Integrals with Parameters and Improper Integrals

4.1 Continuity, Integrability, and Differentiability of Integrals with Respect to Parameters

Let the function \(f : [a,b] \times U \to \mathbb{R}\) (where \(U \subseteq \mathbb{R}^n\)) satisfy the condition that for any \(\mathbf{y} \in U\), the function \(f(\cdot, \mathbf{y}) : [a,b] \to \mathbb{R}\) is Riemann integrable. In this case, we call \(\int_{a}^{b} f(x, \mathbf{y}) dx\) a definite integral depending on a parameter \(\mathbf{y}\).

This integral defines a function of \(\mathbf{y}\) as follows: \[ \begin{aligned} F(\mathbf{y}) = \int_{a}^{b} f(x, \mathbf{y}) dx. \end{aligned} \]

Theorem 4.1 Let the function \(f : [a,b] \times U \to \mathbb{R}\) (where \(U \subseteq \mathbb{R}^n\)) satisfy the following conditions:

  1. For any \(\mathbf{y} \in U\), the Riemann integral \[ \begin{aligned} F(\mathbf{y}) = \int_{a}^{b} f(x, \mathbf{y}) dx \end{aligned} \] exists.

  2. \(f(x, \mathbf{y})\) is continuous with respect to \(\mathbf{y}\) at \(\mathbf{y}_0 \in U\), and this continuity is uniform with respect to the integration variable \(x \in [a, b]\). That is, for any \(\epsilon > 0\), there exists a \(\delta(\mathbf{y}_0, \epsilon) > 0\) such that for any \(x \in [a, b]\), whenever \(\mathbf{y} \in U\) satisfies \(\|\mathbf{y} - \mathbf{y}_0\| < \delta(\mathbf{y}_0, \epsilon)\), we have: \[ \begin{aligned} |f(x, \mathbf{y}) - f(x, \mathbf{y}_0)| < \epsilon. \end{aligned} \]

Then \(F\) is continuous at \(\mathbf{y}_0\), which implies: \[ \begin{aligned} \lim_{\mathbf{y} \to \mathbf{y}_0} \int_{a}^{b} f(x, \mathbf{y}) dx = \int_{a}^{b} f(x, \mathbf{y}_0) dx = \int_{a}^{b} \lim_{\mathbf{y} \to \mathbf{y}_0} f(x, \mathbf{y}) dx. \end{aligned} \]

Theorem 4.2 If \(U \subset \mathbb{R}^n\) is a closed set or an open set, and \(f : [a,b] \times U \to \mathbb{R}\) is continuous, then \(F\) is continuous on \(U\).

Theorem 4.3 Let \(U \subseteq \mathbb{R}^n\) be an open set. Suppose the function \(f : [a,b] \times U \to \mathbb{R}\) satisfies:

  1. For any \(\mathbf{y} \in U\), the Riemann integral \[ \begin{aligned} F(\mathbf{y}) = \int_{a}^{b} f(x, \mathbf{y}) dx \end{aligned} \] exists.

  2. The partial derivatives \(\frac{\partial f}{\partial y_1}, \dots, \frac{\partial f}{\partial y_n}\) are bounded on \([a,b] \times U\).

Then \(F\) is continuous on \(U\).

Theorem 4.4 Let \(U \subseteq \mathbb{R}^n\) be an open set, and let the function \(f : [a,b] \times U \to \mathbb{R}\) satisfy the following conditions:

  1. For any \(\mathbf{y} \in U\), \(f(x, \mathbf{y})\) is Riemann integrable with respect to \(x\) on the interval \([a, b]\).

  2. For any \(x \in [a, b]\), \(f(x, \mathbf{y})\) is differentiable with respect to \(\mathbf{y} \in U\).

  3. For each \(k = 1, 2, \dots, n\), the partial derivative \(\frac{\partial f}{\partial y_k}(x, \mathbf{y})\) is Riemann integrable with respect to \(x\), and is continuous with respect to \(\mathbf{y}\). Furthermore, this continuity is uniform with respect to the integration variable \(x \in [a, b]\). That is, for any \(\mathbf{y}_0 \in U\) and any \(\epsilon > 0\), there exists a \(\delta(\mathbf{y}_0, \epsilon) > 0\) such that for any \(x \in [a, b]\), whenever \(\mathbf{y} \in U\) satisfies \(\|\mathbf{y} - \mathbf{y}_0\| < \delta(\mathbf{y}_0, \epsilon)\), we have: \[ \begin{aligned} \left| \frac{\partial f}{\partial y_k}(x, \mathbf{y}) - \frac{\partial f}{\partial y_k}(x, \mathbf{y}_0) \right| < \epsilon. \end{aligned} \]

