Can you simplify this integral or do you need more background knowledge?

[SemM]

Hi, I have the following integral which I am not confident on how to interpret (solve):\begin{equation}
\alpha \bigg( \int_0^L [\frac{d^3}{dx^3}\phi] \psi dx - \int_0^L [\frac{d^3}{dx^3}\psi] \phi dx \bigg)
\end{equation}

at this stage, I am not sure which rule to use to solve each of the two integrals in the brackets, or simplify them.

Is this possible to be simplified?

Thanks!

 

[mfb]

What are ϕ and ψ? Without further knowledge about them it is hard to say anything. If you can calculate the derivatives explicitly, just do so. Partial integration might help, but it depends on your functions.
It is a bit strange to see x appear both as integration variable and in the derivative.

 

[Buffu]

Meh ... my try :

$$\int \frac{d^3\phi}{dx^3} \psi dx = \psi \dfrac{d^2 \phi}{ dx^2} - \int \dfrac{d\psi}{dx} \dfrac{d^2 \phi}{ dx^2} dx = \psi \dfrac{d^2 \phi}{ dx^2} - \left( \dfrac{d\psi}{dx} \dfrac{d \phi}{ dx}-\int \dfrac{d^2\psi}{dx^2} \dfrac{d \phi}{ dx} dx\right) \\= \psi \dfrac{d^2 \phi}{ dx^2} - \left( \dfrac{d\psi}{dx} \dfrac{d \phi}{ dx} - \left( \dfrac{d^2\psi}{dx^2} \phi - \int \dfrac{d^3\psi}{dx^3} \phi dx \right)\right) $$

 

[SemM]

Hi mfb, these are general functions. I can perform the calculation numerically, however, I wanted to delineate a general principle, and thus simplify these expressions as much as possible.

There is a paper with an analogous derivation:

https://arxiv.org/pdf/quant-ph/0103153.pdf

on p 5, eqn, 12 derives this integral for the derivative d/dx on two functions, $\psi and \hat{\psi}$ . The next last step in that formulation is unclear, and it may be the solution to this question, but I am not sure.

 

[Buffu]

Hi mfb, these are general functions. I can perform the calculation numerically, however, I wanted to delineate a general principle, and thus simplify these expressions as much as possible.

There is a paper with an analogous derivation:

https://arxiv.org/pdf/quant-ph/0103153.pdf

on p 5, eqn, 12 derives this integral for the derivative d/dx on two functions, $\psi and \hat{\psi}$ . The next last step in that formulation is unclear, and it may be the solution to this question, but I am not sure.

Unless I made a sign error I think the best you can do is

$$
\int \frac{d^3\phi}{dx^3} \psi dx + \int \dfrac{d^3\psi}{dx^3} \phi dx = \psi \dfrac{d^2 \phi}{ dx^2} - \dfrac{d\psi}{dx} \dfrac{d \phi}{ dx} + \dfrac{d^2\psi}{dx^2} \phi
$$

 

[eys_physics]

In the mentioned paper, they use partial integrai

Hi mfb, these are general functions. I can perform the calculation numerically, however, I wanted to delineate a general principle, and thus simplify these expressions as much as possible.

There is a paper with an analogous derivation:

https://arxiv.org/pdf/quant-ph/0103153.pdf

on p 5, eqn, 12 derives this integral for the derivative d/dx on two functions, $\psi and \hat{\psi}$ . The next last step in that formulation is unclear, and it may be the solution to this question, but I am not sure.

They use partial integration and that $\psi(0)=\psi(L)=0$.

 

[SemM]

Unless I made a sign error I think the best you can do is

$$
\int \frac{d^3\phi}{dx^3} \psi dx + \int \dfrac{d^3\psi}{dx^3} \phi dx = \psi \dfrac{d^2 \phi}{ dx^2} - \dfrac{d\psi}{dx} \dfrac{d \phi}{ dx} + \dfrac{d^2\psi}{dx^2} \phi
$$

Thanks I will follow this, with a sign correction on $$
\int \frac{d^3\phi}{dx^3} \psi dx - \int \dfrac{d^3\psi}{dx^3} \phi dx = \psi \dfrac{d^2 \phi}{ dx^2} - \dfrac{d\psi}{dx} \dfrac{d \phi}{ dx} - \dfrac{d^2\psi}{dx^2} \phi
$$
?

