# Asymptotic behavior of the Coulomb logarithm

In transport theory in plasma physics, there's an important integral called the Coulomb logarithm, which relates to the scattering cross section off the Yukawa potential. It can be written as $$\ell(\Lambda) = \int_0^\infty \cos^2\left(\int_0^{u^*}\left[1 - 2\frac{u}{\xi}\exp\left(-\frac{\xi}{\Lambda u}\right) - u^2\right]^{-1/2}du\right) \xi d\xi,$$ where $$u^*$$ is the turning point--the positive solution to $$1 - 2u/\xi\exp[-\xi/(\Lambda u)] - u^2=0.$$

Now, for a plasma, we usually have $$\Lambda \gg 1$$. So being the lazy mathematicians that we are, we instead use the Coulomb potential (the limit $$\Lambda \rightarrow \infty$$), which is exactly solvable but the $$\xi$$ integral diverges, then cut off the $$\xi$$ integral at $$\Lambda$$ and say "close enough". This gives $$\ell(\Lambda)\approx \ln \Lambda$$.

Calculating the above integral numerically indeed gives $$\ell(\Lambda) \sim \ln \Lambda$$ as $$\Lambda\rightarrow\infty$$. But I'd like to be able to show this through analytically, and through a somewhat less handwave-y method. Unfortunately, I'm not really sure where to start--that integral is kind of a hot mess. Any ideas how to get to $$\ell(\Lambda) \sim \ln \Lambda$$ from that?

• Could you clarify the condition on $u^*$? Is it: $$1 - 2\frac{u^*}{\xi}\exp\left(-\frac{\xi}{\Lambda u^*}\right) - u^{*2}>0$$ Oct 19, 2018 at 7:19
• If we denote the $u$ integral $I_1$, then is the outer integral $\int_0^\infty \cos (I_1^2) \xi d \xi$ or $\int_0^\infty ( \cos I_1)^2 \xi d \xi$? Oct 19, 2018 at 8:01
• @YuriyS That quantity is zero, since any larger $u^*$ would make the integral have an imaginary part. And it's $(\cos I_1)^2$. Oct 19, 2018 at 10:49
• What do you actually want to calculate? The divergent part was (sort of) obtained by Yuriy. But I feel like you really want to determine the regularized finite part? Nov 15, 2018 at 22:15
• @Diger There is no divergent part. The integral is finite for all $\Lambda >0$. I'm looking for a rigorous way to make the estimate $\ell(\Lambda) \sim \ln \Lambda$ as $\Lambda \rightarrow 0$. Nov 15, 2018 at 22:24

I have tried to expand the integrand up to first order in $$1/ \Lambda$$, however I still obtained a divergent integral.

I will still provide the attempt, as it might be useful.

First, I denote:

$$t=\frac{1}{\Lambda}$$

We are interested in the function

$$f(t)=\int_0^\infty \xi ~\mathrm{d} \xi \cos^2 \int_0^{u^*} \frac{du}{\sqrt{1-u^2-2\frac{u}{\xi} \exp (-\frac{t ~\xi}{u})}}$$

It makes sense to change the variables in both integrals:

$$v=u / \xi \\ \xi^2=w$$

Then:

$$f(t)=\frac{1}{2}\int_0^\infty \mathrm{d} w \cos^2 \left(\sqrt{w} \int_0^{v^*} \frac{dv}{\sqrt{1-w v^2-2v \exp (-\frac{t }{v})}} \right)$$

Where $$v^*$$ is the (smallest positive) root of:

$$1-w v^2-2 v \exp (-\frac{t }{v})=0$$

Note that we can explicitly define the function $$w(v^*)$$.

Now comes the tricky part. The most simple way to evaluate the inner integral for small $$t$$ is expanding the exponential up to first order, then:

$$1-w v^2-2v \exp (-\frac{t }{v}) \approx 1+2t-w v^2-2v$$

We obtained a simple integral, which has an exact expression:

$$\int_0^{v^*} \frac{dv}{\sqrt{1+2t-w v^2-2v}}=\frac{1}{\sqrt{w}} \left(\arcsin \frac{1+w v^*}{\sqrt{1+(1+2t)w}}-\arcsin \frac{1}{\sqrt{1+(1+2t)w}} \right)$$

From the condition:

$$1+2t-w v^{*2}-2v^*=0$$

we obtain:

$$1+w v^*=\sqrt{1+(1+2t)w}$$

So now we have:

$$f(t) \approx \frac{1}{2}\int_0^\infty \mathrm{d} w \cos^2 \left( \frac{\pi}{2}-\arcsin \frac{1}{\sqrt{1+(1+2t)w}} \right)=$$

$$=\frac{1}{2} \int_0^\infty \frac{\mathrm{d} w}{ 1+(1+2t)w}=\frac{1}{2(1+2t)} \ln (1+(1+2t)w) \bigg|^\infty_0$$

This integral diverges logarithmically.

