Hello,


When I read Stein's book of [Singular Integrals](http://rads.stackoverflow.com/amzn/click/0691080798), 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](http://en.wikipedia.org/wiki/Beta_function). 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.](https://mathoverflow.net/questions/72570/high-dimensional-beta-integral-question-following-the-previous-post)



High dimensional beta integral (a typo in Stein's book "singular integrals")
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.


