Ytukyg

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
$$
ell(Lambda) = int_0^infty cos^2left(int_0^{u^*}left[1 - 2frac{u}{xi}expleft(-frac{xi}{Lambda u}right) - u^2right]^{-1/2}duright) xi dxi,
$$
where $u^*$ is the turning point--the positive solution to $1 - 2u/xiexp[-xi/(Lambda u)] - u^2=0.$

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 $Lambdarightarrowinfty$. 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?

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

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

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.

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 $$1-wv^2-2v expleft({-frac{t}{v}}right)=0 tag{0}$$ 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 $expleft(-frac{t}{v}right)$ 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

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

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

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

Strategy:

We start with the following sandwiching for the inner integral

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

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

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

with $epsilon=expleft(-frac{t}{v^*}right) leq 1$.

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

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

for general $epsilon$ and

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

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

begin{align}
&int_0^infty {rm d}w , frac{1}{w+1} \
geq &int_0^infty {rm d} w, {2epsilon} v^* \
geq &int_0^infty {rm d} w, cos^2 left( sqrt{w} int_0^{v^*} frac{{rm d}v}{sqrt{1-wv^2-2v expleft(-t/vright)}} right) \
geq &int_0^infty {rm d} w, frac{epsilon^2}{epsilon^2 + w} tag{4} , .
end{align}

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

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

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

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