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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