# Interesting integral property

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

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

yes exactly

HallsofIvy
Homework Helper
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
$$\int_{-\infty}^\infty g(x)dx= 0$$

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.