I need to evaluate the following function

$p(T,\lambda,\alpha)=\pi\lambda\int_0^{\infty}\exp\left(-\pi\lambda \beta(T,\alpha)-\mu T\sigma^2v^{\alpha/2}\right)\text{d}v$

Note that when $\sigma^2=0$, the function $p(T,\lambda,\alpha)$ becomes independent of $\lambda$.

$\bf{EDIT}$: There is a little correction. I missed the parameter $v$ in the first part of the exponential. Here is the correct form.

$p(T,\lambda,\alpha)=\pi\lambda\int_0^{\infty}\exp\left(-\pi\lambda v\beta(T,\alpha)-\mu T\sigma^2v^{\alpha/2}\right)\text{d}v$

How to perform low noise ($\sigma^2$) approximation?

  • $\begingroup$ If I am not wrong, this seems to be related to the gamma function. $\endgroup$ – Claude Leibovici Feb 9 '17 at 9:58


If I properly understood the problem, it seems that you need first to compute $$I=\int e^{-a-b v^c}\,dv=e^{-a}\,\int e^{-b v^c}\,dv$$ First, change variable $$b v^c=t\implies v=\left(\frac{t}{b}\right)^{\frac{1}{c}}\implies dv=\frac{1}{b c}\left(\frac{t}{b}\right)^{\frac{1}{c}-1}\,dt$$ which makes $$\int e^{-b v^c}\,dv=\frac{b^{-1/c}}{c}\int e^{-t} t^{\frac{1}{c}-1}\,dt$$ where you could recognize the definition of the incomplete gamma function.

I am sure that you can take it from here.


The above was an answer to the initial post which has been changed to $$I=\int_0^\infty e^{-av-b v^c}\,dv$$ The only cases I have been able to solve are for $c=2$ or $c=\frac 12$. Completing the square, we then arrive to some error functions provided that $\Re(b)>0$.

  • $\begingroup$ Thank you very much for your answer. I really appreciate it. Franky speaking, I am not a mathematician. I come across this problem from my wireless problem. I have understood so far. I would request you to do some more steps. $\endgroup$ – George Harnandez Feb 10 '17 at 3:12
  • $\begingroup$ What incomplete Gamma function it is? Upper incomplete Gamma function or lower incomplete Gamma function $\endgroup$ – George Harnandez Feb 10 '17 at 3:19
  • $\begingroup$ @GeorgeHarnandez. Upper incomplete Gamma function for the antiderivative. With the bounds, the result is a simple expression contianing the standard gamma function. $\endgroup$ – Claude Leibovici Feb 10 '17 at 4:39
  • $\begingroup$ Unfortunately, there is an error. I missed the parameter $v$ in the first part of the exponential. $\endgroup$ – George Harnandez Feb 10 '17 at 6:29
  • $\begingroup$ @GeorgeHarnandez. This makes the problem totally different. I suggest you edit the post to precise this change (otherwise, my answer does not mean anything). If $c=2$, completing the square, you should get $$\frac{\sqrt{\pi } e^{\frac{a^2}{4 b}} \text{erfc}\left(\frac{a}{2 \sqrt{b}}\right)}{2 \sqrt{b}}$$ under the condition $\Re(b)>0$. For sure, by symmetry, we can compute for $c=\frac 12$. For other values, I am stuck. $\endgroup$ – Claude Leibovici Feb 10 '17 at 6:40

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