Find $k$ such that $\int_0^{\infty} ky^3 e^{\frac{-y}{2}}dy = 1.$

Question

I'm trying to solve for $k$ given that the integral $$\int_0^{\infty} ky^3 e^{\frac{-y}{2}}dy = 1.$$ I can see that I can pull out k to get $$k \int_0^{\infty} y^3 e^{\frac{-y}{2}}dy = 1.$$ However, I'm having some issues trying to figure out where to go from here. Each time I try to do something using integration by parts or u-substitution, I keep running into equations that don't neatly integrate. For example, taking $y^3 = u$ leads to some severe issues trying to integrate around $u.$ Is there some trick to this that I'm not getting? Is it possible that this integral diverges? The infinity bound on the integral is making me worried that it could be so.

Answer 1

More generally,

$\begin{array}\\ I(a, b) &=\int_0^{\infty} x^a e^{-bx}dx\\ &=\int_0^{\infty} (y/b)^a e^{-y}\frac{dy}{b} \qquad\text{letting } y=bx\\ &=\frac1{b^{a+1}}\int_0^{\infty} y^a e^{-y}dy\\ &=\frac1{b^{a+1}}\Gamma(a+1) \qquad\text{(Gamma function)}\\ &=\frac{a!}{b^{a+1}} \qquad\text{if }a\text{ is an integer}\\ \end{array} $

The OP's integral is $kI(3, \frac12) =k\frac{3!}{(1/2)^4} =6\cdot16k =96k $. To make this $1$, $k = \frac1{96}$.

Answer 2

Integration-by-parts three times is a sure-fire recipe for success. We have:

$$\int y^{3}e^{-\frac{y}{2}}dy=-2y^{3}e^{-\frac{y}{2}}+\int 6y^{2}e^{-\frac{y}{2}}dy=-2y^{3}e^{-\frac{y}{2}}+6\left(-2y^{2}e^{-\frac{y}{2}}+4\int ye^{-\frac{y}{2}}dy \right),$$

so $$\int y^{3}e^{-\frac{y}{2}}dy=-2y^{2}e^{-\frac{y}{2}}\left(y+6 \right)+24\left( -2ye^{-\frac{y}{2}}+2\int e^{-\frac{y}{2}}dy\right)=e^{-\frac{y}{2}}\left(-2y^{3}-12y^{2}-48y-96 \right)$$ Therefore, $$k\int_{0}^{\infty}y^{3}e^{-\frac{y}{2}}dy=k\left(0-(-96)\right)=96k$$ Therefore, $k=\frac{1}{96}$ is necessary to ensure the integral under consideration equals $1$.