Verified answer

You can find a discontinous one by setting

g = a on [0,b] and g = c on [d,e] with

1/2 = ab + c(e-d) = ab^3/3 + c(e^3-d^3)/3.

Given b, d, e, you'll have a and c > 0 if all the Cramer determinants are >=0 namely

(e^3 - d^3)*b > b^3*(e-d), e^3 - d^3 > 3 (e - d) and 3b > b^3.

With b < 1 < d < e, those conditions are automatically satisfied.

Now int(x*y,0≤x<∞) = ab^2/2 + c(e^2 - d^2)/2

<= (b/2)(ab) + (3/2)*c(e^3-d^3)/3*[(e^2 - d^2)/(e^3 - d^3)]

Notice that ab < 1/2, c(e^3-d^3)/3 < 1/2 and

[(e^2 - d^2)/(e^3 - d^3)] = (e+d)/(e^2 + ed + d^2) <= 2/e.

So int(x*y,0≤x<∞) <= b/4 + 3/2e.

You can make this as small as you want.

You can adapt this to have f continuous, replacing characteristic functions of intervals by smooth versions; The only thing you need is that your function lives only when x is small or big, but not in between.

Interesting question. Let's assume y is of the form Ce^kx, with k<0. Then:

Integral(C e^kx dx) = C/k e^kx | 0 to infinity = 0 - C/k

Integral(Cx^2 e^kx dx = C (1/k x^2 - 2/k^2 x + 2/k^3) e^kx | 0 to Inf = 0 - 2C/k^3

So we have -C/k = 1/2, C = -k/2, and

-2C/k^3 = 1/2

1/k^2 = 1/2, k = -sqrt(2)

C = sqrt(2)/2

This gives only one possible value for the function:

Integral (x Ce^kx) = C((1/k x - 1/k^2) e^kx | x=0 to Inf

= 0 - (-C/k^2) = C/k^2 = sqrt(2)/4

But there are many other possibilities - you could try y = P(x) e^kx...

I'll continue with this later, a very interesting problem.

Ed: The answer below (or above, after this change) doesn't set the Integral of (x^2y) = 1/2 too. Here's what you have to do:

This problem requires the use of something called the "Calculus of variations". Basically, you create a function of x, y, y', and y'', and minimize that function. I've never completely understood how to do it, but I'll give it a shot here. Also, here's a link to a page on the calculus of variations:

http://en.wikipedia.org/wiki/Calculus_of_variation...

So - we need L(x,y,y') on 0<=x<=Infinity.

Ed2: Ok - using the Euler-Lagrange equation, we can write the following:

dL/dy - d/dx (dL/d(y')) = 0

Where the derivatives of L are PARTIAL derivatives.

The function w're trying to optimize is:

l(x) = Integral (xy) dx from 0 to infinity, so

L(x,y,y') = xy, and plugging into the Euler-Lagrange equation:

x - d/dx (dL/dy') = 0

x = d/dx (dL/dy')

1/2 x^2 = dL/dy'

L = 1/2 x^2 dy/dx = xy

dy/dx = 2 y/x

dy/y = 2 dx/x

ln |y| = 2 ln x + C = ln x^2 + C

y = k e^(-x^2) or k e^(x^2) - the first one works.

So the form of the answer must be y = ke^(-x^2)

Now, we need to include the boundary conditions created by the integral of x^2 y.

Integral (ke^(-x^2)) dx = k sqrt(Pi)/2 = 1/2

k = sqrt(Pi)/Pi

Integral (kx e^(-x^2) dx) = Integral (k/2 e^-u) du = -k/2 e^(-x^2) = k/2 = sqrt(pi) / 2Pi

{ Note that this is smaller than the solution sqrt(2)/4 above. }

Integral (kx^2 e^(-x^2) dx) = k Sqrt(Pi) / 4 = 1/4... we have to adjust something here. But this is very close - I think I'm off by a factor of sqrt(2) somewhere.

I'll double check this later - it's a really neat problem!

Ed4: Ah - I got it! It wasn't k e^-x^2, it's k e^(-ax^2). It turns out that a has to be 1/2, which makes the integrals equal sqrt(Pi/2). So k = 1/(sqrt(2Pi))

That makes the answer to the function:

y = [sqrt(2Pi) / 2Pi] * e^(-1/2 x^2),

and the minimum value for the integral of xy from 0 to infinity will be sqrt(2Pi) / 2Pi.

Something's still not quite right here... I'll have to come back to this. The solution for y = Ce^-kx is smaller...

Ed5: Now I think it's of the form y = c e^-k sqrt(x). I must've gotten something backwards above. The answer for this form is lower than the others above... is it possible that the real solution approaches 0? I'm not sure anymore... this is tough stuff. The Lagrange function might look like y=0, with a boundary condition of y(0) = 1. This would look like an exponential function with a high degree, but the function above should show it. Not sure why I got an exponential - working...