Then \(F(\mathbf{y}) = \int_{a}^{b} f(x, \mathbf{y}) dx\) is of class \(C^1\) on \(U\), and: \[ \begin{aligned} \frac{\partial F}{\partial y_k}(\mathbf{y}) = \frac{\partial}{\partial y_k} \int_{a}^{b} f(x, \mathbf{y}) dx = \int_{a}^{b} \frac{\partial f}{\partial y_k}(x, \mathbf{y}) dx. \end{aligned} \]

Corollary 4.1 If in the aforementioned theorem, all partial derivatives of \(f\) with respect to \(\mathbf{y}\) up to order \(k\) are continuous with respect to \((x, \mathbf{y})\), then \(F\) is of class \(C^k\) on \(U\). Furthermore, all partial derivative operations of \(F\) up to order \(k\) can be interchanged with the integral operator: \[ \begin{aligned} \frac{\partial^k F}{\partial y_{i_1} \partial y_{i_2} \dots \partial y_{i_k}} (\mathbf{y}) = \int_{a}^{b} \frac{\partial^k f}{\partial y_{i_1} \partial y_{i_2} \dots \partial y_{i_k}} (x, \mathbf{y}) dx. \end{aligned} \]

Theorem 4.5 (Interchanging the Order of Iterated Integration) Let the function \(f : [a,b] \times [c, d] \to \mathbb{R}\) be continuous. Then: \[ \begin{aligned} \int_{a}^{b} \left( \int_{c}^{d} f(x, y) dy \right) dx = \int_{c}^{d} \left( \int_{a}^{b} f(x, y) dx \right) dy. \end{aligned} \]

The above problem can be summarized as: \[ \begin{aligned} \min \int_{a}^{b} L(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) dt, \quad \mathbf{x}(a) = A, \mathbf{x}(b) = B. \end{aligned} \]

\(L(t, \mathbf{x}, \dot{\mathbf{x}})\): Lagrangian function

Embed the curve \(\mathbf{x}(t)\) to be solved into a family of curves \(X(t, s)\) (where for each fixed \(s\), \(X(t, s)\) is a curve) such that: \[ \begin{aligned} X(t, 0) = \mathbf{x}(t), \quad \forall t, \end{aligned} \] and for all \(s\), we have \(X(a, s) = A\) and \(X(b, s) = B\). For example: \(X(t, s) = \mathbf{x}(t) + s\mathbf{y}(t)\), where \(\mathbf{y}(a) = \mathbf{y}(b) = 0\).

Consider: \[ \begin{aligned} F(s) = \int_{a}^{b} L\left(t, X(t, s), \frac{\partial X}{\partial t}(t, s)\right) dt. \end{aligned} \] Then \(F(0) = \int_{a}^{b} L(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) dt\) is the minimum value.

Thus \(F'(0) = 0\). Namely, \[ \begin{aligned} \int_{a}^{b} \left[ L_{\mathbf{x}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) X_s(t, 0) + L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) X_{s,t}(t, 0) \right] dt = 0 \end{aligned} \]

By integration by parts, the second term becomes: \[ \begin{aligned} \int_{a}^{b} L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) d_t X_s(t, 0) &= \left. L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) X_s(t, 0) \right|_{a}^{b} - \int_{a}^{b} \frac{d}{dt} L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) X_s(t, 0) dt \\ &= - \int_{a}^{b} \frac{d}{dt} L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) X_s(t, 0) dt \end{aligned} \]

Therefore, \[ \begin{aligned} \int_{a}^{b} \left[ L_{\mathbf{x}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) - \frac{d}{dt} L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) \right] X_s(t, 0) dt = 0 \end{aligned} \]

Since \(X_s(t, 0)\) is arbitrary, we obtain the Euler-Lagrange equation: \[ \begin{aligned} L_{\mathbf{x}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) - \frac{d}{dt} L_{\dot{\mathbf{x}}}(t, \mathbf{x}(t), \dot{\mathbf{x}}(t)) = 0. \end{aligned} \] This is a second-order ordinary differential equation (ODE) with respect to \(\mathbf{x}(t)\).

