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In "Mechanics" book by Landau and Lifshitz, $\S18$ "Scattering", problem 7 on deducing the form of scattering field given the effective cross-section as a function of scattering angle an equation is obtained:

$$-\pi\text{d}\log w=\text{d}\left(\frac s w\right)\int_0^{s^2/w^2}\frac{\chi'(x)\text{d}x}{\sqrt{\frac{s^2}{w^2}-x}}.$$

Before this, several variables have been introduced: $s=1/r, x=1/\rho^2, w=\sqrt{1-\frac U E}$, where $U>0, U(\infty)=0$.

They then say:

This equation can be integrated immediately if the order of integration on the right-hand side is inverted. Since for $s=0$ (i.e. $r\to\infty$) we must have $w=1$ (i.e. $U=0$), we have, on returning to the original variables $r$ and $\rho$, the following two equivalent forms of the final result: $$w=\exp\left\{\frac1\pi\int_{rw}^\infty\cosh^{-1}\left(\frac{\rho}{rw}\right)\frac{\text{d}\chi}{\text{d}\rho}\text{d}\rho\right\}=\exp\left\{-\frac1\pi\int_{rw}^\infty\frac{\chi(\rho)\text{d}\rho}{\sqrt{\rho^2-r^2w^2}}\right\}.$$

As I understand, they integrate taking $\frac s w$ as variable of integration. This seems OK, and I tried to take integral with the seemingly sensible limits from $0$ to $\frac s w$ (i.e. variable upper limit), but the result for RHS appears quite strange: denoting $\frac s w=\beta$, here's what I get after switching order of integration:

$$\int_0^\beta \text{d}x\chi'(x)\int_0^\beta\frac{\text{d}\beta}{\sqrt{\beta^2-x}}=\int_0^\beta\text{d}x\chi'(x)\left(\log\left(\beta+\sqrt{\beta^2-x}\right)-\log\sqrt{-x}\right).$$

Given that $x=1/\rho^2$, the RHS seems to become complex, which makes me think I do something wrongly. I suspect that I'm taking wrong limits of integration, but L&L don't specify which ones are to be taken! Supposedly the readers could determine this themselves, but I'm not as smart.

So, what are the right limits of integration here?

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I did not checked the computation, but the logarithm in the last equation is related to the inverse hyperbolic cosine http://mathworld.wolfram.com/InverseHyperbolicCosine.html.

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What I did wrong here was switching order of integration. Let's integrate from $0$ to $\beta'$, preserving the variable $\beta=\frac s w$ I introduced in the OP, and priming it to avoid confusion with integration variable. The RHS integral will then look as

$$\int_0^{\beta'}\text{d}\beta\int_0^{\beta^2}\frac{\chi'(x)\text{d}x}{\sqrt{\beta^2-x}}=\mathcal I.$$

Its domain of integration looks like:

enter image description here

Here we integrate over $x$ from $0$ to the curve, then over $\beta$ from $0$ to the top horizontal line. So, to switch order of integration, we should integrate from the curve to the top over $\beta$, then over $x$ from left to right. So, we get:

$$\mathcal I=\int_0^{x(\beta')=(\beta')^2}\text{d}x\int_{\beta(x)=\sqrt x}^{\beta'}\frac{\chi'(x)\text{d}x}{\sqrt{\beta^2-x}}=\\ =\int_0^{(\beta')^2}\text{d}x\chi'(x)\left.\left(\cosh^{-1}\frac \beta {\sqrt x}\right)\right|_\sqrt x^{\beta'}=\\ =\int_0^{(\beta')^2}\text{d}x\chi'(x)\cosh^{-1}\frac{\beta'}{\sqrt x}.$$

After substitution of the previously introduced variables, the result in Landau & Lifshitz is easy to get.

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