integration concerning Fourier transform of homogeneous kernel(of degree 0) Let $m\in C^\infty(\mathbb{R}^d-\{0\})$ be homogeneous of degree zero and has mean zero on the sphere$(S^{d-1})$. Then $m$ defines a tempered distribution and $\partial_j^dm\in S'(\mathbb{R}^d)$ is homogeneous of degree $-d$, which also has mean zero on the sphere. Thus principal value of $\partial_j^d m$ is well defined in the usual integral sense. Let's denote it by (P.V. $\partial_j^d m)$, which is an element of $S'(\mathbb{R}^d)$. I want to show that (P.V. $\partial_j^d m$)^ is a smooth function on $\mathbb{R}^d-\{0\}$.(Here, ^ means a Fourier transform). This is what my textbook says. Choose a smooth, radial, compactly supported bump function $\phi$ that equals 1 on a neighborhood of the origin. Then for all $x\neq0$, we have
(P.V. $\partial_j^d m)$^$(x)=\displaystyle\int_{\mathbb{R}^d}\partial_j^dm(\xi)\phi(\xi)(e^{-2\pi ix\cdot\xi}-1)\,d\xi\\ \displaystyle+\sum_{j=1}^d\frac{x_j}{2\pi i|x|^2}\int_{\mathbb{R}^d}\partial_j\Big(\partial_jm(\xi)(1-\phi(\xi))\Big)e^{-2\pi ix\cdot\xi}\,d\xi$
$\textbf{The question is, I don't know how to get to this equality.}$ Let me show you my try. 
(P.V. $\partial_j^d m)$^$(x)=\displaystyle\int_{\mathbb{R}^d}\partial_j^dm(\xi)e^{-2\pi ix\cdot\xi}\,d\xi\\
\displaystyle=\int_{\mathbb{R}^d}\partial_j^dm(\xi)\phi(\xi)e^{-2\pi ix\cdot\xi}\,d\xi+\int_{\mathbb{R}^d}\partial_j^dm(\xi)(1-\phi(\xi))e^{-2\pi ix\cdot\xi}\,d\xi\\
\displaystyle=\int_{\mathbb{R}^d}\partial_j^dm(\xi)\phi(\xi)(e^{-2\pi ix\cdot\xi}-1)\,d\xi+\int_{\mathbb{R}^d}\partial_j^dm(\xi)(1-\phi(\xi))e^{-2\pi ix\cdot\xi}\,d\xi\\$
The first equality is from the definition of principal value and Fourier transform. Second equality is also obvious. Third equality follows since $\partial_j^dm$ has mean zero on the sphere and $\phi$ is radial(together with which implies cancellation). So we are done with the first term. The second term looks like integration by part. But, when I do this, what I get is
$\displaystyle\frac{1}{2\pi i x_j}\int_{\mathbb{R}^d}\partial_j\Big(\partial_j^dm(\xi)(1-\phi(\xi))\Big)e^{-2\pi ix\cdot\xi}\,d\xi\\$
instead of the expression in the desired result. How come the summation and square term appear here? or did I apply integration by part incorrectly? Could somebody show me the detailed computation concerning the second term? I think I have put enough information in my question, but in case you want to see the original book, see p180 of this link. Thank you very much.
(http://books.google.co.kr/books?id=d7EZBAAAQBAJ&pg=PA180&lpg=PA180&dq=%22we+may+repeat+integration+by+parts+of+the+second+integral+any+number%22&source=bl&ots=G5gl3TKO6i&sig=ss-ODIvZwrPCTp3UBZuUPyUDZ2M&hl=ko&sa=X&ei=u8gqVPDLJYvf8AXblILQDw&ved=0CBwQ6AEwAA#v=onepage&q=%22we%20may%20repeat%20integration%20by%20parts%20of%20the%20second%20integral%20any%20number%22&f=false)
 A: Your computations aren't rigorous. Since $\partial_{j}^{d}$ is only homogeneous of degree $-d$, the integral expressions $$\int_{\mathbb{R}^{d}}\partial_{j}^{d}(\xi)\phi(\xi)e^{-2\pi ix\cdot\xi}d\xi$$
and
$$\int_{\mathbb{R}^{d}}\partial_{j}^{d}m(\xi)(1-\phi(\xi))e^{-2\pi ix\cdot\xi}d\xi$$
a priori don't converge absolutely. In the first expression you have the singularity at the origin, and in the second expression, you don't have sufficient decay at $\infty$--$(1-\phi(\xi))=1$ outside a neighborhood of the origin.

