mathematical-analysis mathematics real-mathematical-analysis vector-mathematical-analysis Parametric Integrals Definition: Parametric Integral
A parametric integral is a real scalar field \(\mathcal{I}: \mathcal{D}_{\mathcal{I}} \subseteq \mathbb{R}^{n-1} \to \mathbb{R}\) , defined via a Riemann integral
\[\mathcal{I}(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n) \overset{\text{def}}{=} \int_{l(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)}^{u(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)} f(x_1, \dotsc, x_n) \, \mathrm{d}x_k,\]
where \(f: \mathcal{D}_f \to \mathbb{R}\) and \(l, u: \mathcal{D}_{\mathcal{I}} \to \mathbb{R}\) are real scalar fields such that the real function \(x_k \to f(x_1, \dotsc, x_k, \dotsc, x_n)\) is Riemann integrable on the closed interval between \(l(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)\) and \(u(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)\) for all \(\begin{bmatrix}x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n \end{bmatrix}^{\mathsf{T}} \in \mathcal{D}_{\mathcal{I}}\) .
Theorem: Continuity of Parametric Integrals
Let \(\mathcal{I}: \mathcal{D}_{\mathcal{I}} \subseteq \mathbb{R}^{n-1} \to \mathbb{R}\) be a parametric integral defined as
\[\mathcal{I}(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n) = \int_{l(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)}^{u(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)} f(x_1, \dotsc, x_n) \, \mathrm{d}x_k\]
for some real scalar fields \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) and \(l, u: \mathcal{D}_{\mathcal{I}} \to \mathbb{R}\) .
If \(f\) , \(l\) and \(u\) are continuous , then so is \(\mathcal{I}\) .
Example: \(F(x) = \int_0^{\uppi} x \sin y \, \mathrm{d}y\) Consider the parametric integral \(F: \mathbb{R} \to \mathbb{R}\) defined as follows:
\[F(x) = \int_0^{\uppi} x \sin y \, \mathrm{d}y\]
In this case, \(l\) and \(u\) are the following real functions \(l, u: \mathbb{R} \to \mathbb{R}\) :
\[l(x) = 0 \qquad u(x) = \uppi\]
For \(f\) , we have the following real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) :
\[f(x, y) = x \sin y\]
We see that \(l\) and \(u\) are continuous on \(\mathbb{R}\) and that \(f\) is continuous on \(\mathbb{R}^2\) . Therefore, \(F\) is continuous on \(\mathbb{R}\) .
Example Consider the parametric integral \(F: \mathbb{R} \to \mathbb{R}\) defined as
\[F(x) = \int_0^{1} f(x,y) \mathrm{d}y,\]
where \(f\) is the real scalar field \(f: [0,1] \times [0,1] \to \mathbb{R}\) defined as follows:
\[f(x,y) = \begin{cases} \frac{y}{x^2} & \text{if} & 0 \le y \lt x \\ \frac{2x - y}{x^2} & \text{if} & x \le y \lt 2x \\ 0 & \text{if} & 2x \le y \end{cases}\]
In this case, \(l\) and \(u\) are the following real functions \(l, u: \mathbb{R} \to \mathbb{R}\) :
\[l(x) = 0 \qquad u(x) = 1\]
Both \(l\) and \(u\) are continuous on \(\mathbb{R}\) . However, \(f\) is not continuous at \(\boldsymbol{0}\) . Therefore, we cannot use the theorem.
Indeed, it turns out that \(F(x)\) is not continuous at \(x = 0\) :
\[F(0) = \int_0^{1} f(0,y) \mathrm{d}y = \int_0^{1} 0 \mathrm{d}y = 0\]
However,
\[\begin{aligned}\lim_{x \to 0} F(x) & = \lim_{x \to 0} \int_0^{1} f(x,y) \mathrm{d}y \\ & = \lim_{x \to 0} \left( \int_0^{x} \frac{y}{x^2} \mathrm{d}y + \int_{x}^{2x} \frac{2x - y}{x^2} \mathrm{d}y + \int_{2x}^{1} 0 \mathrm{d}y \right) \\ & = \lim_{x \to 0} \left( \frac{1}{x^2} \left[ \frac{y^2}{2} \right]_0^x + \frac{1}{x^2} \left[ 2xy - \frac{y^2}{2} \right]_x^{2x} \right) \\ & = \lim_{x \to 0} \left( \frac{1}{x^2} \left( \frac{x^2}{2} \right) + \frac{1}{x^2} \left( \left(4x^2 - 2x^2\right) - \left(2x^2 - \frac{x^2}{2}\right) \right) \right) \\ & = \lim_{x \to 0} \left( \frac{1}{2} + \frac{1}{x^2} \left( 2x^2 - \frac{3x^2}{2} \right) \right) \\ & = \lim_{x \to 0} \left( \frac{1}{2} + \frac{1}{2} \right) \\ & = 1 \ne 0 \end{aligned}\]
Proof TODO
Theorem: Leibniz Integral Rule (Differentiation under the Integral Sign)
Let \(\mathcal{I}: \mathcal{D}_{\mathcal{I}} \subseteq \mathbb{R}^{n-1} \to \mathbb{R}\) be a parametric integral defined as
\[\mathcal{I}(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n) = \int_{l(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)}^{u(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)} f(x_1, \dotsc, x_n) \, \mathrm{d}x_k\]
for some real scalar fields \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) and \(l, u: \mathcal{D}_{\mathcal{I}} \to \mathbb{R}\) .