Reducible Case 1: \(L = L(\mathbf{x}, \dot{\mathbf{x}})\):

The Euler-Lagrange equation is: \[ \begin{aligned} L_x(x, \dot{x}) - \frac{d}{dt} L_{\dot{x}}(x, \dot{x}) = 0. \end{aligned} \]

Multiplying by \(\dot{x}\) and using the chain rule: \[ \begin{aligned} L_x(x, \dot{x})\dot{x} + L_{\dot{x}}(x, \dot{x})\ddot{x} - \frac{d}{dt} \left( L_{\dot{x}}(x, \dot{x})\dot{x} \right) = 0. \end{aligned} \]

Since \(\frac{d}{dt} L(x, \dot{x}) = L_x\dot{x} + L_{\dot{x}}\ddot{x}\) when \(L\) does not explicitly depend on \(t\), we have: \[ \begin{aligned} \frac{d}{dt} \left( L(x, \dot{x}) - L_{\dot{x}}(x, \dot{x})\dot{x} \right) = 0, \end{aligned} \]

Which leads to the Beltrami Identity: \[ \begin{aligned} L(x, \dot{x}) - L_{\dot{x}}(x, \dot{x})\dot{x} = C. \end{aligned} \]

4.2 Improper Integrals with Parameters

For a function \(f : [a, +\infty) \times U \to \mathbb{R}\) (where \(U \subseteq \mathbb{R}^n\)), the integral \(\int_{a}^{+\infty} f(x, \mathbf{y}) dx\) is called an improper integral depending on a parameter \(\mathbf{y}\).

A convergent improper integral with a parameter defines a function with \(\mathbf{y}\) as the independent variable: \[ \begin{aligned} F(\mathbf{y}) = \int_{a}^{+\infty} f(x, \mathbf{y}) dx. \end{aligned} \]

The improper integral \(\int_{a}^{+\infty} f(x, \mathbf{y}) dx\) is said to converge uniformly to \(F(\mathbf{y})\) on \(U\) if for every \(\epsilon > 0\), there exists \(A(\epsilon) > a\) such that for all \(A > A(\epsilon)\) and for every \(\mathbf{y} \in U\), we have: \[ \begin{aligned} \left| \int_{a}^{A} f(x, \mathbf{y}) dx - F(\mathbf{y}) \right| < \epsilon. \end{aligned} \]

Theorem 4.6 Let the function \(f(x, \mathbf{y})\) be Riemann integrable with respect to \(x\) on any bounded interval \(I \subset [a, +\infty)\). The improper integral \(\int_{a}^{+\infty} f(x, \mathbf{y}) dx\) converges uniformly with respect to \(\mathbf{y} \in U\) if and only if it satisfies the uniform Cauchy condition: for any \(\epsilon > 0\), there exists \(A(\epsilon) > a\) such that \[ \begin{aligned} \left| \int_{A_1}^{A_2} f(x, \mathbf{y}) dx \right| < \epsilon, \quad \forall A_2 > A_1 \ge A(\epsilon), \quad \forall \mathbf{y} \in U. \end{aligned} \]

Theorem 4.7 Let the functions \(f(x, \mathbf{y})\) and \(g(x, \mathbf{y})\) satisfy: \[ \begin{aligned} |f(x, \mathbf{y})| \le g(x, \mathbf{y}), \quad \forall x \ge a, \forall \mathbf{y} \in U \end{aligned} \] If the integral \(\int_{a}^{+\infty} g(x, \mathbf{y}) dx\) converges uniformly with respect to \(\mathbf{y} \in U\), then the integral \(\int_{a}^{+\infty} f(x, \mathbf{y}) dx\) converges uniformly (and absolutely) with respect to \(\mathbf{y} \in U\).

Theorem 4.8 (Dirichlet and Abel Tests) Assume the finite-interval integrals exist. The improper integral \(\int_a^\infty f(x,y)g(x,y)\,dx\) converges uniformly for \(y\in U\) under either set of conditions:

  1. Dirichlet’s test:
    • \(F(t,y)=\int_a^t f(x,y)\,dx\) is bounded uniformly in \(t\ge a\) and \(y\in U\).
    • For each \(y\), \(g(x,y)\) is monotone in \(x\).
    • \(g(x,y)\to0\) as \(x\to\infty\), uniformly for \(y\in U\).
  2. Abel’s test:
    • \(\int_a^\infty f(x,y)\,dx\) converges uniformly for \(y\in U\).
    • For each \(y\), \(g(x,y)\) is monotone in \(x\).
    • \(g(x,y)\) is bounded uniformly in \(x\ge a\) and \(y\in U\).

