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?

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.
