# Changing derivative inside an integral

1. Apr 30, 2013

### Denver Dang

1. The problem statement, all variables and given/known data
Hello.

My teacher did something on the blackboard today that I really didn't figure out how was done.

He took this equation:
$${{\mathcal{L}}^{\,\left( \alpha \right)}}=\frac{2{{e}^{2}}\tau }{m}\int_{-\infty }^{\infty }{g\left( \varepsilon \right)\varepsilon {{\left( \varepsilon -\mu \right)}^{\alpha }}\left( -\frac{\partial f}{\partial \varepsilon } \right)d\varepsilon }$$
And then made a change into:
$${{\mathcal{L}}^{\,\left( \alpha \right)}}=\frac{2{{e}^{2}}\tau }{m}\int_{-\infty }^{\infty }{\overbrace{\frac{\partial }{\partial \varepsilon }\left( g\left( \varepsilon \right)\varepsilon {{\left( \varepsilon -\mu \right)}^{\alpha }} \right)}^{H\left( \varepsilon \right)}f\left( \varepsilon \right)d\varepsilon }$$
So he was able to use it in the Sommerfeld expansion.

But I don't know how he was able to change from the derivative of f (The fermi function) to that not being derived but the other expression.

So I was kinda hoping someone could help me figure this out :)

Regards

2. Apr 30, 2013

### Office_Shredder

Staff Emeritus
He just did integration by parts, and the boundary terms were zero

3. Apr 30, 2013

### Denver Dang

That seems fair :)

Thank you very much.