Theorem 4.9 Let the function \(f: [a, +\infty) \times U \to \mathbb{R} \ (U \subseteq \mathbb{R}^m)\) satisfy:

  1. The improper integral \(F(\mathbf{y}) = \int_a^{+\infty} f(x, \mathbf{y}) \text{d}x\) converges uniformly on \(U\);
  2. For any \(x \ge a\), \(f(x, \mathbf{y})\) is continuous with respect to \(\mathbf{y}\) at \(\mathbf{y}_0 \in U\), and this continuity is uniform with respect to \(x\) in any bounded closed interval \([a, b]\): that is, for any \(\epsilon > 0\), there exists \(\delta(\mathbf{y}_0, \epsilon) > 0\), such that for any \(x \in [a, b]\), as long as \(\mathbf{y} \in U\) satisfies \(\|\mathbf{y} - \mathbf{y}_0\| < \delta(\mathbf{y}_0, \epsilon)\), we have \(|f(x, \mathbf{y}) - f(x, \mathbf{y}_0)| < \epsilon\).

Then \(F\) is continuous at \(\mathbf{y}_0\), i.e., \[ \begin{aligned} \lim_{\mathbf{y} \to \mathbf{y}_0} \int_a^{+\infty} f(x, \mathbf{y}) \text{d}x = \int_a^{+\infty} f(x, \mathbf{y}_0) \text{d}x = \int_a^{+\infty} \lim_{\mathbf{y} \to \mathbf{y}_0} f(x, \mathbf{y}) \text{d}x. \end{aligned} \]

Theorem 4.10 Let the continuous function of two variables \(f: [a, +\infty) \times [1, +\infty) \to \mathbb{R}\) satisfy:

  1. The improper integral \(F(y) = \int_a^{+\infty} f(x, y) \text{d}x\) converges uniformly on \([1, +\infty)\);
  2. For any \(x \ge a\), the limit \(\lim_{y \to +\infty} f(x, y) = g(x)\) exists, and this limit is uniform with respect to \(x\) in any bounded closed interval \([a, b]\): that is, for any \(\epsilon > 0\) and any interval \([a, b]\), there exists \(N_{a,b,\epsilon} > 1\), such that for any \(x \in [a, b]\), whenever \(y > N_{a,b,\epsilon}\), we have \(|f(x, y) - g(x)| < \epsilon\).

Then \[ \begin{aligned} \lim_{y \to +\infty} \int_a^{+\infty} f(x, y) \text{d}x = \int_a^{+\infty} g(x) \text{d}x = \int_a^{+\infty} \lim_{y \to +\infty} f(x, y) \text{d}x. \end{aligned} \]

Theorem 4.11 Let \(U \subseteq \mathbb{R}^m\) be an open set, \(\mathbf{y}_0 \in U\), and let the function \(f: [a, +\infty) \times U \to \mathbb{R}\) satisfy:

  1. The improper integral \(\int_a^{+\infty} f(x, \mathbf{y}_0) \text{d}x\) converges;
  2. For each \(k = 1, 2, \dots, m\), \(\frac{\partial f}{\partial y_k}(x, \mathbf{y})\) is continuous, and the improper integral \(\int_a^{+\infty} \frac{\partial f}{\partial y_k}(x, \mathbf{y}) \text{d}x\) converges uniformly with respect to \(\mathbf{y}\) in \(U\).

Then \(\int_a^{+\infty} f(x, \mathbf{y}) \text{d}x\) converges uniformly in a neighborhood of \(\mathbf{y}_0\), is of class \(\mathscr{C}^1\) with respect to \(\mathbf{y}\), and \[ \begin{aligned} \frac{\partial}{\partial y_k} \int_a^{+\infty} f(x, \mathbf{y}) \text{d}x = \int_a^{+\infty} \frac{\partial f}{\partial y_k}(x, \mathbf{y}) \text{d}x, \quad k = 1, 2, \dots, m. \end{aligned} \]

Theorem 4.12 Let \(f : [c, +\infty) \times [a, b] \to \mathbb{R}\) be continuous, and the improper integral \(\int_c^{+\infty} f(x, y)dx\) converges uniformly with respect to \(y\) on \([a, b]\). Then the improper integral \(\int_c^{+\infty} \int_a^b f(x, y)dydx\) converges, and \[ \begin{aligned} \int_a^b \int_c^{+\infty} f(x, y)dxdy = \int_c^{+\infty} \int_a^b f(x, y)dydx. \end{aligned} \]

Theorem 4.13 Let \(f: [c, +\infty) \times [a, +\infty) \to \mathbb{R}\) be continuous, satisfying:

  1. At least one of the iterated integrals \(\int_a^{+\infty} \int_c^{+\infty} |f(x, y)| dx dy\) and \(\int_c^{+\infty} \int_a^{+\infty} |f(x, y)| dy dx\) exists.
  2. The improper integral \(\int_c^{+\infty} f(x, y) dx\) converges uniformly with respect to \(y\) on any bounded closed interval \([a, b]\).
  3. The improper integral \(\int_a^{+\infty} f(x, y) dy\) converges uniformly with respect to \(x\) on any bounded closed interval \([c, d]\).

