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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.

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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
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The subject is well covered in this Wikipedia entry. –  André Nicolas Aug 13 '12 at 23:12
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2 Answers

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

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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 $dg/dz$.

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