Differentiate under integral sign for iterated integral? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:27:32Z http://mathoverflow.net/feeds/question/20092 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20092/differentiate-under-integral-sign-for-iterated-integral Differentiate under integral sign for iterated integral? Mike 2010-04-01T18:20:10Z 2010-04-01T18:51:38Z <p>This is a bit of a trivial question, but as I don't know the answer immediately I thought I'd just ask.</p> <p>Given the integral $\int_{0}^{t} \int_{0}^{t} f(x,x') dx dx'$, what is $\frac{\partial}{\partial t} \int_{0}^{t} \int_{0}^{t} f(x,x') dx dx'$? It looks a bit like differentiating under the integral sign, but I'm not sure how to handle it.</p> http://mathoverflow.net/questions/20092/differentiate-under-integral-sign-for-iterated-integral/20097#20097 Answer by Gerald Edgar for Differentiate under integral sign for iterated integral? Gerald Edgar 2010-04-01T18:51:38Z 2010-04-01T18:51:38Z <p>As usual when differentiating something with respect to a variable that appears twice. The chain rule for partial derivatives.</p> <p>For example, consider function $z = f(u,v)$. Suppose we want $(d/dt)f(t,t)$. Let $u=v=t$ and use $dz/dt = (\partial z/\partial u)(du/dt) + (\partial z/\partial v)(dv/dt)$.</p> <p>Thus... $$\frac{d}{dt}\int_0^t\int_0^t f(x,y)\,dx\,dy = \int_0^t f(t,y)\,dy + \int_0^t f(x,t)\,dx$$</p> <p>By the way, why did you write $\partial/\partial t$ to differentiate a function of the single variable $t$? It's not wrong, just confusing to students.</p>