 

[SemM]

In the mentioned paper, they use partial integraiThey use partial integration and that $\psi(0)=\psi(L)=0$.

In that link posted above, they write:

\begin{equation}
-i\hbar \int_0^L \frac{d}{dx}[\overline{\psi}(x)\phi(x)]dx = -i\hbar[\overline{\psi}(L)\phi(L)-\overline{\psi}(0)\phi(0)]
\end{equation}

This part $[\overline{\psi}(x)\phi(x)]$ seems odd with those brackets. What does it mean?

 

[SemM]

Unless I made a sign error I think the best you can do is

$$
\int \frac{d^3\phi}{dx^3} \psi dx + \int \dfrac{d^3\psi}{dx^3} \phi dx = \psi \dfrac{d^2 \phi}{ dx^2} - \dfrac{d\psi}{dx} \dfrac{d \phi}{ dx} + \dfrac{d^2\psi}{dx^2} \phi
$$

Buffu, it seems you use a symmetric relation between the operations on $\phi$ and $\psi$, in the terms:

$\psi \dfrac{d^2 \phi}{ dx^2}$

and
$\dfrac{d^2\psi}{dx^2} \phi$Was this intentional, and thus avoiding

$\dfrac{d^2 \phi}{ dx^2} \psi$

$\dfrac{d^2\psi}{dx^2} \phi$

?
Which rule did you use to derive this? Sorry, I have not seen this form before. Any literature reference or link is welcome! (Not because I don't trust this, but because I'd like to learn how to derive this)

 

[SemM]

In the mentioned paper, they use partial integraiThey use partial integration and that $\psi(0)=\psi(L)=0$.

It appears they set $\int_0^L dx d/dx[\overline {\psi}\phi] = [\overline {\psi}(L)\phi(L)-[\overline {\psi}(0)\phi(0)]$
but nothing indicates that they actually solved the integral, it appears the d/dx sign just vanished, and the difference between the L and 0 remains to be expressed.

What form of integration is this?

 

[eys_physics]

It appears they set $\int_0^L dx d/dx[\overline {\psi}\phi] = [\overline {\psi}(L)\phi(L)-[\overline {\psi}(0)\phi(0)]$
but nothing indicates that they actually solved the integral, it appears the d/dx sign just vanished, and the difference between the L and 0 remains to be expressed.

What form of integration is this?

I apologize for a mistake in my previous post. You don't need the partial integration. You only need the fundamental theorem of calculus (https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus), i.e.
$\int_a^b f(x) dx=F(b)-F(b),$
where $F(x)$ is the antiderivative of $f(x)$, and that $[\bar{\psi}\phi]$ is the antiderivative of $d/dx[\bar{\psi}\phi]$.

 

[Mark44]

I thought integration by parts was: $\int \frac{d \psi}{dx} \phi dx = \int \frac{d\phi}{dx}\psi dx+\int \frac{d \psi}{dx}{\phi}dx$

That's not integration by parts -- I can't tell what you're doing there.
This is integration by parts:
$\int u~ dv = uv - \int v~ du$

For your integral I get $\int \frac{d \psi}{dx} \phi dx = \int \phi~ d\psi = \phi \psi - \int \psi~ d \phi$

 

[SemM]

I apologize for a mistake in my previous post. You don't need the partial integration. You only need the fundamental theorem of calculus (https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus), i.e.
$\int_a^b f(x) dx=F(b)-F(b),$
where $F(x)$ is the antiderivative of $f(x)$, and that $[\bar{\psi}\phi]$ is the antiderivative of $d/dx[\bar{\psi}\phi]$.

That is fine, but in this case, F is unknown, so this remains elusive.

