Talk:Differentiation under the integral sign

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Consider the function defined by f(x,t) = 0 if t = 0 and by

f(x,t) = \frac{1}{\sqrt{t}} sin \frac{x}{\sqrt{t}}

if 0< t \le 1. This is integrable with respect to t for t \in [0,1] so the function

F(x) = \int_0^1 f(x,t) dt

is at least defined. But \frac{\partial f}{\partial x}(x,t) = \frac{1}{t} cos \frac{x}{\sqrt{t}} isn't integrable, so one of the integrals appearing in the result is undefined. And in fact F is not differentiable.


So at the least we must add the hypothesis that F be differentiable. But I doubt if this will be enough to make the result true. If we allow t to vary over an infinite inteval then there is the following counter example. Let

f(x,t) = x3exp( − x2t)

for all x and for t \in [0,\infty). Then

F(x) = \int_0^\infty f(x,t) dt = x,

so F is differentiable, with F'(0) = 1. But

\int_0^\infty \frac{\partial f}{\partial x}(0,t) dt = \int_0^\infty 0 dt = 0

so the derivative of F at 0 is not given by differentiating under the integral sign.

88.105.188.214 05:23, 1 December 2006 (UTC)
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