Second derivative of an integral

  • Thread starter Pietair
  • Start date
59
0
Good day,

I don't understand the following:

[tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi''(t)[/tex]

All I know is:

[tex]\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[/tex]

Is it allowed to say:

[tex]\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[/tex]?

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

Thank you in advance!
 

Mute

Homework Helper
1,381
10
The result (if correct; I haven't checked) will follow from the Liebniz rule:

[tex]\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}[/tex]

Then you have to do the derivative again, so you would have to use the Liebniz rule on the integral term again.
 
1,786
53
Good day,

I don't understand the following:

[tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi''(t)[/tex]
I think it's

[tex]\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi(t)[/tex]
 

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