How Do You Differentiate an Integral with Variable Limits?

[p75213]

Homework Statement

F is an antiderivative of f, so F’=f.
$$\begin{array}{l}<br /> \int_{g(x)}^{h(x)} {f(t)\,dt} = F\left( {h\left( x \right)} \right) - F\left( {g\left( x \right)} \right) \\ <br /> \frac{d}{{dx}}\int_{g(x)}^{h(x)} {f(t)\,dt} = F'\left( {h\left( x \right)} \right)h'\left( x \right) - F'\left( {g\left( x \right)} \right)g'\left( x \right) \\ <br /> \end{array}$$

Homework Equations

Can somebody show how to find the derivative of the following integral?

$$\frac{d}{{ds}}\int_0^\infty {f\left( t \right){e^{ - st}}dt}$$

The Attempt at a Solution

 

[vela]

Just reverse the order of integration and differentiation. The fundamental theorem doesn't really apply here.

 

[Harrisonized]

Can somebody show how to find the derivative of the following integral?

$$\frac{d}{{ds}}\int_0^\infty {f\left( t \right){e^{ - st}}dt}$$

Use integration by parts. Then d/ds every term. Then write the solution as a sum.

 

[Villyer]

I would think the answer is 0, since the definite integral is just a number, and the derivative of a number if 0.

 

[Harrisonized]

I would think the answer is 0, since the definite integral is just a number, and the derivative of a number if 0.

This isn't true. The definite integral for multivariable functions can be a function of a different variable, and if you actually tried the problem, you'd see that the definite integral becomes a function of s.

 

[Dick]

The limits of your integral don't depend on s. You can ignore them. Just differentiate inside the integral. Look at http://en.wikipedia.org/wiki/Leibniz_integral_rule

 

[p75213]

Leibniz's Integral rule as shown by Dick does the trick:

$$\frac{d}{{ds}}\int_0^\infty {f\left( t \right){e^{ - st}}dt} = \int_0^\infty {\frac{\delta }{{\delta s}}f\left( t \right){e^{ - st}}dt} = \int_0^\infty {f\left( t \right)\left( { - t{e^{ - st}}} \right)dt} = \int_0^\infty { - tf\left( t \right){e^{ - st}}dt}$$