[proofplan] The fundamental solution $\Phi$ is smooth on $\mathbb{R}^n \setminus \{0\}$, so the only obstacle to local integrability is the singularity at the origin. We show $\Phi \in L^1(B(0, R))$ for every $R > 0$ by integrating $|\Phi|$ over the punctured ball $B(0, R) \setminus B(0, \varepsilon)$, converting to polar coordinates, evaluating the resulting one-dimensional integrals explicitly (using [Integration By Parts](/theorems/210) in the logarithmic case), and then sending $\varepsilon \to 0^+$ via the [Monotone Convergence Theorem](/theorems/509). [/proofplan] [step:Reduce the local integrability check to a computation on punctured balls] Fix $R > 0$. Since $\Phi \in C^\infty(\mathbb{R}^n \setminus \{0\})$, the function $|\Phi|$ is [Lebesgue measurable](/page/Measurable%20Functions) on $\mathbb{R}^n$. For each $0 < \varepsilon < R$, define the truncated integrand \begin{align*} f_\varepsilon : \mathbb{R}^n &\to [0, \infty) \\ x &\mapsto |\Phi(x)| \cdot \mathbb{1}_{B(0,R) \setminus B(0,\varepsilon)}(x). \end{align*} As $\varepsilon \to 0^+$, the sequence $f_\varepsilon$ increases pointwise to $|\Phi| \cdot \mathbb{1}_{B(0,R)}$ on $\mathbb{R}^n \setminus \{0\}$ (which differs from $|\Phi| \cdot \mathbb{1}_{B(0,R)}$ only on the $\mathcal{L}^n$-null set $\{0\}$). Each $f_\varepsilon$ is non-negative and measurable, so the [Monotone Convergence Theorem](/theorems/509) gives \begin{align*} \int_{B(0,R)} |\Phi(x)| \, d\mathcal{L}^n(x) = \lim_{\varepsilon \to 0^+} \int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^n(x). \end{align*} It therefore suffices to show that the right-hand side remains bounded as $\varepsilon \to 0^+$. Since $\Phi$ is radial, we evaluate these integrals by converting to [polar coordinates](/page/Spherical%20Integrals%20and%20Radial%20Functions) in the two cases $n = 2$ and $n \geq 3$. [guided] The function $\Phi$ is defined and smooth on $\mathbb{R}^n \setminus \{0\}$ but blows up as $|x| \to 0$. To determine whether this singularity is integrable, we cannot simply evaluate $\int_{B(0,R)} |\Phi| \, d\mathcal{L}^n$ directly -- $\Phi$ is not defined at the origin. Instead, we excise a small ball around the origin, compute the integral on the remaining annular region, and then pass to the limit. Fix $R > 0$. For each $0 < \varepsilon < R$, define the truncated integrand \begin{align*} f_\varepsilon : \mathbb{R}^n &\to [0, \infty) \\ x &\mapsto |\Phi(x)| \cdot \mathbb{1}_{B(0,R) \setminus B(0,\varepsilon)}(x). \end{align*} Each $f_\varepsilon$ is non-negative, measurable, and bounded (since $|\Phi|$ is continuous on the compact annulus $\overline{B}(0,R) \setminus B(0,\varepsilon)$). As $\varepsilon$ decreases, the indicator $\mathbb{1}_{B(0,R) \setminus B(0,\varepsilon)}$ increases pointwise, so $f_\varepsilon \leq f_{\varepsilon'}$ whenever $\varepsilon \geq \varepsilon'$. The pointwise limit as $\varepsilon \to 0^+$ is $|\Phi| \cdot \mathbb{1}_{B(0,R) \setminus \{0\}}$, which agrees with $|\Phi| \cdot \mathbb{1}_{B(0,R)}$ everywhere except at the single point $\{0\}$, a set of $\mathcal{L}^n$-measure zero. We verify the hypotheses of the [Monotone Convergence Theorem](/theorems/509): the functions $f_\varepsilon$ are (i) non-negative, (ii) measurable, and (iii) increasing as $\varepsilon \to 0^+$. The theorem therefore yields \begin{align*} \int_{B(0,R)} |\Phi(x)| \, d\mathcal{L}^n(x) = \lim_{\varepsilon \to 0^+} \int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^n(x). \end{align*} It suffices to show that the right-hand side remains bounded. The integrals over the punctured ball will be evaluated explicitly using polar coordinates, splitting into the cases $n = 2$ (logarithmic singularity) and $n \geq 3$ (power-type singularity). [/guided] [/step] [step:Compute the integral for $n = 2$ via polar coordinates and integration by parts] When $n = 2$, the fundamental solution satisfies $|\Phi(x)| = \frac{1}{2\pi} |\log |x||$. We apply the polar coordinates decomposition for $\mathcal{L}^2$: under the substitution $x = r\omega$ with $r \in (\varepsilon, R)$ and $\omega \in \partial B(0,1)$, the Lebesgue measure transforms as $d\mathcal{L}^2(x) = r \, d\sigma(\omega) \, d\mathcal{L}^1(r)$, where $\sigma$ denotes the [surface measure](/page/Hausdorff%20Measure) on the unit circle $\partial B(0,1) \subset \mathbb{R}^2$. The annular domain $B(0, R) \setminus B(0, \varepsilon)$ corresponds to $(\varepsilon, R) \times \partial B(0,1)$ in polar coordinates. Since $|\log |x|| = |\log r|$ is independent of $\omega$, integrating over $\partial B(0,1)$ yields the factor $\sigma(\partial B(0,1)) = 2\pi$: \begin{align*} \int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^2(x) &= \frac{1}{2\pi} \int_\varepsilon^R \int_{\partial B(0,1)} |\log r| \cdot r \, d\sigma(\omega) \, d\mathcal{L}^1(r) \\ &= \frac{1}{2\pi} \cdot 2\pi \int_\varepsilon^R |\log r| \cdot r \, d\mathcal{L}^1(r) = \int_\varepsilon^R |\log r| \cdot r \, d\mathcal{L}^1(r). \end{align*} We split the integral at $r = 1$, where $\log r$ changes sign. For $r \in (0,1)$, $|\log r| = -\log r$; for $r \in (1, \infty)$, $|\log r| = \log r$. Assuming $R > 1$ (the case $R \leq 1$ is simpler and follows the same method): \begin{align*} \int_\varepsilon^R |\log r| \cdot r \, d\mathcal{L}^1(r) = \int_\varepsilon^1 (-\log r) \cdot r \, d\mathcal{L}^1(r) + \int_1^R (\log r) \cdot r \, d\mathcal{L}^1(r). \end{align*} We evaluate each integral by [Integration By Parts](/theorems/210). For the first integral, set $u = -\log r$ and $dv = r \, d\mathcal{L}^1(r)$, so that $du = -\frac{1}{r} \, d\mathcal{L}^1(r)$ and $v = \frac{r^2}{2}$: \begin{align*} \int_\varepsilon^1 (-\log r) \cdot r \, d\mathcal{L}^1(r) &= \left[-\frac{r^2}{2} \log r\right]_\varepsilon^1 - \int_\varepsilon^1 \frac{r}{2} \, d\mathcal{L}^1(r) \\ &= \frac{\varepsilon^2}{2} \log \varepsilon + \frac{1 - \varepsilon^2}{4}. \end{align*} As $\varepsilon \to 0^+$, $\varepsilon^2 \log \varepsilon \to 0$ (since $\lim_{t \to 0^+} t^2 \log t = 0$ by L'Hopital's rule applied to $\frac{\log t}{t^{-2}}$), so this integral converges to $\frac{1}{4}$. For the second integral, set $u = \log r$ and $dv = r \, d\mathcal{L}^1(r)$, so that $du = \frac{1}{r} \, d\mathcal{L}^1(r)$ and $v = \frac{r^2}{2}$: \begin{align*} \int_1^R (\log r) \cdot r \, d\mathcal{L}^1(r) &= \left[\frac{r^2}{2} \log r\right]_1^R - \int_1^R \frac{r}{2} \, d\mathcal{L}^1(r) \\ &= \frac{R^2}{2} \log R - \frac{R^2 - 1}{4}. \end{align*} This is a finite constant depending only on $R$. Combining both integrals, the limit as $\varepsilon \to 0^+$ yields \begin{align*} \lim_{\varepsilon \to 0^+} \int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^2(x) = \frac{1}{4} + \frac{R^2}{2} \log R - \frac{R^2 - 1}{4} < \infty. \end{align*} [guided] When $n = 2$, the fundamental solution is $\Phi(x) = -\frac{1}{2\pi} \log |x|$, so $|\Phi(x)| = \frac{1}{2\pi} |\log |x||$. The question is whether the logarithmic singularity at the origin is integrable in two dimensions. We convert to polar coordinates. Under the substitution $x = r\omega$ with $r = |x| \in (\varepsilon, R)$ and $\omega = x/|x| \in \partial B(0,1)$, the two-dimensional Lebesgue measure decomposes as \begin{align*} d\mathcal{L}^2(x) = r \, d\sigma(\omega) \, d\mathcal{L}^1(r), \end{align*} where $\sigma$ is the surface measure (arc length measure) on the unit circle $\partial B(0,1) \subset \mathbb{R}^2$, with total mass $\sigma(\partial B(0,1)) = 2\pi$. The annular domain $B(0, R) \setminus B(0, \varepsilon)$ in Cartesian coordinates corresponds to the product domain $(\varepsilon, R) \times \partial B(0,1)$ in polar coordinates. Since $|\Phi(x)| = \frac{1}{2\pi}|\log r|$ depends only on the radial variable $r$, the angular integral factors out: \begin{align*} \int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^2(x) &= \frac{1}{2\pi} \int_\varepsilon^R \int_{\partial B(0,1)} |\log r| \cdot r \, d\sigma(\omega) \, d\mathcal{L}^1(r) \\ &= \frac{1}{2\pi} \cdot \sigma(\partial B(0,1)) \int_\varepsilon^R |\log r| \cdot r \, d\mathcal{L}^1(r) \\ &= \int_\varepsilon^R |\log r| \cdot r \, d\mathcal{L}^1(r). \end{align*} The factor of $\frac{1}{2\pi}$ from $|\Phi|$ cancels exactly with $\sigma(\partial B(0,1)) = 2\pi$. The problem has been reduced to a one-dimensional integral. We split at $r = 1$ where $\log r$ changes sign. For $r \in (0, 1)$, $\log r < 0$, so $|\log r| = -\log r$. For $r \in (1, \infty)$, $\log r > 0$, so $|\log r| = \log r$. Assuming $R > 1$ (if $R \leq 1$, only the first integral appears, and the argument is the same): \begin{align*} \int_\varepsilon^R |\log r| \cdot r \, d\mathcal{L}^1(r) = \int_\varepsilon^1 (-\log r) \cdot r \, d\mathcal{L}^1(r) + \int_1^R (\log r) \cdot r \, d\mathcal{L}^1(r). \end{align*} **First integral.** We apply [Integration By Parts](/theorems/210) with $u = -\log r$ and $dv = r \, d\mathcal{L}^1(r)$. Then $du = -\frac{1}{r} \, d\mathcal{L}^1(r)$ and $v = \frac{r^2}{2}$, giving \begin{align*} \int_\varepsilon^1 (-\log r) \cdot r \, d\mathcal{L}^1(r) &= \left[-\frac{r^2}{2} \log r\right]_\varepsilon^1 - \int_\varepsilon^1 \left(-\frac{1}{r}\right) \cdot \frac{r^2}{2} \, d\mathcal{L}^1(r) \\ &= \left(0 - \left(-\frac{\varepsilon^2}{2} \log \varepsilon\right)\right) - \left(-\int_\varepsilon^1 \frac{r}{2} \, d\mathcal{L}^1(r)\right) \\ &= \frac{\varepsilon^2}{2} \log \varepsilon + \frac{1}{2} \cdot \frac{1 - \varepsilon^2}{2} \\ &= \frac{\varepsilon^2}{2} \log \varepsilon + \frac{1 - \varepsilon^2}{4}. \end{align*} The critical question is whether $\varepsilon^2 \log \varepsilon$ vanishes as $\varepsilon \to 0^+$. Writing $\varepsilon^2 \log \varepsilon = \frac{\log \varepsilon}{\varepsilon^{-2}}$ and applying L'Hopital's rule (since both numerator and denominator diverge), we get $\frac{1/\varepsilon}{-2\varepsilon^{-3}} = -\frac{\varepsilon^2}{2} \to 0$. Therefore $\varepsilon^2 \log \varepsilon \to 0$, and the first integral converges to $\frac{1}{4}$. This is the key point: the logarithmic singularity grows as $|\log r| \sim |\log \varepsilon|$ near the origin, but the polar coordinates Jacobian factor $r$ provides enough decay to make the integral finite. The product $r \cdot |\log r|$ is integrable at $r = 0$ because $r |\log r| \to 0$ as $r \to 0^+$. **Second integral.