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

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.

• Just integrate by parts 3 times starting with $u=y^{3}$, $v' = \exp \bigg(\frac{-y}{2}\bigg)$. – Mattos Nov 20 '15 at 2:37
• I think you need to use the udv=uv-vdu method. – cr001 Nov 20 '15 at 2:37
• Also say hello to the Gamma function – stochasticboy321 Nov 20 '15 at 2:38
• You can get a hint from Wolfram Alpha: enter your integral, click "step by step" and you get the first step of solution. No subscription needed to see the first step. – user147263 Nov 20 '15 at 2:38

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 $1$, $k = \frac1{96}$.

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 $1$.

• I get a different answer. – marty cohen Nov 20 '15 at 3:53
• Aha! I see your error. All those other terms, besides the 96, are 0 at 0 and $\infty$. – marty cohen Nov 20 '15 at 3:55
• Good spot, @martycohen – Sinister Cutlass Nov 20 '15 at 15:43