- #1

Santilopez10

- 81

- 8

## Homework Statement

The following is a problem from

*"*

**Applied Complex Variables for Scientists and Engineers"**It states:

The following integral occurs in the quantum theory of collisions:

$$I=\int_{-\infty}^{\infty} \frac {sin(t)} {t}e^{ipt} \, dt$$

where p is real. Show that

$$I=\begin{cases}0 & \text{if } |p| \geq 1 \\\pi & \text{if } |p| < 1\end{cases}$$

Discuss the cases when ##p= \pm 1##

## Homework Equations

ML inequality and Cauchy theorem

## The Attempt at a Solution

First, I consider this integral in the complex plane:

$$ \oint_C \frac {e^{iz}} {z} e^{ipz} \, dz$$

where C is the contour defined as a semicircular path of radius R with a cut at the origin, as a semicircular path of radius ##\epsilon##.

Okey, first, we see that no singularity is within the contour, so the integral as a whole is just 0. Then, I proceed to write the integral as different line integrals.

$$ \oint_C \frac {e^{iz}} {z} e^{ipz} \, dz = \int_{\Gamma}\ + \int_{-R}^{\epsilon} \frac {e^{it}} {t} e^{ipt} \, dt + \int_{\gamma}\ + \int_{\epsilon}^{R} \frac {e^{it}} {t} e^{ipt} \, dt = 0$$

Where ##\Gamma## is the larger semicircle, and ##\gamma## is the smaller. I expressed the integrals with variable ##t## because they lie in the real axis.

I start with the smaller semicircle, it can be parametrized as ##z=\epsilon e^{i\phi}, dz=i \epsilon e^{i\phi}## and it goes from ##\phi = \pi## to ##\phi = 0##

There is a theorem that states that the value of the integral is just ##-i\pi Residue(f(z))## and in this case the residue its just 1, so that integral is done.

Now here is where I am stuck with 2 problems:

1) How do I change variables in the Real axes integrals so that I can get back the original integral?

2) This is just trying to know if my estimation of the upper bound of the large semicircle is correct:

I begin using the common half circle parametrization so that I end up with:

$$\int_{0}^{\pi} i e^{i(p+1)R e^{i\theta}}\, d\theta$$

and so I use absolute values to find an upper bound for the integral:

$$|\int_{0}^{\pi} i e^{i(p+1)R e^{i\theta}}\, d\theta|=\int_{0}^{\pi} |i e^{i(p+1)R e^{i\theta}}|\, d\theta$$

then:

$$\int_{0}^{\pi} |i| |e^{i(p+1)R e^{i\theta}}|\, d\theta$$

and

$$\int_{0}^{\pi} |e^{i(p+1)R e^{i\theta}}|\, d\theta = \int_{0}^{\pi}|e^{i(p+1)R (cos(\theta)+isin(\theta)}|\, d\theta =\int_{0}^{\pi} |e^{i(p+1)R cos(\theta)}||e^{-R(p+1)sin(\theta)}| $$

and then

$$\int_{0}^{\pi} |e^{-R(p+1)sin(\theta)}|\, d\theta$$

then by using the fact the integral is simmetric respect to ##\frac {\pi}{2}##:

$$2\int_{0}^{\frac{\pi}{2}} |e^{-R(p+1)sin(\theta)}|\, d\theta$$

and because ##-sin(\theta)\leq -\frac{2\theta}{\pi}, \theta \in (0,\frac{\pi}{2})## we can write the integral as:

$$2\int_{0}^{\frac{\pi}{2}} |e^{-R(p+1)\frac{2\theta}{\pi}}|\, d\theta$$

we can drop the absolute values, and the result of the integral is:

$$\frac{\pi(1-e^{-R(p+1)})}{R(p+1)}$$

which I suppose approaches different values depending on the value of p, still need to check that (if it does not, better, it means it goes to 0 for any value of p and simplifies everything). But still, I would be grateful if someone could tell me if the estimation is correct.

As it seems, the thing that does not let me continue is the way to return to the original integral.