In the discussed case

$\int_0^L d/dx[\overline {\psi}\phi]dx = [\overline {\psi}(L)\phi(L)-[\overline {\psi}(0)\phi(0)]$

psi and phi is the same before and after integration. It appears they are symbolically calling phi and psi before and after integration as the same

 

[SemM]

That's not integration by parts -- I can't tell what you're doing there.
This is integration by parts:
$\int u~ dv = uv - \int v~ du$

For your integral I get $\int \frac{d \psi}{dx} \phi dx = \int \phi~ d\psi = \phi \psi - \int \psi~ d \phi$

I am sorry, I mixed derivation of uv with the integration. Let me rectify that in order to avoid confusion for other readers.

 

[Mark44]

I am sorry, I mixed derivation of uv with the integration. Let me rectify that in order to avoid confusion for other readers.

You also had the sign wrong on the right side. In your work it looked like you were confusing the product rule in differentiation with integration by parts

 

[SemM]

You also had the sign wrong on the right side. In your work it looked like you were confusing the product rule in differentiation with integration by parts

Yes, as I said, I did that.

Can you comment on this strange form?$\int_0^L d/dx[\overline {\psi}\phi]dx = [\overline {\psi}(L)\phi(L)-[\overline {\psi}(0)\phi(0)]$

Is it a plain imprecise formality to use F(x) = psi and f(x)=psi and the same for phi?

 

[fresh_42]

This is integration by parts: $\int u~ dv = uv - \int v~ du$

Just as an add on for casual readers on how this formula can be easily memorized:

The product rule / Leibniz rule / definition of a derivation reads:
$$ \dfrac{d}{dx}(u\cdot v) = \dfrac{d}{dx}u \cdot v + u\cdot\dfrac{d}{dx}v$$ and integrating this gives us directly the formula for integration by parts so there isn't even another formula to memorize.

 

[SemM]

Can we conclude that a general form of integration with arbitrary limits can be:

\begin{equation}
\int_0^L \frac{d \psi}{dx} \phi dx = [{\Psi}(L)\Phi(L)- {\Psi}(0)\Phi(0)]
\end{equation}This is apparently what was done in the paper quoted above.

Instead of:

\begin{equation}
\int_0^L \frac{d \psi}{dx} \phi dx = [\Psi \Phi]_0^L - \int_0^L \Psi \frac{d \Phi}{dx} dx
\end{equation}to be plain, I think the form given above (in that paper) doesn't seem written correctly. The latter is correct, although less nice.

 

[Mark44]

Yes, as I said, I did that.

Can you comment on this strange form?$\int_0^L d/dx[\overline {\psi}\phi]dx = [\overline {\psi}(L)\phi(L)-[\overline {\psi}(0)\phi(0)]$

The integral can be written more simply as $\int_0^L d[\overline {\psi}\phi]$, and is equal to the expression you show on the right.

Is it a plain imprecise formality to use F(x) = psi and f(x)=psi and the same for phi?

Not only imprecise, but it's wrong. What you're saying is that a function and its antiderivative are equal, which isn't generally true.

 

[SemM]

Not only imprecise, but it's wrong. What you're saying is that a function and its antiderivative are equal, which isn't generally true.

The paper I quote appears to do so,

 

[SemM]

The integral can be written more simply as $\int_0^L d[\overline {\psi}\phi]$, and is equal to the expression you show on the right.

Thanks, I have the impression eqn nr 4 in my prev post is more accurate.

 

[Mark44]

Can we conclude that a general form of integration with arbitrary limits can be:

\begin{equation}
\int_0^L \frac{d \psi}{dx} \phi dx = [{\Psi}(L)\Phi(L)- {\Psi}(0)\Phi(0)]
\end{equation}

No. The two terms on the right are what you get from the first expression on the right in the equation below. The above is correct only if the integral on the right side below happens to be zero.

This is apparently what was done in the paper quoted above.

