This post follows the request of proving that
\[\lim_{y \rightarrow 0^+} \frac{y}{\pi} \int_{x=-\infty}^{+\infty} \frac{f(x)}{x^2 + y^2} dx = f(0) \ ,\]
for \(f(x)\) continuous and bounded in \(\mathbb{R}\), i.e. that the function \(\frac{y}{\pi (x^2 + y^2)}\) acts as an approximation of the Dirac’s delta as \(y \rightarrow 0^+\).
Method 1: Rigorous Domain Splitting (\(\varepsilon\)-\(\delta\) / Squeeze Method)
1. Algebraic Transformation
Factoring \(y^2\) out of the denominator and using \(d\left(\frac{x}{y}\right) = \frac{1}{y} \, dx\):
\[ \frac{y}{\pi} \int_{-\infty}^{+\infty} \frac{f(x)}{x^2 + y^2} \, dx = \frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{f(x)}{1 + \left(\frac{x}{y}\right)^2} \, d\left(\frac{x}{y}\right) \]
2. Domain Partitioning
For a fixed parameter \(a > 0\), split the integration domain \((-\infty, +\infty)\) into three regions:
\[ (-\infty, -a] \cup [-a, a] \cup [a, +\infty) \]
Let the corresponding integral be decomposed as:
\[ I(y) = I_1(y) + I_2(y) + I_3(y) \]
3. Evaluating the integrals
3.1 Outer Integrals (\(I_1\) and \(I_3\))
For \(I_3 = \frac{1}{\pi} \int_{a}^{+\infty} \frac{f(x)}{1 + (x/y)^2} \, d\left(\frac{x}{y}\right)\), let \(m_3 = \inf_{x \ge a} f(x)\) and \(M_3 = \sup_{x \ge a} f(x)\):
\[ \frac{m_3}{\pi} \int_{a}^{+\infty} \frac{d(x/y)}{1 + (x/y)^2} \le I_3 \le \frac{M_3}{\pi} \int_{a}^{+\infty} \frac{d(x/y)}{1 + (x/y)^2} \]
Evaluating the primitive:
\[ \int_{a}^{+\infty} \frac{d(x/y)}{1 + (x/y)^2} = \left[ \arctan\left(\frac{x}{y}\right) \right]_{x=a}^{x=+\infty} = \frac{\pi}{2} - \arctan\left(\frac{a}{y}\right) \]
Taking the limit as \(y \to 0^+\) (for fixed \(a > 0\)):
\[ \lim_{y \to 0^+} \arctan\left(\frac{a}{y}\right) = \frac{\pi}{2} \implies \left( \frac{\pi}{2} - \arctan\left(\frac{a}{y}\right) \right) \to 0 \]
Thus, \(I_3(y) \to 0\) as \(y \to 0^+\). By symmetric argument, \(I_1(y) \to 0\) as \(y \to 0^+\).
3.2 Middle Integral (\(I_2\)) Around the Origin
For \(I_2 = \frac{1}{\pi} \int_{-a}^{a} \frac{f(x)}{1 + (x/y)^2} \, d\left(\frac{x}{y}\right)\), let \(m_2(a) = \inf_{x \in [-a, a]} f(x)\) and \(M_2(a) = \sup_{x \in [-a, a]} f(x)\):
\[ \frac{m_2(a)}{\pi} \int_{-a}^{a} \frac{d(x/y)}{1 + (x/y)^2} \le I_2 \le \frac{M_2(a)}{\pi} \int_{-a}^{a} \frac{d(x/y)}{1 + (x/y)^2} \]
Evaluating the primitive:
\[ \int_{-a}^{a} \frac{d(x/y)}{1 + (x/y)^2} = \arctan\left(\frac{a}{y}\right) - \arctan\left(-\frac{a}{y}\right) = 2 \arctan\left(\frac{a}{y}\right) \]
So:
\[ \frac{2 m_2(a)}{\pi} \arctan\left(\frac{a}{y}\right) \le I_2 \le \frac{2 M_2(a)}{\pi} \arctan\left(\frac{a}{y}\right) \]
Taking \(y \to 0^+\) for fixed \(a > 0\), since \(\frac{2}{\pi} \arctan\left(\frac{a}{y}\right) \to 1\):
\[ m_2(a) \le \liminf_{y \to 0^+} I(y) \le \limsup_{y \to 0^+} I(y) \le M_2(a) \]
Finally, taking \(a \to 0^+\), by the continuity of \(f(x)\) at \(x = 0\), we have
\[\lim_{a \to 0^+} m_2(a) = \lim_{a \to 0^+} M_2(a) = f(0)\ , \]
and by the squeeze theorem
\[ \lim_{y \to 0^+} \frac{y}{\pi} \int_{-\infty}^{+\infty} \frac{f(x)}{x^2 + y^2} \, dx = f(0) \ .\]
Method 2: Change of Variables (Dominated Convergence Theorem)
todo Treat convergence types (pointwise, and uniform), and consequences on limits, limits and derivatives, limits and integrals.
1. Substitution
Substitute \(t = \frac{x}{y}\), so \(x = ty\) and \(dx = y \, dt\):
\[ \lim_{y \to 0^+} \frac{y}{\pi} \int_{-\infty}^{+\infty} \frac{f(x)}{x^2 + y^2} \, dx = \lim_{y \to 0^+} \frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{f(ty)}{1 + t^2} \, dt \]
2. Passing the Limit Inside
Assuming \(f\) is bounded (\(|f(x)| \le M\)), the integrand is dominated by an integrable function:
\[ \left| \frac{f(ty)}{1 + t^2} \right| \le \frac{M}{1 + t^2} \in L^1(\mathbb{R}) \]
By Lebesgue’s Dominated Convergence Theorem, we interchange the limit and the integral:
\[ \frac{1}{\pi} \int_{-\infty}^{+\infty} \lim_{y \to 0^+} \frac{f(ty)}{1 + t^2} \, dt = \frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{f(0)}{1 + t^2} \, dt \]
3. Evaluation
\[ = \frac{f(0)}{\pi} \Big[ \arctan(t) \Big]_{-\infty}^{+\infty} = \frac{f(0)}{\pi} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = f(0) \]