Let $\phi$ be a compactly supported function, smooth, nonnegative, radial function which equals $1$ on the unit ball. Write the distribution $u:=\text{P.V.}\partial_{j}^{d}m$ as the sum of the distributions $u_{0}:=\phi\text{P.V.}\partial_{j}^{d}m$ and $u_{\infty}:=(1-\phi)\text{P.V.}\partial_{j}^{d}m$. Note that $u_{0}$ is a distribution with compact support which coincides with the $C^{\infty}$ function $\phi\partial_{j}^{d}m$ on $\mathbb{R}^{n}\setminus\left\{0\right\}$. And $u_{\infty}$ coincides with the $C^{\infty}$ function $(1-\phi)\partial_{j}^{d}m$ since $(1-\phi)$ vanishes on the unit ball and $(\partial_{j}^{d}m$ decays like $\left|x\right|^{-d}$_.
Let $f\in\mathcal{S}(\mathbb{R}^{d})$ be a test function. Then by linearity of the Fourier transform,
\begin{align*}
\langle{\widehat{u},f}\rangle&=\langle{\widehat{u_{0}},f}\rangle+\langle{\widehat{u_{\infty}},f}\rangle\\
&=\langle{u_{0},\widehat{f}}\rangle+\langle{u_{\infty},\widehat{f}}\rangle\\
&=\langle{\text{P.V.}\partial_{j}^{d}m,\phi \widehat{f}}\rangle+\int_{\mathbb{R}^{d}}\partial_{j}^{d}m(\xi)(1-\phi(\xi))\widehat{f}(\xi)d\xi\\
&=\lim_{\epsilon\downarrow 0}\int_{\left|\xi\right|\geq\epsilon}\partial_{j}^{d}(\xi)\phi(\xi)\widehat{f}(\xi)d\xi+\int_{\mathbb{R}^{d}}\partial_{j}^{d}m(\xi)(1-\phi(\xi))\widehat{f}(\xi)d\xi\tag{1}\\
\end{align*}
For the first term in (1), note that since $\phi$ is radial and $m$ has mean zero over $\mathbb{S}^{d-1}$, it follows that $\int_{\left|\xi\right|\geq\epsilon}(\partial_{j}^{d}m)\phi=0$. So using Fubini's theorem,
\begin{align*}
\int_{\left|\xi\right|\geq\epsilon}\partial_{j}^{d}(\xi)\phi(\xi)\widehat{f}(\xi)d\xi=\int_{\mathbb{R}^{d}}f(x)dx\int_{\left|\xi\right|\geq\epsilon}\partial_{j}^{d}m(\xi)\phi(\xi)\left[e^{-2\pi i x\cdot\xi}-1\right]d\xi
\end{align*}
For $x$ fixed, Taylor's theorem gives
$$\left|e^{-2\pi ix\cdot\xi}-1\right|\leq 2\pi n\left|x\cdot\xi\right|\leq 2\pi n\left|x\right|\left|\xi\right|$$
Since $\partial_{j}^{d}m$ is homogeneous of degree $-d$, we see that
\begin{align*}
\int_{\left|\xi\right|\geq\epsilon}\left|\partial_{j}^{d}m(\xi)\right|\left|\phi(\xi)\right|\left|e^{-2\pi ix\cdot\xi}d-1\right|d\xi&\lesssim_{n,m}\int_{\left|\xi\right|\geq\epsilon}\left|\xi\right|^{-d}\left|\phi(\xi)\right|\left|\xi\right|\left|x\right|d\xi\\
&\leq\left|x\right|\int_{\mathbb{R}^{d}}\left|\xi\right|^{-d+1}\left|\phi(\xi)\right|d\xi
