0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If I am not wrong, this seems to be related to the gamma function. $\endgroup$ – Claude Leibovici Feb 9 '17 at 9:58
1
$\begingroup$

Hint

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.

Edit

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 $\Re(b)>0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.