Interesting integral property

[CPL.Luke]

so I'm having trouble with an integral. However I noticed an interesting property to it.int{gdu}=c*int{gd^2u}

with the integral being from -infinity to positive infinity.

The problem arose while I was trying to solve the schroedinger equation in the presence of a potential, I'm trying to solve it through the use of a Fourier transform, but in order to do that I need to know the Fourier transform of the potential.

can anybody lend a hand?

 

[HallsofIvy]

I don't understand what you mean by "d^2 u" inside the integral.

 

[CPL.Luke]

I found the property from integration by parts, the d^2u represents the differential of du.techincally the c came out of the g function when I integrated it, however I don't know whether or not it can be used to develop other properties of the integral.

if it helps at all g is equal to e^ikx and u is an unknown function which is the product of some known potential, and an unknown function psi.

 

[Hurkyl]

The differential of du is zero. Did you mean something like u''(x) dx?

 

[CPL.Luke]

yes exactly

 

[HallsofIvy]

I'd be interested in knowing HOW you got that.

You are saying
\int_{-\infty}^\infty g(x)u'(x)dx= c \int_{-\infty}^\infty g(x) u"(x)dx
for any g(x) and u(x)? I doubt that is true. In particular, if you take u= x, that turns into
\int_{-\infty}^\infty g(x)dx= 0

 

[CPL.Luke]

no, as I said it was a property of a specific integral I was doing, where u is an unknown function, and g is equal to e^ikx, however running it through integration by parts I was able to deduce that that property exhisted, (because my function u goes to zero at -infinity and positive infinity, which is known because of other known properies of u)

the integral came up as I was experimenting with Fourier transforms of the schroedinger equation, I'm hoping that there may be some trick that would simplify the problem.