Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The derivative I would like to find is

$$\frac{\partial}{\partial z} \int_{-z_0}^z (z - z^{\prime}) \; f(z^{\prime}) \; dz^{\prime}$$

where $f(z^{\prime})$ is some arbitrary function of $z^{\prime}$. So $z$ actually appears twice: once in the upper limit of integration, and once in the integrand.

share|cite|improve this question
Note that the variable in the integral is $z'$. The letter $z$ is to be treated as constant until after the integral is done, and you try to differentiate with respect to $z$. BUT: Look at the Fundamental Theorem of Calculus, which deals with this situation directly. – Lubin Aug 13 '12 at 22:08
The subject is well covered in this Wikipedia entry. – André Nicolas Aug 13 '12 at 23:12
up vote 4 down vote accepted

Similar questions are this one and this one.

The integral depends on the variable $z$ only, because $z'$ is a dummy variable. So you want to find the derivative of the integral and not its partial derivative. This is a special case of the following more general rule (Leibniz rule):

If $$I(z)=J(u(z),v(z),z)=\int_{u(z)}^{v(z)} f(z,z')dz',\tag{1}$$

i.e. $$\ J(a,b,z)=\int_{a}^{b} f(z,z')dz',\qquad \text{ with }a=u(z),b=v(z),$$

then, under suitable conditions, we have

$$I^{\prime }(z)=\displaystyle\int_{u(z)}^{v(z)}\dfrac{\partial f(z,z')}{\partial z}dz'+f(z,v(z))v^{\prime }(z)-f(z,u(z))u^{\prime }(z).\tag{2}$$

It is a consequence of

$$\frac{dI}{dz}=\frac{\partial J}{\partial z}+\frac{\partial J}{\partial u} \frac{du}{dz}+\frac{\partial J}{\partial v}\frac{dv}{dz}.$$

The first term is the differentiation under the integral sign

$$\frac{\partial J}{\partial z}=\int_{u(z)}^{v(z)}\dfrac{\partial f(z,z')}{\partial z}dz'$$ and the other two are a consequence of the Fundamental Theorem of Calculus:

$$\frac{\partial J}{\partial v} =f(z,v(z)),\qquad\frac{\partial J}{\partial u} =-f(z,u(z)).$$

In your case you want do find $$\frac{d }{d z}\int_{-z_{0}}^{z}g(z,z')dz',$$ where $$g(z,z^{\prime })=(z-z^{\prime })f(z^{\prime }).$$ Since the lower limit of integration is constant, we have two terms only

$$\frac{d }{d z}\int_{-z_{0}}^{z}g(z,z^{\prime })dz'=\int_{-z_{0}}^{z}\frac{\partial g(z,z^{\prime })}{\partial z}dz^{\prime }+g(z,z)=\int_{-z_{0}}^{z}f(z')dz^{\prime },$$

because $g(z,z)=0$ and $\frac{\partial g(z,z^{\prime })}{\partial z}=f(z').$

share|cite|improve this answer

Introduce a second variable $y$ and set $g(x,y) = \int_{z_o}^{x}(y-z')f(z')dz'$. Calculate $dg$, $$ dg = \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial y}dy $$ Then set $x=y=z$ and find $dg/dz$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.