# Thread: int exp(-1/2 x^2) dx

1. ## int exp(-1/2 x^2) dx

hi there, I'm reading a book on probability, but having trouble with following some equations. It's like this;
$\displaystyle \int_0^\infty \! \text{exp}\left( - \displaystyle \frac{1}{2}y^2 \right)\, dy \cdot\displaystyle \int _0^\infty \! \text{exp}\left( - \displaystyle \frac{1}{2}z^2 \right) \. dz \\ =\int_0^\infty\!\!\int_0^\infty \text{exp} \left\{- \frac{1}{2}(y^2+z^2) \right\}dydz$

2. $e^a \cdot e^b = e^{a+b}$

3. thanks ,but that's what i understand(I should have asked more clearly...)

what confuses me is that
$\int f(y)dy \cdot \int f(z)dz = \int \int {f(y)f(z)}dydz.$

would you help a little more? Thank you very much.

4. This is taken directly from Stewart's book

$\int_c^d \int_a^b g(x)h(y)dxdy = \int_c^d \left[\int_a^b g(x)h(y)dx\right]dy$

In the inner integral, $y$ is constant, so $h(y)$ is constant and we can write

$\int_c^d \left[h(y)\left(\int_a^b g(x)dx\right)\right]dy = \int_a^b g(x)dx \int_c^d h(y)dy$

since $\int_a^b g(x)dx$ is constant.

5. Originally Posted by joll
thanks ,but that's what i understand(I should have asked more clearly...)

what confuses me is that
$\int f(y)dy \cdot \int f(z)dz = \int \int {f(y)f(z)}dydz.$

would you help a little more? Thank you very much.
Fubini's theorem - Wikipedia, the free encyclopedia