Convergence of Improper Integral for any polynomials

Question

Let $f, g$ be real polynomials. How can I show that the improper integral $\int_{0}^{\infty}\frac{f(x)g(x)}{e^x}dx$ converges absolutely for any $f$ and $g$? I think some sort of bounding argument should be used where $f$ and $g$ can be replaced by explicit polynomials that bound $f$ and $g$, but I'm not really sure how to approach this.

Answer 1

Lemma. The integral $\int_0^\infty x^ne^{-x}\,dx$ converges for all $n$.

Proof. In the case where $n=0$, we have $-e^{-x}$ as an antiderivative, which clearly evaluates to 1 when you plug in the limits.

Now assume that the claim holds for all powers of $x$ up to $n-1$. Using integration by parts, we see $$\int x^ne^{-x}\,dx = -x^ne^{-x}+n\int x^{n-1}e^{-x}\,dx.$$ When we plug in limits the $-x^ne^{-x}$ term disappears, yielding $$\int_0^\infty x^ne^{-x}\,dx = n\int_0^\infty x^{n-1}e^{-x}\,dx,$$ that latter integral being finite by assumption. This completes the proof of the lemma.

Now on to the proof of the statement. Write $f(x)g(x) = a_0+a_1x+\cdots+a_nx^n$. We have $|f(x)g(x)| \leq |a_0|+|a_1|x+\cdots+|a_n|x^n$ for all $x \geq 0$. This yields $$\int_0^\infty |f(x)g(x)|e^{-x}\,dx \leq \sum_{k=0}^n|a_k|\int_0^\infty x^ke^{-x}\,dx.$$ Every term in the sum is finite, by the lemma.

Answer 2

I think you could try letting f and g be finite degree polynomials so, $f(x)=\sum_{k=0}^{k_{max}} a_kx^k$ and $g(x)=\sum_{j=0}^{j_{max}} a_jx^j$ then $f(x)g(x)=\sum_{k=0}^{k_{max}} ( \sum_{j=0}^{j_{max}} a_k a_j x^{j+k})$, this is the distributive law for sums. Then your integral can actually be written as a series of integrals where every term has the form $\int_{0}^{ \infty} \frac{x^{j+k}}{e^x}dx$ with some constants out front.

Answer 3

Note that, if $f$ is of degree $n$ and $g$ is of degree $m$, then $$\lim_{x\to\infty}\frac{|f(x)|}{x^{n+1}}=\lim_{x\to\infty}\frac{|g(x)|}{x^{m+1}}=0.$$ Therefore, there exists $M>0$ such that, if $x\geq M$, then $|f(x)|\leq x^{n+1}$ and $|g(x)|\leq x^{m+1}$. Also, $$\lim_{x\to\infty}\frac{x^{m+n+4}}{e^x}=0,$$ so there exists $M_0>M$ such that, if $x\geq M_0$, then $x^{m+n+4}\leq e^x$. Therefore, for $x\geq M_0$, $$\frac{|f(x)g(x)|}{e^x}\leq\frac{x^{m+1}x^{n+1}}{e^x}=\frac{1}{x^2}\frac{x^{m+n+4}}{e^x}\leq\frac{1}{x^2}.$$ But, the improper integral $\int_{M_0}^{\infty}\frac{1}{x^2}\,dx$ converges, so the original integral converges as well.