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?
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 $\Re(b)>0$.