# Integrals in the derivation of the Optical Theorem

#### maverick280857

Hi everyone,

I'm trying to work out all the intricate details of the limits involved in computing the integrals leading to the Optical Theorem in nonrelativistic Quantum Mechanics. In doing so, I have arrived at the following integrals which are to be evaluated in the $r \rightarrow \infty$ regime:

$$r^{2}\int_{0}^{2\pi}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta)d\phi = I_{1}$$
$$r^{2}\int_{0}^{2\pi}\int_{-1}^{+1}f^{*}(\theta)e^{-ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)+\frac{1}{r^2}\right)d(\cos\theta)d\phi = I_{2}$$

Modulo the $\phi$ integral, the expression for $I_{1}$ is

$$r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta) = r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)\right)d(\cos\theta) - \int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}d(\cos\theta)$$

I am stuck at these integrals. I know what the "final" term in the optical theorem is, but without using the asymptotic solution, I don't know the right direction to proceed in. Specifically, if I consider $f(\theta)(1+\cos\theta)$ as the "first" function in the first integral, then integrating by parts I get

$$\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}(1+\cos\theta)d(\cos\theta) = \left[f(\theta)(1+\cos\theta)\frac{e^{ikr(1-\cos\theta)}}{-ikr}\right]_{-1}^{+1} - \int_{-1}^{+1}\frac{e^{ikr(1-\cos\theta)}}{-ikr}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)} + f(\theta)\right)d(\cos\theta)$$

Therefore,

$$r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)\right)d(\cos\theta) = ikr\left[\mbox{b.d.}\right] + \int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)} + f(\theta)\right)d(\cos\theta)$$

where b.d. denotes the boundary term in the second last equation. Put together, these equations give

$$r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta) = ikr\left[\mbox{b.d.}\right] + \int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)}\right)d(\cos\theta)$$

Clearly the boundary term tends to zero as $r \rightarrow \infty$, so my original question reduces to the following question:

How does one evaluate the following integrals

$$\int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)}\right)d(\cos\theta)$$
$$\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}d(\cos\theta)$$

in the $r \rightarrow \infty$ regime? I want to get some more insight into the physics as $r \rightarrow \infty$ so I'm hoping that an explicit evaluation of all these terms will be helpful.

Thanks.

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#### Count Iblis

Doesn't the saddle point method work? If r is very large, then the contribution to the integral will come from a narrow range of theta near the point where the aregument of the exponential is extremal.

So, you expand i k r (1-cos(theta)) around its extremal values and then, for large r, you end up with Gaussian integrals.

#### maverick280857

If r is very large, then the contribution to the integral will come from a narrow range of theta near the point where the aregument of the exponential is extremal.
That seems intuitive. But I don't know how to rigorously justify it. Is this the saddle point method?

(I can plug in the asymptotic form of the spherical bessel functions in

$$e^{ikr\cos\theta} = \sum_{l=0}^{\infty}(2l+1)i^{l}j_{l}(kr)P_{l}(\cos\theta)$$

and get the limit.)

"Integrals in the derivation of the Optical Theorem"

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