Ngô Quốc Anh

September 14, 2018

Leibniz rule for proper integral with parameter whose limits also depends on the parameter

Filed under: Uncategorized — Ngô Quốc Anh @ 21:24

The following Leibniz integral rule is well-known

Theorem. Let $f(x, t)$ be a function such that both $f(x, t)$ and its partial derivative $f_x(x, t)$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) \leqslant t \leqslant b(x)$, $x_0 \leqslant x \leqslant x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 \leqslant x \leqslant x_1$. Then, for $x_0 \leqslant x \leqslant x_1$, $\displaystyle \frac {d}{dx}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)=f\big (x,b(x)\big )b'(x)-f\big (x,a(x)\big) a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial f }{\partial x}}(x,t)\,dt.$

The purpose of this note is to show that, in fact, it is not  is not necessary to assume the function $f$ to be continuous. We note that this is indeed the case in which the limits of the integral $\int_{a(x)}^{b(x)}f(x,t)\,dt$ do not depend on the parameter $x$. For convenience, it is routine to assume the continuity, which immediately implies that all integrals are well-defined.

As mentioned above, we want to show that this is also the case for integrals of the form above.

To this purpose, we first establish the following result:

Lemma. Let $f(x, t)$ be an integrable function such that $f_x(x, t)$ is continuous in both $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) \leqslant t \leqslant A$, $x_0 \leqslant x \leqslant x_1$. Also suppose that the function $a(x)$ is continuous and has continuous derivative for $x_0 \leqslant x \leqslant x_1$. Then, for $x_0 \leqslant x \leqslant x_1$, $\displaystyle \frac {d}{dx}\left(\int _{a(x)}^A f(x,t)\,dt\right)= -f\big (x,a(x)\big) a'(x)+\int _{a(x)}^A \frac {\partial }{\partial x}f(x,t)\,dt.$

Here is a proof for this lemma.

Proof of Lemma. By definition, it is necessary to consider the quotient $\displaystyle \frac 1h \left(\int _{a(x+h)}^A f(x+h,t)\,dt - \int _{a(x)}^A f(x,t)\,dt\right).$

Fixing $x$ and letting $h$ to zero, it is suffices to show that the above tends to the right hand side of the identity in the lemma. To this end, we write $\displaystyle \frac 1h \left(\int _{a(x+h)}^A f(x+h,t)\,dt - \int _{a(x)}^A f(x,t)\,dt\right)=\frac 1h \int _{a(x+h)}^{a(x)} f(x+h,t)\,dt +\frac 1h \left(\int _{a(x)}^A \big[ f(x+h,t)- f(x,t) \big] \,dt\right).$

We observe that the second term on the right hand side of the preceding identity is a good term as it tends to $\displaystyle \int _{a(x)}^A \frac {\partial }{\partial x}f(x,t)\,dt$

as $h \to 0$. For the first term, we shall make use of the mean value theorem in the following way. We write $\displaystyle \frac 1h \int _{a(x+h)}^{a(x)} f(x+h,t)\,dt= \frac 1h \int _{a(x+h)}^{a(x)} \big[ f(x+h,t) - f(x,t) \big]\,dt + \frac 1h \int _{a(x+h)}^{a(x)} f(x,t)\,dt.$

For the first term on the right hand side, we apply the Lagrange theorem, for each fixed $t$ and fixed $h$ to get $\displaystyle \frac 1h \int _{a(x+h)}^{a(x)} \big[ f(x+h,t) - f(x,t) \big]\,dt=\int _{a(x+h)}^{a(x)} f_x(x+\theta (h,t),t)\,dt,$

where $\theta (h,t)$ lies between $h$ and $0$. Since $f_x$ is continuous in the region $a(x) \leqslant t \leqslant A$, $x_0 \leqslant x \leqslant x_1$, it is bounded in it. Hence, there is some constant $C > 0$ such that $|f_x(x,t)| \leqslant C\quad$

for all $(x,t)$. In particular, we estimate $\displaystyle \left|\frac 1h \int _{a(x+h)}^{a(x)} \big[ f(x+h,t) - f(x,t) \big]\,dt \right|\leqslant C |a(x+h)-a(x)|$

to get $\displaystyle \lim_{h \to 0} \frac 1h \int _{a(x+h)}^{a(x)} \big[ f(x+h,t) - f(x,t) \big]\,dt = 0$

due to the continuity of $a$ at $x$. For the second term, we apply the mean value theorem to get $\displaystyle \frac 1h \int _{a(x+h)}^{a(x)} f(x,t)\,dt = \frac {a(x)-a(x+h)}h f(x, \theta (x,h))$

for some $\theta (x,h)$ lying between $a(x+h)$ and $a(x)$. Since $f_x$ exists, $f$, as a function of $x$, is continuous. Hence taking the limit as $h \to 0$ gives $\displaystyle \lim_{h \to 0}\frac 1h \int _{a(x+h)}^{a(x)} f(x,t)\,dt = -a'(x) f\big(x, a(x)\big),$

thanks to the fact that $\theta (x,h) \to a(x)$ as $h \to 0$. The present proof is complete.

We now apply the above lemma to conclude the Leibniz rule in the following way. If we want to prove that the integral $\Phi (x) := \int _{a(x)}^{b(x)}f(x,t)\,dt$ is differentiable at the point $x=x_0$ and its derivative $\Phi'(x_0)$ is nothing but $\displaystyle {\frac {d}{dx}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)=f{\big (}x,b(x){\big )}b'(x)-f\big (x,a(x)\big) a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt,$

we decompose $(\Phi(x)-\Phi(x_0))/(x-x_0)$ as follows $\displaystyle \frac {\Phi(x)-\Phi(x_0)}{x-x_0}=\frac 1{x-x_0} \int _{a(x)}^{a(x_0)}f(x,t)\,dt + \int _{a(x_0)}^{b(x_0)} \frac {f(x,t) - f(x_0,t)}{x-x_0} \,dt - \frac 1{x-x_0} \int _{b(x)}^{b(x_0)}f(x,t)\,dt.$

From this and after taking the limit as $x \to x_0$, it is not hard to realize that $\displaystyle \frac {d}{dx}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right) \bigg|_{x=x_0}=f\big (x_0,b(x_0)\big )b'(x_0)-f\big (x_0,a(x_0)\big) a'(x_0)+\int _{a(x_0)}^{b(x_0)}{\frac {\partial f}{\partial x}}(x_0,t)\,dt$

as claimed.

I thank Dr. Trinh Viet Duoc for useful discussion on this rule.

1 Comment »

1. Thank you so much for your hard work and the beautiful proof presented here!

Comment by Tiến — September 15, 2018 @ 6:28

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