If \(f\) , \(l\) and \(u\) are continuously partially differentiable with respect to \(x_j\) (\(j \ne k\) ) variable, then so is \(\mathcal{I}\) and its partial derivative is the following:
\[\partial_{x_j} \mathcal{I} = \int_{l(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)}^{u(x_1, \dotsc, x_{k-1}, x_{k+1}, \dotsc, x_n)} \partial_{ x_j} f(x_1, \dotsc, x_n) \, \mathrm{d}x_k + f\big|_{x_k=u} \partial_{x_j} u - f\big|_{x_k=l} \partial_{x_j} l\]
Example: \(F(x) = \int_{x^2}^{3-x} \sin(xy) \,\mathrm{d}y\) Consider the parametric integral \(F: \mathbb{R} \to \mathbb{R}\) defined as follows:
\[F(x) = \int_{x^2}^{3-x} \sin(xy) \,\mathrm{d}y\]
In this case, \(l\) and \(u\) are the following real functions \(l, u: \mathbb{R} \to \mathbb{R}\) :
\[l(x) = x^2 \qquad u(x) = 3-x\]
For \(f\) , we have the following real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) :
\[f(x, y) = \sin (xy)\]
We see that \(l\) and \(u\) are continuously differentiable on \(\mathbb{R}\) and that \(f\) is continuously partially differentiable on \(\mathbb{R}^2\) w.r.t. \(x\) . Therefore, \(F\) is continuously differentiable on \(\mathbb{R}\) with the following derivative :
\[\begin{aligned} F'(x) & = \int_{x^2}^{3-x} \partial_x f(x,y) \,\mathrm{d}y + f(x,u(x)) u'(x) - f(x, l(x)) l'(x) \\ & = \int_{x^2}^{3-x} y \cos(xy) \,\mathrm{d}y + \sin(x(3-x)) \cdot (3-x)' - \sin(x(x^2)) \cdot (x^2)' \\ & = \int_{x^2}^{3-x} y \cos(xy) \,\mathrm{d}y + \sin(3x - x^2) \cdot (-1) - \sin(x^3) \cdot (2x) \\ & = \int_{x^2}^{3-x} y \cos(xy) \,\mathrm{d}y - \sin(3x - x^2) - 2x \sin(x^3) \end{aligned}\]
For \(x \ne 0\) , we get:
\[\begin{aligned} F'(x) & = \int_{x^2}^{3-x} y \cos(xy) \,\mathrm{d}y - \sin(3x - x^2) - 2x \sin(x^3) \\ & = \left.\left( \frac{y}{x} \sin(xy) + \frac{1}{x^2} \cos(xy) \right)\right\vert_{y=x^2}^{y=3-x} - \sin(3x - x^2) - 2x \sin(x^3) \\ & = \left( \frac{3-x}{x} \sin(3x-x^2) + \frac{1}{x^2} \cos(3x-x^2) \right) - \left( x \sin(x^3) + \frac{1}{x^2} \cos(x^3) \right) - \sin(3x - x^2) - 2x \sin(x^3) \\ & = \left( \frac{3-x}{x} - 1 \right) \sin(3x - x^2) + \frac{1}{x^2} \cos(3x - x^2) - (x + 2x) \sin(x^3) - \frac{1}{x^2} \cos(x^3) \\ & = \frac{3-2x}{x} \sin(3x - x^2) + \frac{1}{x^2} \cos(3x - x^2) - 3x \sin(x^3) - \frac{1}{x^2} \cos(x^3) \end{aligned}\]
For \(x = 0\) , we get:
\[\begin{aligned} F'(0) & = \int_{0}^{3} y \cos(0) \,\mathrm{d}y - \sin(0) - 0 \\ & = \int_{0}^{3} y \,\mathrm{d}y \\ & = \left. \frac{1}{2}y^2 \right\vert_{0}^{3} = \frac{9}{2}\end{aligned}\]