Then \[ \begin{aligned} \int_a^{+\infty} \int_c^{+\infty} f(x, y) dx dy = \int_c^{+\infty} \int_a^{+\infty} f(x, y) dy dx. \end{aligned} \]

\[ \begin{aligned} \Gamma(\alpha) = \int_{0}^{+\infty} x^{\alpha-1} \mathrm{e}^{-x} \mathrm{d}x \end{aligned} \]

It is a \(\mathscr{C}^\infty\) function for \(\alpha > 0\), satisfying:

\[ \begin{aligned} \Gamma(1) &= 1 \\ \Gamma(\alpha+1) &= \alpha\Gamma(\alpha), \quad \forall \alpha > 0 \end{aligned} \]

Thus, the \(\Gamma\) function is a generalization of the factorial of natural numbers, embedding the factorial sequence into a \(\mathscr{C}^\infty\) function. Additionally, the \(\Gamma\) function possesses the log-convex property:

\[ \begin{aligned} \ln \Gamma\left(\frac{\alpha + \beta}{2}\right) \le \frac{\ln \Gamma(\alpha) + \ln \Gamma(\beta)}{2}, \quad \forall \alpha > 0, \beta > 0 \end{aligned} \]

\[ \begin{aligned} B(\alpha, \beta) = \int_{0}^{1} x^{\alpha-1}(1 - x)^{\beta - 1} \mathrm{d}x, \quad \alpha > 0, \beta > 0 \end{aligned} \]

It is a \(\mathscr{C}^\infty\) function, satisfying:

\[ \begin{aligned} B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \end{aligned} \]

4.3 Integral Transforms

By an integral transform, we refer to a transformation of the form:

\[ \begin{aligned} \mathscr{T}: f \mapsto \mathscr{T}(f) = g, \quad g(y) = \int_{-\infty}^{+\infty} K(x, y) f(x) \mathrm{d}x \end{aligned} \]

A convolution is of the form:

\[ \begin{aligned} (K \star f)(y) = \int_{-\infty}^{+\infty} K(y - x) f(x) \mathrm{d}x \end{aligned} \]

We denote \(K \star f\) as the convolution of \(K\) and \(f\).

If \(K(t)\) is a differentiable function, then (provided that the following improper integral converges uniformly):

\[ \begin{aligned} (K \star f)'(y) = \int_{-\infty}^{+\infty} K'(y - x) f(x) \mathrm{d}x = (K' \star f)(y) = (K \star f')(y). \end{aligned} \]

For example, the probability density function (PDF) of a normal distribution \(N(\mu, \sigma^2)\) is:

\[ \begin{aligned} f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}. \end{aligned} \]

\[ \begin{aligned} \mathscr{L}[f](p) = \int_{0}^{+\infty} \mathrm{e}^{-px} f(x) \mathrm{d}x = F(p) \end{aligned} \]

\(F\) is called the Laplace transform of \(f\).

If \(f\) satisfies \(|f(x)| \le M \mathrm{e}^{\alpha x}\), then for any \(\delta > 0\), the aforementioned improper integral and the improper integral

\[ \begin{aligned} \int_{0}^{+\infty} \mathrm{e}^{-px} x^k f(x) \mathrm{d}x \end{aligned} \]

converge uniformly for \(p\) on the interval \([\alpha + \delta, +\infty)\). Consequently, \(F\) is a \(\mathscr{C}^\infty\) function on the interval \((\alpha, +\infty)\).

In addition to linearity, the Laplace transform also satisfies:

\[ \begin{aligned} \mathscr{L}[f(\alpha x)](p) &= \frac{1}{\alpha} \mathscr{L}[f]\left(\frac{p}{\alpha}\right) \\ \mathscr{L}[\mathrm{e}^{\alpha x} f(x)](p) &= \mathscr{L}[f](p - \alpha) \\ \mathscr{L}[f'](p) &= -f(0) + p \mathscr{L}[f](p) \\ \frac{\partial^k}{\partial p^k} \mathscr{L} \left[ \frac{f(x)}{x^k} \right](p) &= (-1)^k \mathscr{L}[f](p) \end{aligned} \]