Let y = x^n e^(-x) , where n ∈ mathbbN . 1° Show that…

Let $y = x^n e^{-x}$, where $n \in \mathbb{N}$. 1° Show that $x(y'+y) = ny$. 2° Show that $z = \dfrac{d^n}{dx^n}\bigl(x^n e^{-x}\bigr)$ satisfies $xz''+(x+1)z'+(n+1)z=0$. 3° Show that $w(x)=e^x z(x)$ is a polynomial of degree $n$ satisfying Laguerre's equation $xw''+(1-x)w'+nw=0$. Finally, evaluate $I = \displaystyle\int_0^\infty x^k\,\frac{d^n}{dx^n}\bigl(e^{-x}x^n\bigr)\,dx$ for $k\lt n$ and for $k=n$.

Step 1: 1° — The first-order identity

With $y=x^ne^{-x}$, $y' = nx^{n-1}e^{-x} - x^ne^{-x} = e^{-x}\bigl(nx^{n-1}-x^n\bigr)$.

Hence $y'+y = nx^{n-1}e^{-x}$, and multiplying by $x$: \[x(y'+y) = nx^ne^{-x} = ny. \tag{E}\]

Step 2: 2° — A recursion for the higher derivatives

Write (E) as $xy'+xy-ny=0$ and differentiate once, applying the product rule to $xy'$: \[xy''+y'+xy'+y-ny' = 0 \quad\Longrightarrow\quad xy''+(x+1-n)y'+y=0. \tag{$E_1$}\]

Now apply $D^k$ (differentiate $k$ times) to $(E_1)$. By the Leibniz rule, $D^k(x\,u)=x\,u^{(k)}+k\,u^{(k-1)}$ for any $u$; applying this to the term $xy''$ (with $u=y''$) and to $xy'$ inside $(x+1-n)y'$ (with $u=y'$) gives

\[ D^k\bigl(xy''\bigr) = xy^{(k+2)}+ky^{(k+1)}, \qquad D^k\bigl[(x+1-n)y'\bigr] = (x+1-n)y^{(k+1)}+ky^{(k)}, \qquad D^k(y) = y^{(k)}. \]

Summing these three pieces and collecting terms produces the general recursion

\[ x\,y^{(k+2)} + (x+1-n+k)\,y^{(k+1)} + (k+1)\,y^{(k)} = 0. \tag{$E_{k+1}$} \]

Setting $k=n-1$ (so that $x+1-n+k=x$): \[x\,y^{(n+1)} + x\,y^{(n)} + n\,y^{(n-1)} = 0. \tag{$E_n$}\]

Applying the recursion once more, now with $k=n$: \[x\,y^{(n+2)} + (x+1)\,y^{(n+1)} + (n+1)\,y^{(n)} = 0. \tag{$E_{n+1}$}\]

Setting $z = y^{(n)} = \dfrac{d^n}{dx^n}\bigl(x^ne^{-x}\bigr)$, so that $z'=y^{(n+1)}$ and $z''=y^{(n+2)}$, this reads exactly

\[ xz'' + (x+1)z' + (n+1)z = 0. \]

Step 3: 3° — $w=e^xz$ is a polynomial obeying Laguerre's equation

By the Leibniz rule applied directly to the product $x^ne^{-x}$, \[z = \frac{d^n}{dx^n}\bigl(x^ne^{-x}\bigr) = \sum_{j=0}^{n}\binom{n}{j}\Bigl[D^jx^n\Bigr]\Bigl[D^{n-j}e^{-x}\Bigr] = \sum_{j=0}^{n}\binom{n}{j}\frac{n!}{(n-j)!}(-1)^{n-j}\,x^{n-j}\,e^{-x}.\]

Therefore $w=e^xz$ is the same sum without the exponential factor, $w(x) = \sum_{j=0}^{n}\binom{n}{j}\frac{n!}{(n-j)!}(-1)^{n-j}\,x^{n-j}$.

a genuine polynomial in $x$; its $j=0$ term, $(-1)^nx^n$, is nonzero, so $w$ has degree exactly $n$.

From $z=e^{-x}w$ one gets $z'=e^{-x}(w'-w)$ and $z''=e^{-x}(w''-2w'+w)$. Substituting into $xz''+(x+1)z'+(n+1)z=0$ and cancelling the common factor $e^{-x}$: $x(w''-2w'+w) + (x+1)(w'-w) + (n+1)w = 0$.

Expanding, the $xw'$ and $-xw'$ terms cancel, as do $xw$ and $-xw$, leaving

\[ xw'' + (1-x)w' + nw = 0, \]

which is precisely Laguerre's equation of order $n$.

Step 4: The integral $I$: case $k\lt n$

Write $z=\dfrac{d^n}{dx^n}(e^{-x}x^n)$; since $x^ne^{-x}$ and all its derivatives up to order $n-1$ vanish at both $x=0$ and $x\to\infty$, every boundary term below drops out. Integrating by parts, each time differentiating the power of $x$ and integrating the remaining derivative of $z$'s antecedent down by one order:

\[ I = \int_0^\infty x^k\,z\,dx = \Bigl[x^k\,z^{(-1)}\Bigr]_0^\infty - k\int_0^\infty x^{k-1}z^{(-1)}\,dx = -k\int_0^\infty x^{k-1}z^{(-1)}\,dx, \] where $z^{(-1)}$ denotes an antiderivative of $z$, i.e.\ a derivative of $x^ne^{-x}$ of order $n-1$, which again vanishes at both endpoints. Repeating this $k$ times ($k\lt n$) reduces $x^k$ to the constant $k!$, leaving

\[ I = (-1)^k k! \int_0^\infty \frac{d^{\,n-k}}{dx^{\,n-k}}\bigl(x^ne^{-x}\bigr)\,dx = (-1)^k k!\, \Bigl[\tfrac{d^{\,n-k-1}}{dx^{\,n-k-1}}\bigl(x^ne^{-x}\bigr)\Bigr]_0^\infty. \]

Since $n-k-1 \le n-1$, this derivative order still vanishes at $x=0$ and $x=\infty$, so

\[ I = 0 \qquad (k\lt n). \]

Step 5: The integral $I$: case $k=n$

The same integration by parts, now carried out $n$ times on $x^n$, again has every boundary term vanish, and after $n$ steps $\dfrac{d^n}{dx^n}(x^n)=n!$, a constant. Hence

\[ I = (-1)^n n! \int_0^\infty e^{-x}x^n\,dx = (-1)^n n!\cdot n! = (-1)^n (n!)^2, \]

using the classical value $\int_0^\infty x^ne^{-x}dx=n!$.

Step 6: Verification

Parts 2° and 3° were checked symbolically for $n=1,2,3,4$: computing $z=d^n(x^ne^{-x})/dx^n$ explicitly and substituting into $xz''+(x+1)z'+(n+1)z$ gives residual $0$ in every case; likewise $w=e^xz$ is confirmed to be a degree-$n$ polynomial, and substituting it into $xw''+(1-x)w'+nw$ gives residual $0$ in every case. The integral was checked by direct symbolic integration for $n=1,2,3$ and all admissible $k$: $I=0$ whenever $k\lt n$, and at $k=n$, e.g.\ $n=1$: $I=-1=(-1)^1(1!)^2$; $n=2$: $I=4=(-1)^2(2!)^2$; $n=3$: $I=-36=(-1)^3(3!)^2$ — matching $(-1)^n(n!)^2$ exactly in every case.

\[I = 0 \ \ (k\lt n), \qquad\qquad I = (-1)^n(n!)^2 \ \ (k=n)\]