We can also verify that \(F'\) is continuous :
\[\begin{aligned} \lim_{x \to 0} F'(x) & = \lim_{x \to 0} \left( \frac{3-2x}{x} \sin(3x - x^2) + \frac{\cos(3x - x^2) - \cos(x^3)}{x^2} - 3x \sin(x^3) \right) \\ & = \lim_{x \to 0} \left( (3-2x)(3-x) \frac{\sin(3x - x^2)}{3x - x^2} \right) + \lim_{x \to 0} \left( \frac{\cos(3x - x^2) - \cos(x^3)}{x^2} \right) - 0 \\ & = 9 \cdot 1 + \lim_{x \to 0} \frac{-(3-2x)\sin(3x - x^2) + 3x^2\sin(x^3)}{2x} \quad \text{(using L'Hôpital's rule)} \\ & = 9 + \lim_{x \to 0} \left( -\frac{(3-2x)(3-x)}{2} \frac{\sin(3x - x^2)}{3x - x^2} + \frac{3}{2} x \sin(x^3) \right) \\ & = 9 - \frac{9}{2} \cdot 1 + 0 \\ & = \frac{9}{2} \end{aligned}\]
Proof TODO
Fubini's Theorem
Let \(f: [a_1, b_1] \times \cdots \times [a_n, b_n] \subset \mathbb{R}^n \to \mathbb{R}\) be a real scalar field .
If \(f\) is continuous , then iterated integrals of \(f\) are invariant for all permutations \(\sigma: \{1, \dotsc, n\} \to \{1, \dotsc, n\}\) :
\[\int_{a_1}^{b_1} \cdots \int_{a_n}^{b_n} f(x_1, \dotsc, x_n) \,\mathrm{d}x_n \cdots \mathrm{d}x_1 = \int_{a_{\sigma(1)}}^{b_{\sigma(1)}} \cdots \int_{a_{\sigma(n)}}^{b_{\sigma(n)}} f(x_1, \dotsc, x_n) \,\mathrm{d}x_{\sigma(n)} \cdots \mathrm{d}x_{\sigma(1)}\]
Example: \(f(x, y) = 2x\mathrm{e}^y\) Consider the real scalar field \(f: [0,1] \times [0,1]\) defined as follows:
\[f(x, y) = 2x\mathrm{e}^y\]
We have:
\[\begin{aligned}\int_0^1 \int_0^1 2x\mathrm{e}^y \,\mathrm{d}y \,\mathrm{d}x & = \int_0^1 \left(\left. 2x \mathrm{e}^y\right\vert_{y=0}^{y=1}\right) \,\mathrm{d}x \\ & = \int_0^1 (2x \mathrm{e}^1 - 2x \mathrm{e}^0) \, \mathrm{d}x \\ & = \int_0^1 2x(\mathrm{e} - 1) \, \mathrm{d}x \\ & = (\mathrm{e} - 1) \int_0^1 2x \,\mathrm{d}x \\ & = \left.(e-1)x^2 \right\vert_{0}^1 \\ & = \mathrm{e} - 1\end{aligned}\]
Also:
\[\begin{aligned}\int_0^1 \int_0^1 2x\mathrm{e}^y \,\mathrm{d}x \,\mathrm{d}y & = \int_0^1 \left(\left. x^2 \mathrm{e}^y\right\vert_{x=0}^{x=1}\right) \,\mathrm{d}y \\ & = \int_0^1 (\mathrm{e}^y - 0) \, \mathrm{d}y \\ & = \int_0^1 \mathrm{e}^y \, \mathrm{d}y \\ & = \left. \mathrm{e}^y \right\vert_{0}^1 \\ & = \mathrm{e} - 1\end{aligned}\]
Therefore,
\[\int_0^1 \int_0^1 2x\mathrm{e}^y \,\mathrm{d}y \,\mathrm{d}x = \int_0^1 \int_0^1 2x\mathrm{e}^y \,\mathrm{d}x \,\mathrm{d}y,\]
which is what we expect because \(f\) is continuous .
Proof TODO
April 3, 2026 April 3, 2026