\end{align*}
Since $\left|x\right|\left|f(x)\right|$ decays rapidly, we can apply dominated convergence to conclude
$$\lim_{\epsilon\downarrow 0}\int_{\left|\xi\right|\geq\epsilon}\partial_{j}^{d}(\xi)\phi(\xi)\widehat{f}(\xi)d\xi=\int_{\mathbb{R}^{d}}f(x)\left(\int_{\mathbb{R}^{d}}\partial_{j}^{d}m(\xi)\phi(\xi)\left[e^{-2\pi i x\cdot\xi}-1\right]d\xi\right)dx \tag{2}$$
For the second term in (1), the trick is to write
\begin{align*}
\widehat{f}(\xi)=\int_{\mathbb{R}^{d}}\left(\sum_{k=1}^{d}\dfrac{x_{k}^{2}}{\left|x\right|^{2}}\right)f(x)e^{-2\pi ix\cdot\xi}dx&=\sum_{k=1}^{d}\int_{\mathbb{R}^{d}}\dfrac{x_{k}}{(-2\pi i)\left|x\right|^{2}}(-2\pi ix_{j})f(x)e^{-2\pi ix\cdot\xi}dx\\
&=\sum_{k=1}^{d}\dfrac{\partial}{\partial_{j}\xi}\int_{\mathbb{R}^{d}}\dfrac{x_{k}}{(-2\pi i)\left|x\right|^{2}}f(x)e^{-2\pi ix\cdot\xi}dx,
\end{align*}
where we can differentiate inside in the integral because $x_{k}/\left|x\right|^{2}$ doesn't blow up too quickly at zero and $f$ is Schwartz. So integration by parts yields that the second integral expression in (1) equals
$$-\sum_{k=1}^{d}\int_{\mathbb{R}^{d}}\partial_{j}(\partial_{j}^{d}m(1-\phi))(\xi)\left(\int_{\mathbb{R}^{d}}\dfrac{x_{k}}{(-2\pi i)\left|x\right|^{2}}f(x)e^{-2\pi ix\cdot\xi}dx\right)d\xi$$
Observe that $\partial_{j}(\partial_{j}^{d}m(1-\phi))=\partial_{j}^{d+1}m\phi+\partial_{j}^{d}m\partial_{j}(1-\phi)$. The first function is the product of a $C^{\infty}$ function vanishing in a neighborhood of the origin and decaying like $\left|x\right|^{-d-1}$ at infinity. Since $1-\phi\equiv 1$ outside a neighborhood of the origin, $\partial_{j}(1-\phi)$ vanishes outside an annulus around the origin. Whence, the second function is $C_{c}^{\infty}$, and we conclude that $\left|\partial_{j}(\partial_{j}^{d}m(1-\phi))\right|\leq\left|\xi\right|^{-d-1}$ outside a neighborhood of the origin. By Fubini's theorem,
we obtain that
$$\sum_{k=1}^{d}\int_{\mathbb{R}^{d}}f(x)\dfrac{x_{k}}{2\pi i\left|x\right|^{2}}\left(\int_{\mathbb{R}^{d}}\partial_{j}(\partial_{j}^{d}m(1-\phi))(\xi)e^{-2\pi i x\cdot\xi}d\xi\right)dx=\int_{\mathbb{R}^{d}}f(x)\left(\sum_{k=1}^{d}\dfrac{x_{k}}{2\pi i\left|x\right|^{2}}\int_{\mathbb{R}^{d}}\partial_{j}(\partial_{j}^{d}m(1-\phi))(\xi)e^{-2\pi i x\cdot\xi}d\xi\right)dx\tag{3}$$
Since $\phi$ was an arbitrary test function, we obtain the formula.
