


I'm gonna take a wild stab at this-does the integrand get any simpler if we convert to polar coordinates like we do in the old "integrate $e ^{-{x^2}}$ as a dummy variable double integral" trick? Except hopefully in this case,it's even easier because it actually IS a double integral. Consider $\gamma_1$ = $ r \cos\theta_1$ and $\gamma_2$ = $ r \cos\theta_2$. Now let's substitute and see if it simplifies the integrand.

$$F(r \cos\theta_1)= A \,\exp(-a \, (r \cos\theta_1)^ {1/2})+B \, r \cos\theta_1^{-1/2}\left(1-\exp(-br \cos\theta_1^ {1/4})\right)$$

$$F(r \cos\theta_2)= A \,\exp(-a \, (r \cos\theta_2)^ {1/2})+B \, r \cos\theta_2^{-1/2}\left(1-\exp(-br \cos\theta_2^ {1/4})\right)$$

Boy,that looks even uglier. The idea is when things are multiplied, hopefully,if the Gods are kind, terms will drop out. Make the appropriate polar substitutions and then plug the integrand into MATLAB. See if the resulting integral is simpler.

Wish I had time to take a crack at it.

  • $\begingroup$ I tried setting the computation up by hand-it doesn't seem to make it any simpler, the resulting integrand is at least as complicated. If there's any simplifying change of variable or substitution,I can't see it. : ( $\endgroup$ Feb 15, 2015 at 22:30
  • $\begingroup$ You could possibly combine this idea with mine below: let $\gamma_1 = (r \cos θ)^4$ to kill the fractional powers. $\endgroup$ Jun 24, 2016 at 16:01
  • $\begingroup$ You might also try $t = \gamma_2 - \gamma_1$ to make the region rectangular $\endgroup$ Jun 24, 2016 at 16:07

Substituting $\gamma = u^4$ at least gives you integer powers everywhere. That gives you the middle integral:

$$\int F(\gamma) d \gamma = \int \left( A \,\exp(-a \, u^2)+\frac{B}{u^2}\left(1-\exp(-b u)\right) \right)4 u^3 du$$ This evaluates to an elementary function, which you can convert back into $\gamma$ if you like. Call it $H(\gamma)$.

Let $g$ be the bit inside the exponential in the function $G$. Integration by parts sucks up $F(\gamma_2)$ into $\exp(H)$, which helps a bit. Then if you can integrate $\exp(H - g) \frac{\partial g}{\partial \gamma_2}$, then you can reduce the problem to a single integral.


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