Proving that a function related to the Fourier transform of the unit ball is in $L^1(\mathbb{R}^3)$ Let $B = \{x \in \mathbb{R^3}\ : ||x|| \leq 1\}$ be a unit ball in $\mathbb{R^3}$. Having computed the Fourier Transform of $B$'s characteristic function $\hat{f}(k)$, where:
$\hat{f}(k) = \int_{B} e^{-2 \pi i \langle k,x \rangle}dx, k \in \mathbb{R^3}$, I have difficulty proving that  $\frac{\hat{f}(k)}{||k||^2} \in L^1(\mathbb{R^3})$.
I know that $|\hat{f}(k)| \leq \hat{f}(0)$, so $\frac{\hat{f}(k)}{||k||^2} \leq \frac{\hat{f}(0)}{||k||^2}$, and $\frac{\hat{f}(0)}{||k||^2}$ is integrable on any ball. But I don't think that thats enough to conclude that $\frac{\hat{f}(k)}{||k||^2}$ is integrable on any ball. Concerning integration on the outside of the ball, I am unsure how to proceed. Is there a general method of dealing with these types of problems?
 A: You have correctly argued that 
$$
\int_{\|k\|\le 1} \frac{|\hat f (k)|}{\|k\|^2} \,dk < \infty
$$ 
A bit more detail: a measurable function bounded by an integrable function is itself integrable. Loosely speaking, anything defined by an explicit formula is going to be Lebesgue integrable. To have a precise proof, note that $\hat f$ is continuous (since it's the Fourier transform of an $L^1$ function). Therefore, $\hat f(k)/\|k\|^2$ is continuous on $\mathbb{R}^n\setminus \{0\}$, hence is measurable there. Adding one point to the domain does not change measurability, since one point is a set of measure zero.
On the exterior of the ball, the Cauchy-Schwarz inequality does the job as both $\hat f$ and $1/\|k\|^2$ are in $L^2$ on that domain: 
$$
\int_{\|k\|\ge 1} \frac{|\hat{f}(k)|}{\|k\|^2} \,dk \le 
\left( \int_{\|k\|\ge 1} |\hat{f}(k)|^2 \,dk \right)^{1/2} \left( \int_{\|k\|\ge 1} \frac{1}{\|k\|^4} \,dk \right)^{1/2}
<\infty
$$ 
A: A Simple Bound
To bound the integral, let $f(x)=\big[\|x\|\le1\big]$. Then
$$
\begin{align}
\left\|\,\lower{1pt}{\hat{f}}\,\right\|_\infty
&=\left\|\,f\,\right\|_1\\
&=\frac{4\pi}3\tag1
\end{align}
$$
Normally $\left\|\,\lower{1pt}{\hat{f}}\,\right\|_\infty
\le\left\|\,f\,\right\|_1$, but $\hat{f}(0)=\int_{\mathbb{R}^3}f(x)\,\mathrm{d}x=\frac{4\pi}3$. Furthermore, Plancherel says
$$
\begin{align}
\left\|\,\lower{1pt}{\hat{f}}\,\right\|_2
&=\left\|\,f\,\right\|_2\\
&=\sqrt{\frac{4\pi}3}\tag2
\end{align}
$$
Hölder says $\int_Xf(x)g(x)\,\mathrm{d}x\le\|f\|_\infty\|g\|_1$; therefore,
$$
\begin{align}
\int_{\|x\|\le1}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x
&\le\left\|\,\hat{f}\,\right\|_\infty\int_{\|x\|\le1}\frac1{\|x\|^2}\,\mathrm{d}x\\
&=\frac{4\pi}3\cdot4\pi\tag3
\end{align}
$$
Hölder says $\int_Xf(x)g(x)\,\mathrm{d}x\le\|f\|_2\|g\|_2$; therefore,
$$
\begin{align}
\int_{\|x\|\gt1}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x
&\le\left\|\,\lower{1pt}{\hat{f}}\,\right\|_2\left(\int_{\|x\|\gt1}\frac1{\|x\|^4}\,\mathrm{d}x\right)^{1/2}\\
&=\sqrt{\frac{4\pi}3}\cdot\sqrt{4\pi}\tag4
\end{align}
$$
Thus,
$$
\begin{align}
\int_{\mathbb{R}^3}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x
&\le\frac{16\pi^2}3+\frac{4\pi}{\sqrt3}\\
&\approx59.893\tag5
\end{align}
$$

A Better Bound
If we split the cases at $\|x\|=r$, we get the bound
$$
\begin{align}
\int_{\mathbb{R}^3}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x
&\le\frac{16\pi^2}3r+\frac{4\pi}{\sqrt3}r^{-1/2}\\
&=\left(\frac{4\pi}{\sqrt3}\right)^{4/3}\left[\left(\frac{4\pi}{\sqrt3}\right)^{2/3}r+\left(\frac{4\pi}{\sqrt3}\right)^{-1/3}r^{-1/2}\right]\tag6
\end{align}
$$
Since $x^2+\frac1x\ge\left(\frac{27}4\right)^{1/3}$ and $x^2+\frac1x=\left(\frac{27}4\right)^{1/3}$ when $x=\left(\frac12\right)^{1/3}$, $(6)$ gives the bound
$$
\begin{align}
\int_{\mathbb{R}^3}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x
&\le\left(192\pi^4\right)^{1/3}\\
&\approx26.544\tag7
\end{align}
$$
