# I Verifying derivative of multivariable integral equation

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1. Sep 7, 2016

### transmini

I had posted a question earlier which this is related to, but a different equation.
$$\frac{d}{dt} \int_0^t H(t,s)ds = H(t,t) + \int_0^t \frac{\partial H}{\partial t}(t,s)ds$$

This was another formula needed in a proof however I don't see how this one holds either. I tried following a proof of the formula from http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf (bottom of page 13) but it seemed like it contradicted itself by passing the partial derivative through the integral even though the limits aren't independent of the variable. That and replacing dI/da(t) with dI/da even though a in the second operation is just a constant.

Could someone explain how to get from the LHS of the equation to the right? Thanks in advance.

2. Sep 7, 2016

### Krylov

This follows directly from Leibniz' rule.

3. Sep 13, 2016

### Ssnow

You must see $\int_{a(t)}^{b(t)}f(t,s)ds=I(t,a(t),b(t))$, so $\int_{0}^{t}H(t,s)ds= I(t,a(t),b(t))$ where $a(t)=0$ and $b(t)=t$, the partial derivative is

$\frac{\partial}{\partial t}I(t,a(t),b(t))=\frac{\partial}{\partial t}I(t,a(t),b(t))\frac{d t}{d t}+\frac{\partial}{\partial a}I(t,a(t),b(t))\frac{da(t)}{dt}+\frac{\partial}{\partial b}I(t,a(t),b(t))\frac{d b(t)}{d t}$

where we used the derivative of the composition... , by the fact that $a(t)=0$ this is

$=\int_{0}^{t}\frac{\partial}{\partial t}H(t,s)ds\cdot \frac{dt}{dt}- 0 +H(t,t)\cdot \frac{dt}{dt},$

now $\frac{dt}{dt}=1$ so your formula.