2
$\begingroup$

Fundamential theorem of calculus says $\frac{d}{dx} \int_a^x f(t) dt = f(x).$

I was wondering if $\frac{d}{dx} \int_a^x f(x,t) dt = f(x,x)?$

Is this mentioned in Baby Rudin book or somewhere else?

Thanks and regards!

  • 0
    calculate $\int_a^x f(x,t) dt = \int_a^x x dt = x (x-a)$. Then you see that $\partial_x \int_a^x f(x,t) dt = \partial_x x (x-a) = 2x -a \neq f(x,x) = x.$2011-04-11

1 Answers 1

6

See the wikipedia page on Differentiation under the integral sign - we have that $\frac{d}{dx} \int_a^x f(x,t) dt = f(x,x)\cdot1-f(x,a)\cdot 0 + \int_a^x\frac{\partial}{\partial x} f(x,t)dt=f(x,x) + \int_a^x\frac{\partial}{\partial x} f(x,t)dt$

so that $\frac{d}{dx} \int_a^x f(x,t) dt = f(x,x)$ only holds for all $x$ if and only if $\frac{\partial}{\partial x} f(x,t)=0$ a.e. (I think?), which (metaphorically) means $f$ was really only a function of $t$ all along.