12
Dec 21

Leibniz integral rule

Leibniz integral rule

This rule is about differentiating an integral that has a parameter in three places: the lower and upper limits of integration and in the integrand.

Let l(z),u(z) be the lower and upper limits of integration and

\Phi \left( z\right) =\int_{l\left( z\right) }^{u\left( z\right) }f\left(  z,t\right) dt.

Then

\Phi ^{\prime }\left( z\right) =f\left( z,u\left( z\right) \right)  u^{\prime }\left( z\right) -f\left( z,l\left( z\right) \right) l^{\prime  }\left( z\right) +\int_{l\left( z\right) }^{u\left( z\right) }\frac{\partial  f\left( z,t\right) }{\partial z}dt.

Special cases

In fact, these special cases allow one to see how the Leibniz rule is obtained.

Case 1

l(z)=a,\ u\left( z\right) =z,\ f depends only on t. Then

\frac{d}{dz}\int_{a}^{z}f\left( t\right) dt=f\left( z\right)

(the upper limit goes into the argument of f).

Slope of tangent is limit of slopes of secants

Chart 1. Slope of tangent (violet) is limit of slopes of secants (green). Source http://faculty.wlc.edu/buelow/CALC/nt2-6.html

Intuition. By definition, f^{\prime }\left( z\right) is the limit of \frac{f\left( z+h\right) -f\left( z\right) }{h} when h\rightarrow 0. (From Chart 1, in which a=z, it is seen that this ratio is almost a slope at point z.) Hence,

\frac{d}{dz}\int_{a}^{z}f\left( t\right) dt

is the limit of

\left[ \int_{a}^{z+h}f\left( t\right) dt-\int_{a}^{z}f\left( t\right) dt  \right] /h=\int_{z}^{z+h}f\left( t\right) dt/h \approx f\left( z\right)  \int_{z}^{z+h}dt/h=f\left( z\right) \left[ z+h-z\right] /h=f\left( z\right) .

When h\rightarrow 0, this approximation becomes better and better and in the limit

\frac{d}{dz}\int_{a}^{z}f\left( t\right) dt=f\left( z\right) .

Case 2

l(z)=z, u\left( z\right) =b, f depends only on t.

Then

\frac{d}{dz}\int_{z}^{b}f\left( t\right) dt=-f\left( z\right)

(the sign is opposite to Case 1).

Intuition. One of the properties of integral is that \int_{z}^{b}f\left( t\right) dt=-\int_{b}^{z}f\left( t\right) dt.

To the last integral we can apply the intuition for Case 1.

Case 3

l\left( z\right) =a,\ u\left( z\right) =b,\ f depends on (z,t).

Then

\frac{d}{dz}\int_{a}^{b}f\left( z,t\right) dt=\int_{a}^{b}\frac{\partial  f\left( z,t\right) }{\partial z}dt

(only the integrand is differentiated).

Intuition.

\left[ \int_{a}^{b}f\left( z+h,t\right) dt-\int_{a}^{b}f\left( z,t\right) dt  \right] /h=\int_{a}^{b}\frac{f\left( z+h,t\right) -f\left( z,t\right) }{h}dt

which leads to differentiation under the integral sign.

Putting it all together

In the general case we can denote

\Psi \left( a,b,z\right)  =\int_{a}^{b}f\left( z,t\right) dt

so that

\Phi \left( z\right)  =\int_{l\left( z\right) }^{u\left( z\right) }f\left( z,t\right) dt=\Psi  \left( l\left( z\right) ,u\left( z\right) ,z\right) .

By the chain rule

\Phi ^{\prime }\left( z\right) =\frac{\partial \Psi \left( l\left( z\right)  ,u\left( z\right) ,z\right) }{\partial a}l^{\prime }\left( z\right) +\frac{  \partial \Psi \left( l\left( z\right) ,u\left( z\right) ,z\right) }{\partial  b}u^{\prime }\left( z\right) +\frac{\partial \Psi \left( l\left( z\right)  ,u\left( z\right) ,z\right) }{\partial z}

giving us the general Leibniz rule.

Note that Case 1 implies

f_{X}\left( x\right) =\frac{d}{dx}\int_{-\infty}^{x}f(t)dt=F_{X}^{\prime }\left( x\right)

(the density is found from the distribution function).

Want more information?

Leave a Reply

You must be logged in to post a comment.