# Squaring an Integral of a function

1. Nov 23, 2014

### occh

In the proof of the closed form of the Gaussian Integral the expression$${(\int_{-\infty}^\infty f(x)dx)}^2= \int_0^\infty f(x)dx+\int_0^\infty f(y)dy$$ appears. I have seen it multiple places and understand this step is justified, but I cannot find a theorem for it anywhere. Can anybody explain the reasoning behind this identity?

2. Nov 23, 2014

### Orodruin

Staff Emeritus
It is not true the way you have written it down. Did you mean to write the following?
$$\left(\int_{-\infty}^\infty f(x) dx \right)^2 = \left(\int_{-\infty}^\infty f(x) dx \right)\cdot \left(\int_{-\infty}^\infty f(y) dy \right)$$

3. Nov 23, 2014

### occh

Yeah definitely multiplied, but I'm nearly positive that the RHS bottom limits are 0.
Regardless, my question still holds, where does this identity come from (in whatever form it actually has)

4. Nov 23, 2014

### Orodruin

Staff Emeritus
Well, the integral is just a number:
$$I = \int_{-\infty}^\infty f(x) dx$$
which you can square, which results in a multiplication $I\cdot I$. The variable $x$ is just a dummy variable which is being integrated over and you can call it $y$ in the second term. If you call it $x$ in the second term you may run into trouble because you just named two different dummy variables the same thing, if you name it $y$, everything in the first integral is constant wrt $y$ and you can move it inside the second integral without problem.

5. Nov 23, 2014

### occh

Great, thanks!

6. Nov 23, 2014

### mathman

If the integrand is symmetric (Gaussian is), then the integral starting at 0 is 1/2 the total.

7. Nov 23, 2014

### Orodruin

Staff Emeritus
However, it should be noted that the standard computation of the Gaussian integral uses all of $\mathbb R^2$ as the integration domain and does the evaluation in polar coordinates in the $xy$-plane. This results in a range for the radial coordinate $r$ which is 0 to $\infty$, which I suspect is the source of the confusion in this case.