If, as the OP said, we "cut off" the integral at $$w=\Lambda^2=1/t^2$$, we get:

$$f(t) \approx \frac{1}{2(1+2t)} \ln (1+(1+2t)w) \bigg|^{1/t^2}_0 \approx \frac{1}{(1+2t)} \ln \frac{1}{t} \approx \ln \frac{1}{t} = \ln \Lambda$$

However, this is just the same trick with a more complicated preliminaries.

I believe, we will have to use more terms in the expansion of the exponential to get a convergent integral, if we even can do it that way. Maybe play with the limits of the $$u$$ integral somehow.

• typo: $v=u/\xi$. Nov 15, 2018 at 22:20
• @Diger, fixed, thanks Nov 15, 2018 at 22:27

I want to build on what Yuriy already established. First of all let me emphasize that I think at no order in the small $$t$$ approximation you will get a finite integral. This is because the equation $$1-wv^2-2v \exp\left({-\frac{t}{v}}\right)=0 \tag{0}$$ has an approximate solution $$v^* \approx \frac{1}{\sqrt{w}} \tag{1}$$ in the large $$w$$ limit which for any finite $$t>0$$ will have $$\frac{t}{v^*} \approx \sqrt{w} \,t >>1$$ and the exponential will be suppressed correspondingly. Since $$\exp\left(-\frac{t}{v}\right)$$ does not have a power series expansion about $$v=0$$ (meaning all the expansion coefficients in the power series expansion being $$0$$) for large enough $$w$$ indeed (1) holds to all orders as the next order is exponentially small and we have

$$\sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2-2v\exp\left(-t/v\right)}} = \frac{\pi}{2} + {\cal O}\left( \exp\left(-\sqrt{w} \, t\right) \right) \, .$$

But this is not what happens in the small $$t$$ approximation if we cut the expansion at order $$n>2$$: The large $$w$$ dependence of the solution will behave like $$v^* \approx (-1)^{n} \left(\frac{2 \, t^{n-1}}{(n-1)! \, w}\right)^{\frac{1}{n}} \, .$$

In Yuriy's case $$n=2$$ this expansion is $$v^* = \sqrt{\frac{2t+1}{w}} - \frac{1}{w} + {\cal O}\left(w^{-3/2}\right)$$ which coincidentally matches (1) at $$t=0$$, but the next order is only $$1/w$$ which leads to

$$\sqrt{w}\int_0^{v^*} \frac{1}{\sqrt{1-wv^2-2v \exp\left(-\frac{t}{v}\right)}} \approx \sqrt{w}\int_0^{v^*} \frac{1}{\sqrt{1-wv^2}} = \frac{\pi}{2} + {\cal O}\left(w^{-1/2}\right)$$ for $$w$$ very large, which order-wise is not far away enough from $$\pi/2$$ for the $$\cos^2$$ function to vanish appropriately.

Strategy:

\begin{align} \left. \begin{array}{l} \sqrt{w} \int_0^{-\frac{1}{w} + \sqrt{ \frac{1}{w^2} + \frac{1}{w} }} \frac{{\rm d}v}{\sqrt{1-wv^2-2v}} \\ \sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2}} \end{array} \right\} &\leq \sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2-2v \exp\left(-t/v\right)}} \\ &\leq \sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2-2v \exp\left(-t/v^*\right)}} \tag{2} \end{align}

where $$v^*$$ is the solution to (0) which has the following obvious limits

$$-\frac{1}{w} + \sqrt{ \frac{1}{w^2} + \frac{1}{w} } \leq v^* \leq \frac{1}{\sqrt{w}} \, . \tag{3}$$

Note that we can implicitly express the solution of (0) as

$$v^*=-\frac{\epsilon}{w} + \sqrt{ \left( \frac{\epsilon}{w} \right)^2 + \frac{1}{w} } \tag{4}$$

with $$\epsilon=\exp\left(-\frac{t}{v^*}\right) \leq 1$$. The lower left inequality and the right inequality are obvious by comparison of the integrand. The upper left inequality follows by writing it as \begin{align} \int_0^{-\frac{1}{w} + \sqrt{ \frac{1}{w^2} + \frac{1}{w} }} \frac{{\rm d}v}{\sqrt{1-wv^2-2v}} \leq \int_0^{-\frac{\epsilon}{w} + \sqrt{ \left( \frac{\epsilon}{w} \right)^2 + \frac{1}{w} }} \frac{{\rm d}v}{\sqrt{1-wv^2-2v \exp\left(-t/v\right)}} \tag{5} \end{align} and the observation that the RHS of (4) as a function of $$\epsilon$$ is decreasing.

