On page 73, it's written that $-\int_{|r|}^\infty s^{n-1} d_s \Phi(s\xi) = \Psi_\xi(r)$, and beneath that it's written that:
$$(*)\int_{-\infty}^\infty \Psi_\xi(r)dr = 2\int_0^\infty r^n d\Phi(r\xi) = 2n\int_0^\infty r^{n-1}\Phi(r\xi)dr$$
I don't understand how do you get the first equality, I get the following:
$$\int_{-\infty}^\infty \Psi_\xi(r)dr = \int_{-\infty}^\infty (- \int_{|r|}^\infty s^{n-1} d_s \Phi(s\xi)) dr$$
Now, by integration by parts we get:
$$\int_{-\infty}^\infty ([-s^{n-1}\Phi(s\xi)]_{|r|}^\infty + \int_{|r|}^\infty (n-1)s^{n-2}\Phi(s\xi)ds) dr$$
Now the second term must somehow vanish (why?), and the first term transforms to:
$2\int_0^\infty r^{n-1}\Phi(r\xi) dr$ since $r\mapsto |r|^{n-1}\Phi(r\xi)$ is an even function.
Am I right? if I am wrong do correct me here.
p.s There's a preview of this page in google books: https://books.google.co.il/books?id=ljcOSMK7t0EC&printsec=frontcover#v=onepage&q&f=false
This is how I would do the computation. I'm not exactly sure what Stein's doing. It seems like his hypotheses run into technical issues in the details of the computation, which he is glossing over.
Assume first that $\Phi$ is right-continuous, compactly supported, nonnegative, and decreasing. This is to avoid any technical issues in the computations below. In particular, it lets us use Lebesgue-Stieltjes integration by parts.
Observe that
$$\int_{-\infty}^{\infty}\Psi_{\xi}(r)dr=\int_{-\infty}^{\infty}\left(-\int_{\left|r\right|}^{\infty}s^{n-1}d_{s}\Phi(s\xi)\right)dr=2\int_{0}^{\infty}\left(-\int_{r}^{\infty}s^{n-1}d_{s}\Phi(s\xi)\right)dr \tag{1}$$
Let $a>0$ be sufficiently large so that $\text{supp}(\Phi)$ is contained the ball $B_{0}(a)$ centered at the origin of radius $a$. By Fubini's theorem,
\begin{align*} 2\int_{0}^{a}\left(-\int_{r}^{\infty}s^{n-1}d_{s}\Phi(s\xi)\right)dr&=2\int_{0}^{a}\left(-\int_{0}^{\infty}\chi_{[r,\infty)}(s)s^{n-1}d_{s}\Phi(s\xi)\right)dr\\ &=-2\int_{0}^{\infty}s^{n-1}d_{s}\Phi(s\xi)\left(\int_{0}^{\min\left\{a,s\right\}}dr\right)\\ &=-2\int_{0}^{\infty}\min\left\{a,s\right\}s^{n-1}d_{s}\Phi(s\xi)\\ &=-2\int_{0}^{\infty}s^{n}d_{s}\Phi(s\xi), \tag{2} \end{align*}
where the last equality follows from our choice of $a$. Integrating by parts, we see that (2) equals
$$-2\left[s^{n}\Phi(s\xi)\right]_{s=0}^{a}+2\int_{0}^{\infty}ns^{n-1}\Phi(s\xi)ds=2\int_{0}^{\infty}ns^{n-1}\Phi(s\xi)ds,$$
where the first term vanishes by our choice of $a$ together with $\Phi$ compactly supported.
For the general $\Phi$, we approximate $\Phi$ from below by a sequence $\left\{\Phi_{n}\right\}$ of functions satisfying the conditions assumed above. This is possible since $\Phi(s\xi)$ is decreasing for $\xi$ fixed, hence continuous except at countably many points. Since $\int_{0}^{\infty}s^{n-1}\Phi(s\xi)ds$ exists for a.e. $\xi$ from Fubini's theorem together with the hypothesis that $\Phi$ is integrable, we obtain the desired result.