# 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?

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Did you check to see if there is an on-line errata list for this book? – Gerald Edgar Aug 10 '11 at 13:40
@Gerald Edgar, not yet. Let me find it... – Anand Aug 10 '11 at 13:44
By the way, do you know the where is the on-line errata list? Thank you very much! – Anand Aug 10 '11 at 13:46

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.

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Thanks Peter Luthy. I changed my typos where $x$ should be $1$ in the one dimensional case since I integrate from 0 to 1. And I also changed $\Gamma$ to $\gamma$ since they are different. I mistook for the same function. Your proof of Setin's result is very nice. In his book, he just mentions that this identity is a consequence of the semigroup property that you use. You fill up the missing step. Thanks. – Anand Aug 10 '11 at 7:41
You're welcome. I can totally understand the confusion about the domain of integration since Stein mentions the beta integral which only integrates on [0,1] and makes a typo that $x=1$! Then again, one often learns more from correcting mistakes in books than from simply reading. There is something to be said for getting one's hands dirty and doing some hard work. – Peter Luthy Aug 10 '11 at 8:49

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

$$\int_{|y|>1}|1-y|^{-1+\alpha}|y|^{-1+\beta} d y = \left(\frac{\sqrt{\pi}\Gamma(\frac{1-\alpha-\beta}{2})}{\Gamma(\frac{1-\alpha}{2})\Gamma(\frac{1-\beta}{2})}-1\right) \frac{\Gamma(\frac{\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{\alpha+\beta}{2})}\:.$$

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