** We apply [Integration By Parts](/theorems/210) with $u = \log r$ and $dv = r \, d\mathcal{L}^1(r)$. Then $du = \frac{1}{r} \, d\mathcal{L}^1(r)$ and $v = \frac{r^2}{2}$: \begin{align*} \int_1^R (\log r) \cdot r \, d\mathcal{L}^1(r) &= \left[\frac{r^2}{2} \log r\right]_1^R - \int_1^R \frac{r}{2} \, d\mathcal{L}^1(r) \\ &= \frac{R^2}{2} \log R - \frac{R^2 - 1}{4}. \end{align*} This is a finite quantity depending only on $R$. Combining both integrals: \begin{align*} \lim_{\varepsilon \to 0^+} \int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^2(x) = \frac{1}{4} + \frac{R^2}{2} \log R - \frac{R^2 - 1}{4} < \infty. \end{align*} [/guided] [/step] [step:Compute the integral for $n \geq 3$ via polar coordinates] When $n \geq 3$, the fundamental solution is non-negative: $\Phi(x) = C_n |x|^{-(n-2)}$, where $C_n := \frac{1}{n(n-2)\alpha(n)} > 0$. We apply the polar coordinates decomposition for $\mathcal{L}^n$: under the substitution $x = r\omega$ with $r = |x| \in (\varepsilon, R)$ and $\omega = x/|x| \in \partial B(0,1)$, the Lebesgue measure transforms as \begin{align*} d\mathcal{L}^n(x) = r^{n-1} \, d\sigma(\omega) \, d\mathcal{L}^1(r), \end{align*} where $\sigma$ denotes the surface measure on $\partial B(0,1) \subset \mathbb{R}^n$ with total mass $\sigma(\partial B(0,1)) = n\alpha(n)$. The annular domain $B(0,R) \setminus B(0,\varepsilon)$ corresponds to $(\varepsilon, R) \times \partial B(0,1)$. Since $|\Phi(x)| = C_n r^{-(n-2)}$ depends only on $r$, the angular integral factors out: \begin{align*} \int_{B(0,R) \setminus B(0,\varepsilon)} \Phi(x) \, d\mathcal{L}^n(x) &= C_n \int_\varepsilon^R \int_{\partial B(0,1)} r^{-(n-2)} \cdot r^{n-1} \, d\sigma(\omega) \, d\mathcal{L}^1(r) \\ &= C_n \cdot n\alpha(n) \int_\varepsilon^R r^{n-1-(n-2)} \, d\mathcal{L}^1(r) \\ &= C_n \cdot n\alpha(n) \int_\varepsilon^R r \, d\mathcal{L}^1(r). \end{align*} The exponent simplifies: $r^{n-1-(n-2)} = r^1 = r$. The singularity $r^{-(n-2)}$ from $\Phi$ is exactly cancelled by all but one power of the Jacobian factor $r^{n-1}$, leaving the integrable function $r$. Evaluating: \begin{align*} C_n \cdot n\alpha(n) \int_\varepsilon^R r \, d\mathcal{L}^1(r) = \frac{1}{n(n-2)\alpha(n)} \cdot n\alpha(n) \cdot \frac{R^2 - \varepsilon^2}{2} = \frac{R^2 - \varepsilon^2}{2(n-2)}. \end{align*} As $\varepsilon \to 0^+$, this converges to $\frac{R^2}{2(n-2)} < \infty$. [guided] When $n \geq 3$, the fundamental solution has a power-type singularity: $\Phi(x) = C_n |x|^{-(n-2)}$ with $C_n = \frac{1}{n(n-2)\alpha(n)} > 0$. The singularity grows like $|x|^{-(n-2)}$ as $|x| \to 0$, which is weaker than $|x|^{-n}$ (the threshold for non-integrability in $\mathbb{R}^n$). We make this precise by converting to polar coordinates. Under the substitution $x = r\omega$ with $r = |x| \in (\varepsilon, R)$ and $\omega = x/|x| \in \partial B(0,1)$, the $n$-dimensional