# Second derivative of an integral

1. Dec 4, 2011

### Pietair

Good day,

I don't understand the following:

$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi''(t)$$

All I know is:

$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\frac{d^{2}}{dt^{2}}\int_{0}^{t}t \cdot \phi (\epsilon)d\epsilon-\frac{d^{2}}{dt^{2}}\int_{0}^{t}\epsilon \cdot \phi (\epsilon)d\epsilon$$

Is it allowed to say:

$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}t \cdot \phi (\epsilon)d\epsilon=\frac{d^{2}}{dt^{2}}\cdot t\int_{0}^{t}\phi (\epsilon)d\epsilon$$?

And if so, why is this correct? This is only correct when $$t\neq f(\epsilon )$$, right? But I am not sure whether this is the case...

2. Dec 4, 2011

### Mute

The result (if correct; I haven't checked) will follow from the Liebniz rule:

$$\frac{d}{dt}\int_{\alpha(t)}^{\beta(t)} du~f(u,t) = \int_{\alpha(t)}^{\beta(t)}du~ \frac{\partial}{\partial t}f(u,t) + f(t,t)\frac{d\beta}{dt} - f(t,t)\frac{d\alpha}{dt}$$

Then you have to do the derivative again, so you would have to use the Liebniz rule on the integral term again.

3. Dec 5, 2011

### jackmell

I think it's

$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi(t)$$