1. ## Evaluating Improper integral

1. The calculation of the probability of excitation of an atom originally in the
ground state to an excited state, involves the contour integral
INTEGRAL(-INF TO +INF) [s exp(iwt)/(t2 + s2)2dt]
Evaluate the above integral.

2. Originally Posted by GAGS
1. The calculation of the probability of excitation of an atom originally in the
ground state to an excited state, involves the contour integral
INTEGRAL(-INF TO +INF) [s exp(iwt)/(t2 + s2)2dt]
Evaluate the above integral.
$\int_{-\infty}^{\infty} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt$

Is that the integral?

Yes sir this is that integral

4. $\int_{-\infty}^{\infty} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt = \int_{-\infty}^{0} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt + \int_0^{\infty} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt$

$= \lim_{n \to -\infty} \int_{n}^{0} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt+ \lim_{m \to \infty} \int_{0}^{m} \frac{s e^{(i \omega t)}}{(t^2 + s^2)^2} dt$

Does this help?

Yes upto a bit but we have just changed notations. How to solve this intergal?

6. First solve the integrals, as you would normally, and then take the limits of these functions, then the sum of the limits.

7. I think what we're trying to say is that we don't know how to solve the integral (I don't think the function converges!), and that the only advice we can offer is general and vague advice on how to solve general improper integrals using limits...

8. Originally Posted by GAGS
1. The calculation of the probability of excitation of an atom originally in the ground state to an excited state, involves the contour integral
$\int_{-\infty}^{\infty}\frac{s \exp(i\omega t)}{(t^2 + s^2)^2}dt$
Evaluate the above integral.
I'll assume that $\omega>0$. Integrate the function $\frac{s \exp(i\omega z)}{(z^2 + s^2)^2}$ round the contour consisting of (a long interval on) the real axis plus a large semicircle in the upper half-plane. The result will be equal to 2πi times the residue at the double pole at z=is.

The residue is $\frac d{dz}\Bigl(\frac{se^{i\omega z}}{(z+is)^2}\Bigr)\Bigr|_{z=is}$. If I have done the calculation correctly then $\int_{-\infty}^{\infty}\frac{s \exp(i\omega t)}{(t^2 + s^2)^2}dt = \frac{\pi(s+\omega)e^{-\omega s}}{2s^2}$.

If $\omega<0$ then the answer will be the same except that you have to replace $\omega$ by $|\omega|$.