Instead of:

\begin{equation}
\int_0^L \frac{d \psi}{dx} \phi dx = [\Psi \Phi]_0^L - \int_0^L \Psi \frac{d \Phi}{dx} dx
\end{equation}

to be plain, I think the form given above (in that paper) doesn't seem written correctly. The latter is correct, although less nice.

"Correct" is always to be preferred over "less nice."

 

[SemM]

No. The two terms on the right are what you get from the first expression on the right in the equation below. The above is correct only if the integral on the right side below happens to be zero.

"Correct" is always to be preferred over "less nice."

Thanks I will use the form in eqn 4 then. Cheers Mark

 

[Mark44]

@SemM, if you are having difficulty why this is wrong
$\int_0^L \frac{d \psi}{dx} \phi dx = [{\Psi}(L)\Phi(L)- {\Psi}(0)\Phi(0)]$
and this is right
$\int_0^L \frac{d \psi}{dx} \phi dx = [\Psi \Phi]_0^L - \int_0^L \Psi \frac{d \Phi}{dx} dx$, and how to carry out integration by parts, these doesn't inspire much confidence in your ability to be successful in comprehending the material you're trying to tackle. Integration by parts is a technique that is presented to freshman calculus students.

 

[SemM]

@SemM, if you are having difficulty why this is wrong
$\int_0^L \frac{d \psi}{dx} \phi dx = [{\Psi}(L)\Phi(L)- {\Psi}(0)\Phi(0)]$
and this is right
$\int_0^L \frac{d \psi}{dx} \phi dx = [\Psi \Phi]_0^L - \int_0^L \Psi \frac{d \Phi}{dx} dx$, and how to carry out integration by parts, these doesn't inspire much confidence in your ability to be successful in comprehending the material you're trying to tackle. Integration by parts is a technique that is presented to freshman calculus students.

Your communication is a blend of pedagogism and iteration of errors. I have asked a question about a different matter in the original post. Underway, I have posted a form from a paper that is not proving very accurate. If I have done so, it means that I have indeed seen an inclarity in this, so what is the problem?

Physics forum has everything to gain in being pleasant and having more humor and light-headedness in things than the inverse.

I think spotting the original deficiency with $\int_0^L \frac{d \psi}{dx} \phi dx = [{\Psi}(L)\Phi(L)- {\Psi}(0)\Phi(0)]$ and presenting $\int_0^L \frac{d \psi}{dx} \phi dx = [\Psi \Phi]_0^L - \int_0^L \Psi \frac{d \Phi}{dx} dx$ shows that calculus is fine. However, careful consideration and discussion/questions are made on the former, as it is from a publication. About mixing derivation and integration in a flash was a slip. Don't you ever slip? I do daily, and I learn much more from that than from grumpy notes. So let's be reasonable and friendly here, shall we?

Cheers

 

[Mark44]

Your communication is a blend of pedagogism and iteration of errors. I have asked a question about a different matter in the original post. Underway, I have posted a form from a paper that is not proving very accurate. If I have done so, it means that I have indeed seen an inclarity in this, so what is the problem?

The problem is that you are attempting to tackle some fairly difficult mathematics without having the requisite knowledge of elementary calculus.
In post #1 you asked a question in which there were two integrals with third derivatives. As pointed out shortly, without knowing more about the functions in the integrands, the integrals couldn't be evaluated.
In post #8 you said this (which you later deleted):

I thought integration by parts was: $\int \frac{d \psi}{dx} \phi dx = \int \frac{d\phi}{dx}\psi dx+\int \frac{d \psi}{dx}{\phi}dx$

That's not even close. First-year calculus students are expected to have competence with integration by parts.

In post #12 you showed an integral (Equation 12) from a paper you had linked to in a previous post.

$-i\hbar \int_0^L \frac{d}{dx}[\overline{\psi}(x)\phi(x)]dx = -i\hbar[\overline{\psi}(L)\phi(L)-\overline{\psi}(0)\phi(0)]$

This part $[\overline{\psi}(x)\phi(x)]$ seems odd with those brackets. What does it mean?

