# Integral, asymptotic approximation and simple pole

How can I use the asymptotic approximation for integrand functions with simple poles?

Let's consider the integral $$\int_{\infty e^ {\pi 3/4}}^{\infty e^{\pi/4}} \frac {f(s)}{s-s_0}e^{-\Omega{(s^2-2s+1)}}ds$$ let $s_0=\rho_0 e^{j\pi /4}$.

The sadle point is in $s=1.$

I have to include the residue so $$I=-2\pi jf(s_0)e^{-(s_0+1)^2}+\int_{-\infty}^{\infty}\frac{f(s+1)}{s-s_0}e^{-\Omega s^2}$$

But what can I do now? Can I use the Q-integral? How?

Many thanks

• Are there $j$s missing in the limits of the originally given integral? Just all in all it might be clearer to write the complex integral in terms of a limit instead of having those infinities in there. – Ian Jun 5 '17 at 14:32
• @Ian Sorry, I haven't understood your comment. This is the integral that I have.... – sunrise Jun 5 '17 at 14:50
• My guess is that $\int_{\infty e^{3\pi/4}}^{\infty e^{\pi/4}}$ was meant to be $\lim_{R \to \infty} \int_{R e^{3 j \pi/4}}^{R e^{j \pi/4}}$. – Ian Jun 5 '17 at 15:56
• @Ian Ok, I can close the contour in the region of convergence, but then.. what can I do? I can find the steepest descent path (y=0) but how can I manage an integral with a pole? – sunrise Jun 5 '17 at 16:45