High dimensional beta integral (a typo in Stein's book "singular integrals") Hello,
When I read Stein's book of Singular Integrals, at p. 118, there is an obvious mistake:
$$
\int_{R^n} |x-y|^{-n+\alpha} |y|^{-n+\beta}=\frac{\gamma(\alpha)\gamma(\beta)}{\gamma(\alpha+\beta)},\quad 0<\alpha, 0<\beta,
$$
with $\alpha+\beta < n$ and 
$$
\gamma(a) = \pi^{n/2} 2^a \frac{\Gamma(a/2)}{\Gamma((n-a)/2)}.
$$
Clearly, in one dimensional case,
$$
\int_{0}^1 |1-y|^{-1+\alpha} |y|^{-1+\beta}=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)},\quad 0<\alpha, 0<\beta,
$$
with $\alpha+\beta < 1$;see wikipedia. However,
$$
\int_{R^1} |1-y|^{-1+\alpha} |y|^{-1+\beta}= \infty.
$$
So we need properly reduce the integral domain from $R^n$ to some balls. Does anyone know this correct integration domain?
Thank you very much!
Anand
EDIT:
(1) Whether does any one know an online errata of this book?
(2) here is a following question of this post.
 A: How is that integral infinite?  If $x\neq 0$, then the function is locally integrable near $y=x$ and locally integrable near $y=0$ and it is integrable at $\infty$ since the integrand behaves like $|y|^{-2+\alpha+\beta}$ which decays like $y^{1+\epsilon}$.  So the integral is not infinite unless $x=0$.
There is also the issue that the integral clearly depends on $x$, while the right side does not.  I believe the typo is that $x$ should be 1.
If one carefully writes out the identity $I_\alpha(I_\beta(f))(z)=I_{\alpha+\beta}(f)(z)$ and sets $z=0$, one finds that
$\displaystyle{\int\left(\int |x-y|^{-n+\alpha}|y|^{-n+\beta}dy\right)f(x)dx=\int\left(\frac{\gamma(\alpha)\gamma(\beta)}{\gamma(\alpha+\beta)}|x|^{-n+\alpha+\beta}\right)f(x)dx}$.
Since this holds for any Schwartz function $f$, it must be that the two functions in parenthesis are equal.  Thus in particular for $x=1$, you get the (corrected) identity in Stein's book.
A: Thanks Peter Luthy for his answer. Here is an interesting consequence:
$$
\int_{|y|>1}|1-y|^{-1+\alpha}|y|^{-1+\beta} d y = \int_{R}|1-y|^{-1+\alpha}|y|^{-1+\beta} d y - \int_{|y|<1}|1-y|^{-1+\alpha}|y|^{-1+\beta} d y 
$$
which gives
$$
\int_{|y|>1}|1-y|^{-1+\alpha}|y|^{-1+\beta} d y = \left(\frac{\sqrt{\pi}\Gamma(\frac{1-\alpha-\beta}{2})}{\Gamma(\frac{1-\alpha}{2})\Gamma(\frac{1-\beta}{2})}-1\right) \frac{\Gamma(\frac{\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{\alpha+\beta}{2})}\:.
$$