Lebesgue measure decomposes as \begin{align*} d\mathcal{L}^n(x) = r^{n-1} \, d\sigma(\omega) \, d\mathcal{L}^1(r), \end{align*} where $\sigma$ is the surface measure on the unit sphere $\partial B(0,1) \subset \mathbb{R}^n$. The total surface area is $\sigma(\partial B(0,1)) = n\alpha(n)$, where $\alpha(n) = \frac{\pi^{n/2}}{\Gamma(n/2 + 1)}$ is the volume of the unit ball. The annular domain $B(0,R) \setminus B(0, \varepsilon)$ corresponds to the product $(\varepsilon, R) \times \partial B(0,1)$ in polar coordinates. Since $\Phi(x) = C_n r^{-(n-2)}$ depends only on the radial variable, the angular and radial integrals separate: \begin{align*} \int_{B(0,R) \setminus B(0,\varepsilon)} \Phi(x) \, d\mathcal{L}^n(x) &= C_n \int_\varepsilon^R \int_{\partial B(0,1)} r^{-(n-2)} \cdot r^{n-1} \, d\sigma(\omega) \, d\mathcal{L}^1(r) \\ &= C_n \cdot \sigma(\partial B(0,1)) \int_\varepsilon^R r^{n-1-(n-2)} \, d\mathcal{L}^1(r) \\ &= C_n \cdot n\alpha(n) \int_\varepsilon^R r \, d\mathcal{L}^1(r). \end{align*} The crucial algebraic simplification is the exponent: $n - 1 - (n - 2) = 1$. The Jacobian factor $r^{n-1}$ from polar coordinates contributes $n - 1$ powers of $r$, and the singularity $r^{-(n-2)}$ consumes $n - 2$ of them, leaving a single power $r^1$. This is the geometric reason the singularity is integrable: in $n$ dimensions, the volume element grows fast enough (as $r^{n-1}$) to compensate for the decay of $\Phi$ (as $r^{-(n-2)}$), with a margin of two powers. Computing the remaining one-dimensional integral: \begin{align*} C_n \cdot n\alpha(n) \int_\varepsilon^R r \, d\mathcal{L}^1(r) &= \frac{1}{n(n-2)\alpha(n)} \cdot n\alpha(n) \cdot \frac{R^2 - \varepsilon^2}{2} = \frac{R^2 - \varepsilon^2}{2(n-2)}. \end{align*} As $\varepsilon \to 0^+$, this converges to $\frac{R^2}{2(n-2)}$, which is finite for every fixed $R > 0$. Note the contrast with the function $|x|^{-n}$: the polar coordinates computation would produce $\int_\varepsilon^R r^{n-1-n} \, d\mathcal{L}^1(r) = \int_\varepsilon^R r^{-1} \, d\mathcal{L}^1(r) = \log(R/\varepsilon) \to \infty$. The singularity $|x|^{-n}$ is therefore not locally integrable, confirming that $|x|^{-(n-2)}$ lies strictly below the critical threshold. [/guided] [/step] [step:Combine the case estimates and conclude $\Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$] Combining the computations from the preceding two steps: for every $R > 0$ and every $n \geq 2$, the integral $\int_{B(0,R) \setminus B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{L}^n(x)$ remains bounded as $\varepsilon \to 0^+$, with the explicit values \begin{align*} \int_{B(0,R)} |\Phi(x)| \, d\mathcal{L}^n(x) = \begin{cases} \dfrac{1}{4} + \dfrac{R^2}{2} \log R - \dfrac{R^2 - 1}{4} & \text{if } n = 2, \\ \dfrac{R^2}{2(n-2)} & \text{if } n \geq 3. \end{cases} \end{align*} By the [Monotone Convergence Theorem](/theorems/509) applied in the first step, $\int_{B(0,R)} |\Phi(x)| \, d\mathcal{L}^n(x) < \infty$ for every $R > 0$. Since every [compact](/page/Compact%20Space) set $K \subset \mathbb{R}^n$ is contained in some ball $B(0,R)$, it follows that $\int_K |\Phi(x)| \, d\mathcal{L}^n(x) < \infty$ for every compact $K$, which is the definition of $\Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. [/step]