1
$\begingroup$

I want to integrate integral $A$, $$A=\int_{-\frac{\mu}{\beta}}^{\infty}\exp\left[-\left(\frac{\beta}{2}+1\right)z-e^{-z}\right]\textrm{d}z,$$ where $\mu,\beta,z>0$. Unfortunately, so far I have been unable to find a close form solution.

So far I have tried substituting $t=z+\frac{\mu}{\beta}$, which yields, $$A=\int_{0}^{\infty}\exp\left[-\left(\frac{\beta}{2}+1\right)t+\frac{\mu}{2}+\frac{\beta}{2}-e^{-t+\frac{\mu}{\beta}}\right]\textrm{d}t.$$

Again, I could not find a closed form solution.

Any ideas on how I could proceed with this? Could you suggest another substitution perhaps?

  • 1
    When asking for a closed form of an integral, particularly of an indefinite integral but even for a definite integral, you should say why you think it may have a closed form in terms of elementary functions (or whatever domain you mean when you say "closed form"). Most integrals will not have a closed form, so for this question to be interesting, you have to say why you think this one may have one. – 2017-02-21
  • 0
    Also, $z$ can't be more than zero on the entire integral, since the lower llimit is less than zero if $\mu$ abd $\beta$ are positive. – 2017-02-21
  • 0
    This integral is a part of a larger equation for the error probability of a system. I want to find a closed-form solution so that I can analyse the performance of the system. – 2017-02-22
  • 0
    Yes, sorry. $-\frac{\mu}{\beta} < z < inf$. – 2017-02-22

1 Answers 1

0

The equation can be rearranged as, $$A=\exp\left(\frac{\mu}{2}+\frac{\mu}{\beta}\right)\int_{0}^{\infty}\exp\left[-\left(\frac{\beta}{2}+1\right)t-e^{\frac{\mu}{\beta}}e^{-t}\right]\textrm{d}t.$$ The integral is in the form, $\int_{0}^{\infty}\exp\left(-Ax-Be^{-x}\right)dx$, where the integral solution can be found in [Gradshteyn2007, eq. (3.331-1)], in the form, $B^{-A}\gamma\left(A,B\right)$ conditioned on $\left[\textrm{Re}A>0\right]$.

Hence the closed-form solution can be given as, $$A=\exp\left(\frac{\mu}{2}+\frac{\mu}{\beta}\right)\exp\left(\frac{\mu_{Q}}{\beta_{Q}}\right)^{-\left(\frac{\beta_{Q}}{2}+1\right)}\gamma\left(\frac{\beta_{Q}}{2}+1,\exp\left(\frac{\mu_{Q}}{\beta_{Q}}\right)\right).$$