We now first calculate the RHS and lower LHS of (2) by using the formula

$$\sqrt{w} \int_0^{-\frac{\epsilon}{w} + \sqrt{ \left( \frac{\epsilon}{w} \right)^2 + \frac{1}{w} }} \frac{{\rm d}v}{\sqrt{1-wv^2-2v \epsilon}} = \frac{\pi}{2} - \arcsin\left(\frac{\epsilon}{\sqrt{\epsilon^2 + w}}\right)$$

for general $$\epsilon$$ and

$$\sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2}} = \arcsin\left(\sqrt{w} \, v^*\right) \, .$$

When applying $$\cos^2$$ the inequalities in (2) reverse since the result is $$\in (0,\pi/2)$$ and we get

\begin{align} \left. \begin{array}{l} \int_0^\infty {\rm d}w \, \frac{1}{w+1} \\ \int_0^\infty {\rm d} w\, {2\epsilon} v^* \end{array} \right\} &\geq \int_0^\infty {\rm d} w\, \cos^2 \left( \sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2-2v \exp\left(-t/v\right)}} \right) \\ &\geq \int_0^\infty {\rm d} w\, \frac{\epsilon^2}{\epsilon^2 + w} \tag{6} \, . \end{align}

The upper integral on the LHS diverges: we'll return to it later. The other integral on the LHS can be approximated by the following sequence

$$\int_0^\infty {\rm d} w\, {2\epsilon} v^* \leq 2 \int_0^\infty {\rm d}w \, \frac{{\rm e}^{-\sqrt{w} \, t}}{\sqrt{w}} = \frac{4}{t}$$

where use of the RHS of (3) was made, which unfortunately is not quite the upper bound we are looking for. Likewise for the RHS

$$\int_0^\infty {\rm d} w\, \frac{\epsilon^2}{\epsilon^2 + w} \geq \int_0^\infty {\rm d} w\, \frac{{\rm e}^{-2t \, \left(1+\sqrt{1+w}\right)}}{1 + w} = 2 \, {\rm e}^{-2t} \, {\rm E}_1(2t) = -2 \left\{ \log(2t) + \gamma \right\} + {\cal O}(t)$$

where use of the LHS of (3) was made.

Going back to the LHS we split the $$w$$-integral according to

$$\int_0^{w_0} {\rm d}w + \int_{w_0}^{\infty} {\rm d}w$$

and notice that (2) or (6) are valid $$\forall w>0$$, so we can use the upper quantity on the LHS of (6) or (2) for the first integral up to $$w_0$$ and the lower quantity on the LHS for the second integral from $$w_0$$ to $$\infty$$. It then follows readily that $$\int_{0}^{w_0} {\rm d}w \, \frac{1}{w+1} + \int_{w_0}^{\infty} {\rm d}w \, 2\epsilon v^* \leq \int_{0}^{w_0} {\rm d}w \, \frac{1}{w+1} + 2\int_{w_0}^{\infty} {\rm d}w \, \frac{{\rm e}^{-\sqrt{w}\,t}}{\sqrt{w}} \\ = \log (w_0+1) + \frac{4}{t} \, {\rm e}^{-\sqrt{w_0} \, t}$$ and by choosing $$w_0={t^{-2\delta-2}}$$ we have $$=-(2\delta+2) \log(t) + \log\left(1+t^{2\delta+2}\right) + \frac{4}{t} \, {\rm e}^{-t^{-\delta}}$$ $$\forall \delta >0$$.

The result therefore is \bbox[lightyellow] { \eqalign{ &-(\delta+1) \log(t) + \frac{1}{2} \log\left(1+t^{2\delta+2}\right) + \frac{2}{t} \, {\rm e}^{-t^{-\delta}} \cr \geq &\frac{1}{2}\int_0^\infty {\rm d} w\, \cos^2 \left( \sqrt{w} \int_0^{v^*} \frac{{\rm d}v}{\sqrt{1-wv^2-2v \exp\left(-t/v\right)}} \right) \cr \geq &-\left\{ \log(2t) + \gamma \right\} + {\cal O}(t) \, . } }