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?

  • $\begingroup$ Could you clarify the condition on $u^*$? Is it: $$1 - 2\frac{u^*}{\xi}\exp\left(-\frac{\xi}{\Lambda u^*}\right) - u^{*2}>0$$ $\endgroup$
    – Yuriy S
    Oct 19, 2018 at 7:19
  • $\begingroup$ If we denote the $u$ integral $I_1$, then is the outer integral $\int_0^\infty \cos (I_1^2) \xi d \xi$ or $\int_0^\infty ( \cos I_1)^2 \xi d \xi$? $\endgroup$
    – Yuriy S
    Oct 19, 2018 at 8:01
  • $\begingroup$ @YuriyS That quantity is zero, since any larger $u^*$ would make the integral have an imaginary part. And it's $(\cos I_1)^2$. $\endgroup$ Oct 19, 2018 at 10:49
  • $\begingroup$ 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? $\endgroup$
    – Diger
    Nov 15, 2018 at 22:15
  • $\begingroup$ @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 $\Lambda \rightarrow 0$. $\endgroup$ Nov 15, 2018 at 22:24

2 Answers 2


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:


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:

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


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

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.

  • 1
    $\begingroup$ typo: $v=u/\xi$. $\endgroup$
    – Diger
    Nov 15, 2018 at 22:20
  • $\begingroup$ @Diger, fixed, thanks $\endgroup$
    – Yuriy S
    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.


We start with the following sandwiching for the inner integral

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


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