Evaluating that integral is simple -- the antiderivative of the derivative of something is just that something. This too is first-year calculus stuff.

I think spotting the original deficiency with $\int_0^L \frac{d \psi}{dx} \phi dx = [{\Psi}(L)\Phi(L)- {\Psi}(0)\Phi(0)]$ and presenting $\int_0^L \frac{d \psi}{dx} \phi dx = [\Psi \Phi]_0^L - \int_0^L \Psi \frac{d \Phi}{dx} dx$ shows that calculus is fine.

Where do you see this integral in the paper you linked to? $\int_0^L \frac{d \psi}{dx} \phi dx$
Although the integral just above and the one in Eqn. 12 of the paper appear similar, different techniques are needed for each one.

Equation 12, the one you asked about, includes this integral: $-i\hbar \int_0^L \frac{d}{dx}[\overline{\psi}(x)\phi(x)]dx$. Again, this is a very straightforward integral to evaluate.

However, careful consideration and discussion/questions are made on the former, as it is from a publication. About mixing derivation and integration in a flash was a slip. Don't you ever slip? I do daily, and I learn much more from that than from grumpy notes.

Sure. Like most humans, I make mistakes, but the things that I point out here aren't mere slips -- they are large gaps in your knowledge of calculus. Like I said, it appears that you are attempting to bite off more than you can chew, an opinion that is also shared by several other mentors who have participated in some of your other threads.

So let's be reasonable and friendly here, shall we?

I don't think I have been unreasonable at all. Mostly I'm trying to be realistic -- studying complicated material requires a solid understanding of the more basic foundation topics.

 

[SemM]

The problem is that you are attempting to tackle some fairly difficult mathematics without having the requisite knowledge of elementary calculus.
In post #1 you asked a question in which there were two integrals with third derivatives. As pointed out shortly, without knowing more about the functions in the integrands, the integrals couldn't be evaluated.
In post #8 you said this (which you later deleted):

That's not even close. First-year calculus students are expected to have competence with integration by parts.

In post #12 you showed an integral (Equation 12) from a paper you had linked to in a previous post.

Evaluating that integral is simple -- the antiderivative of the derivative of something is just that something. This too is first-year calculus stuff.Where do you see this integral in the paper you linked to? $\int_0^L \frac{d \psi}{dx} \phi dx$
Although the integral just above and the one in Eqn. 12 of the paper appear similar, different techniques are needed for each one.

Equation 12, the one you asked about, includes this integral: $-i\hbar \int_0^L \frac{d}{dx}[\overline{\psi}(x)\phi(x)]dx$. Again, this is a very straightforward integral to evaluate.

Sure. Like most humans, I make mistakes, but the things that I point out here aren't mere slips -- they are large gaps in your knowledge of calculus. Like I said, it appears that you are attempting to bite off more than you can chew, an opinion that is also shared by several other mentors who have participated in some of your other threads.
I don't think I have been unreasonable at all. Mostly I'm trying to be realistic -- studying complicated material requires a solid understanding of the more basic foundation topics.

I am not sure there is any point in repeating myself. Physicsforum is to ask questions, should one always expect personal inquisitions, then no real point of asking.
Lets keep this impersonal , shall we?

 

[Mark44]

Physicsforum is to ask questions, should one always expect personal inquisitions, then no real point of asking.

This isn't a "personal inquisition."
We welcome questions, but the answers we give have to be tailored to the level of knowledge of the person asking the question. In your case, the questions you're asking about QM presume a knowledge of mathematics and physics that you don't appear to have. What I'm saying is that you ought to put your current studies on a back burner, and fill in the gaps you have in mathematics and physics.

At least three mentors have commented about your questions, and will no longer take part in your threads. Here's what one of them said in our private mentor area:

I gave up on SemH. SemH is reading Kreyszig's nice, introductory book on functional analysis, but SemH does not have the background or mathematical maturity to benefit from this.

A Science Advisor said this:

"Pulled out of thin air" in this case is simply "at a level of mathematics that the reader is supposed to be familiar with".