1. ## Integrate PDF

Random variable $\displaystyle (X,Y)$ has PDF $\displaystyle f(x,y)=\frac{C}{\pi^2(x^2+20)(y^2+45)}$. What is $\displaystyle C$?
I set up the equation, but don't know the integration part.
$\displaystyle \displaystyle{ \int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y) dy dx = 1 = \frac{1}{\pi^2}\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{C}{(x^2+20)(y^2+45)} dy dx }$

$\displaystyle C$ should be 30.

2. Originally Posted by courteous
I set up the equation, but don't know the integration part.
$\displaystyle \displaystyle{ \int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y) dy dx = 1 = \frac{1}{\pi^2}\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{C}{(x^2+20)(y^2+45)} dy dx }$

$\displaystyle C$ should be 30.
The double integral is seperable using a simple corrollary to Fubini's stronger theorem:

$\displaystyle \displaystyle \int_{-\infty}^\infty \frac{1}{x^2+20} \, dx \int_{-\infty}^\infty \frac{1}{y^2+45} \, dy$.

Each integral involves the inverse tan function.

3. $\displaystyle \displaystyle{ \int_{-\infty}^{\infty}\frac{1}{y^2+45}dy\overbrace{=}^{s ub} \int_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}\frac{\sqrt{4 5}\frac{1}{cos^2(\theta)}}{45(\tan^2(\theta)+1)}d\ theta = \frac{1}{\sqrt{45}}\left[\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{\sqrt{45 }} }$

Similar result for $\displaystyle x$: $\displaystyle \displaystyle{ \frac{\pi}{\sqrt{20}}$.

Which finally gives $\displaystyle \displaystyle{ 1=\frac{C}{\pi^2}\frac{\pi}{\sqrt{20}}\frac{\pi}{\ sqrt{45}}=\frac{C}{30} \Rightarrow C=30 }$ A piece of ... I'm so ignorant.

Why does it work? Wikipedia doesn't say much (to me) under strong versions.
I'm really asking whether there is a natural learning path that step-by-step makes the whole picture clear (seeing forest instead of trees)?

Let's take calculus. Would this be anywhere near a step-by-step "learning signpost"?
1. Calculus (C)
2. Multivariable C
3. Vector C
4. Analytic geometry
5. Topology
6. ?

Do these various theorems ever start "falling in place"?