# Integration by parts of derivative of expectation value problem

## Homework Statement

I don't know how the writer of the book took integral of the first statement and got the second statement? Can anybody clarify on this?

## Homework Equations

Given in the photo

## The Attempt at a Solution

When I took the integral I just ended up with the exact same statement but without the negative sign behind the (ihbar/2m)

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tiny-tim
Homework Helper
Welcome to PF!

Hi Normalization! Welcome to PF!
When I took the integral I just ended up with the exact same statement but without the negative sign …
But that negative sign is always there in integration by parts.

(see the last equation in your photo)

Oh w8 (facepalm) they haven't actually integrated yet? Wow... I feel so stupid right now. I thought they already integrated both products inside the brackets in which case -$\int$$\frac{\partial\Psi^*}{\partial\ x}$$\Psi$dx-$\int$$\frac{\partial\Psi}{\partial\ x}$×-$\Psi^*$dx Which would be the exact same thing. Alright thanks I guess...

tiny-tim
Homework Helper
I'm not convinced you've got this.

They have integrated that big bracket, and so they also differentiate the x (to get 1) …

the […] term is 0 (so they haven't written it), and that only leaves the ∫ term, which always has a minus in front of it.

I'm not convinced you've got this.

They have integrated that big bracket, and so they also differentiate the x (to get 1) …

the […] term is 0 (so they haven't written it), and that only leaves the ∫ term, which always has a minus in front of it.
Yes,yes,yes I know but I thought they had already taken the integral of -int(df/dx g dx)

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Which by nature should be equal to the left side of the equation. So it's fine I was just being daft