# Finding “c” of this joint distribution

A textbook example asks me to solve for $c$ in the joint distribution function;

$$f(x,y) = c(y^2-x^2)e^{-y} \ \ \ -y \le x \le y, \ \ 0 < y <\infty$$

The answer given involves integrating the function from $-y$ to $y$ with respect to $x$, and getting $\frac{4}{3}cy^3e^{-y}$. Then, integrate this new function with respect to $y$ from $0$ to $\infty$.

The textbook gives the following steps: $$\frac{4}{3}c\int_0^\infty y^3e^{-y}dy = 4c\int_0^\infty y^2e^{-y}dy = 8c\int_0^\infty ye^{-y}dy = 8c = 1$$

My main issue is - how did they do these steps? How is one equal to the next?

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Integration by parts. Thrice. – Did Oct 18 '12 at 17:13
I see it now. I've never seen it done that way before. I prefer using the tabular method. Thanks! – CodyBugstein Oct 18 '12 at 17:33
The "tabular method" is nothing more than a bookeeping device for integration by parts. – David Mitra Oct 18 '12 at 17:34
It doesn't have to be integration by parts. Let $J_k = \int_0^\infty y^k e^{-y}\ dy$. Using the change of variables $y = at$ for $a > 0$, $a^{-k-1} J_k = \int_0^\infty t^k e^{-at}\ dt$. Take the derivative with respect to $a$: $$-(k+1) a^{-k-2} J_k = - \int_0^\infty t^{k+1} e^{-at}\ dt = - a^{-k-2} J_{k+1}$$ so $J_{k+1} = (k+1) J_k$. – Robert Israel Oct 18 '12 at 18:00
...or just change variables to $u=(x+y)$, $v=(y-x)$, giving a separable integrand that immediately produces $1/c$ = $(\int_0^\infty u\exp(-u/2)du)^2 / 2$. – whuber Oct 18 '12 at 20:19