I have the following integral where the negative exponent on natural e is making me question my approach.
Problem:
\begin{eqnarray*} \int_0^{\infty} x^3e^{-x} dx. \end{eqnarray*}
My approach so far:
I am stuck here. Any advice would be greatly appreciated. Also, a confirmation of my approach would be helpful. Am I missing any technique due to the negative exponent?
The direct way to do this is to integrate by parts 3 times. Each time reducing the exponent of $x$ by $1$
A little sneaker is to say let $I_n = \int_0^\infty x^n e^{-x}\ dx$ then by integration by parts show that $I_{n+1} = (n+1)I_n,$ therefore $I_n = n!\int_0^{\infty} e^{-x}\ dx = n!.$
If you hate integration by parts, you can use differentiation under the integral sign, aka "Feynman's trick"
let $f(s) = \int_0^\infty e^{-sx} \ dx = \frac {1}{s}$
$\frac {d}{ds} f(s) = \int_0^\infty \frac {d}{ds} e^{-sx} \ dx = \int_0^\infty -xe^{-sx} \ dx = \frac {d}{ds} \frac {1}{s} = - s^{-2}\\ \frac {d^n}{ds^n} f(s) = \int_0^\infty (-1)^nx^ne^{-sx} \ dx = (-1)^n n! s^{-(n+1)}$
Evaluate at $s= 1$
$\int_0^\infty x^ne^{-x} \ dx = n!$
And eventually you will learn about the $\Gamma$ function and the Laplace Transform...
Using the substitution $x=-\log y$ the integral $I= \int_0^\infty x^3 \; e^{-x} \, dx$ transforms to $$I=-\int_0^1 \; (\log y)^3 \; dy$$ However the most obvious route to evaluating this integral still requires integration by parts 3 times plus a lot of additional algebra
The integral of $(\log x)^3$ in terms of the integral of $(\log x)^2$ $$\int (\log x)^3 \,dx=\log x \int (\log x)^2 \,dx\,-\int \frac{1}{x} \, \int (\log x)^2 \, dx \,dx\tag{1}$$ The integral of $(\log x)^2$ in terms of the integral of $\log x$ $$\int (\log x)^2 \,dx=\log x \int \log x \,dx\,-\int \frac{1}{x} \, \int \log x \, dx \,dx\tag{2}$$ The integral of $\log x$ in terms of the integral of $1$ $$\int \log x \,dx=\log x\int\,1\,dx-\int \frac{1}{x}\int\,1\,dx\,dx=x \log x -x\tag{3}$$
The full calculation is made in this question Integral of $(\log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v) plus answers along the lines already investigated.
One more method (when you don't want to do multiple IBP): $$\int x^3e^{-x} dx=C_1x^3e^{-x}+C_2x^2e^{-x}+C_3xe^{-x}+C_4e^{-x}+C_5 \iff \\ (C_1x^3e^{-x}+C_2x^2e^{-x}+C_3xe^{-x}+C_4e^{-x}+C_5)'=x^3e^{-x} \iff \\ \color{green}{3C_1x^2e^{-x}}-\color{red}{C_1x^3e^{-x}}+\color{blue}{2C_2xe^{-x}}-\color{green}{C_2x^2e^{-x}}+\color{brown}{C_3e^{-x}}-\color{blue}{C_3xe^{-x}}-\color{brown}{C_4e^{-x}}=x^3e^{-x} \iff \\ \begin{cases}\color{red}{-C_1}=1\\ \color{green}{3C_1-C_2}=0\\ \color{blue}{2C_2-C_3}=0\\ \color{brown}{C_3-C_4}=0\end{cases} \Rightarrow C_1=-1,C_2=-3,C_3=-6, C_4=-6.\\ $$
$$I=\int_0^{\infty}x^3e^{-x}dx$$ There are two main ways we can approach this, firstly is the gamma function. It is known that: $$n!=\Gamma(n+1)=\int_0^{\infty}x^ne^{-x}dx$$ $$\therefore\,I=\Gamma(4)=3!=6$$ The second way is indeed integration by parts. To recap: $$\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$$ Firstly for our integral it may be easier to make a substitution to remove the minus sign to avoid confusion: $$I=\int_0^{\infty}x^3e^{-x}dx$$ $w=-x\,,dx=-dw$ $$\therefore\,I=\int_0^{-\infty}w^3e^wdw=\left[w^3e^w\right]_0^{-\infty}-3\int_0^{-\infty}w^2e^wdw=-3\left(\left[w^2e^w\right]_0^{-\infty}-2\int_0^{-\infty}we^wdw\right)=6\int_0^{-\infty}we^wdw=6\left(\left[we^w\right]_0^{-\infty}-\int_0^{-\infty}e^wdw\right)=-6\int_0^{-\infty}e^wdw=6\int_{-\infty}^0e^wdw=6\left[e^w\right]_{-\infty}^0=6(1-0)=6$$
$$\int x^3e^{-x}dx=-x^3e^{-x}+\int3x^2e^{-x}dx=-x^3e^{-x}-3x^2e^{-x}+6\int xe^{-x}dx=$$ $$=-x^3e^{-x}-3x^2e^{-x}-6xe^{-x}+6\int e^{-x}dx=-x^3e^{-x}-3x^2e^{-x}-6xe^{-x}-6e^{-x}+C.$$ Can you end it now?
I got the answer: $6$.
