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We're learning Riemann integrals, and I'm trying to see whteher the following indefinite integral $\int _{1}^{\infty}\frac{p(x)}{e^x}$ converges, when p(x) is a polynomial. I don't know where to start this question - does it depends on p(x)? What theorem should I use?

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6 Answers 6

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It converges for any $p(x)$. We can say that the "asymptotic growth" of $e^x$ is greater than that of any polynomial, but we can prove it formally. It suffices to prove that $\int_1^\infty x^n e^{-x}\:\mathrm dx$ converges for any $n = 0, 1, 2, \dots$, and I'll do that by the finite induction principle.

For $n = 0$, it's easy: $\int_1^\infty e^{-x}\:\mathrm dx = \left.-e^{-x}\right|_1^\infty = \frac 1e$.

Now, consider that it's true for some $n$, i.e. $\int_1^\infty x^n e^{-x}\:\mathrm dx$ converges. To prove it for $n+1$, we expand the integral by parts: $$\int_1^\infty x^ne^{-x}\:\mathrm dx = \left.-x^ne^{-x}\right|_1^\infty + \int_1^\infty (n+1)x^ne^{-x}\:\mathrm dx = \frac1e + (n+1)\int_1^\infty x^ne^{-x}\:\mathrm dx$$

Since the right hand part is an integral that converges plus a finite value, the left hand part converges. QED

Now, why does it suffice to show that $\int_1^\infty x^ne^{-x}\:\mathrm dx$ converges? Because it's the building block of any polynomial: $p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$, so: $$\int_1^\infty p(x)e^{-x}\:\mathrm dx = \int_1^\infty \sum_{i=1}^n a_ix^ie^{-x}\:\mathrm dx = \sum_{i=1}^n a_i \int_1^\infty x^ie^{-x}\:\mathrm dx$$ which is a sum of convergent integrals, so it does converge.

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  • $\begingroup$ You have a typo $(x^n)'=nx^{n-1}$ ... $\endgroup$
    – rtybase
    May 1, 2017 at 19:32
  • $\begingroup$ @rtybase Thanks! Corrected! $\endgroup$
    – JoaoBapt
    May 1, 2017 at 19:34
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You just have to use two theorems :

  • growth comparison between a power function and an exponential one

  • linearity of the integral

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  • $\begingroup$ Absolute convergence implies convergence might also be useful here. I like your minimal approach, +1. $\endgroup$
    – zhw.
    May 1, 2017 at 19:23
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    $\begingroup$ This answer is correct, but it may impervious to a beginner. $\endgroup$
    – mlc
    May 1, 2017 at 19:23
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A polynomial with degree $n$ fulfills $\frac{d^{n+1}}{d^{n+1}x}p(x)=0$. You may easily check that the derivative of $$-\left[p(x)+p'(x)+\ldots+p^{(n)}(x)\right]e^{-x} = -q(x) e^{-x}$$ is exactly $p(x)e^{-x}$, hence $$ \int_{1}^{M} p(x)e^{-x}\,dx = q(1)e^{-1}-q(M)e^{-M}\tag{1} $$ by the Fundamental theorem of Calculus. Since $q(x)$ is a polynomial with degree $n$, $$ \lim_{M\to +\infty} q(M) e^{-M} = 0 \tag{2} $$ and $$ \int_{1}^{+\infty}p(x)e^{-x}\,dx = \color{red}{\frac{q(1)}{e}}.\tag{3} $$

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The other answer gives the right intuition. I will try to formalize a bit more.

Let our polynomial $p(x) = \sum_{i=0}^n a_i x^i$. Then, as stated in the other answer, you can break the integral using linearity: $$ \int_1^\infty \frac{p(x)}{e^x} = \int_1^\infty \frac{\sum_{i=0}^n a_i x^i}{e^x} = \sum_{i=0}^n \int_1^\infty \frac{a_i x^i}{e^x} = \sum_{i=0}^n a_i\int_1^\infty \frac{x^i}{e^x} \tag{1} $$ Now let us focus on $\int_1^\infty \frac{x^i}{e^x}$. We know that for $e^x =\sum_{i=0}^\infty \frac{x^i}{i!}$. In particular if we assume $x \geq 0$, (and in the case of our integral this holds), we can bound below $e^x$ by retaining only one term in the summation that defines the exponential. For example we can say $e^x \geq x^2/2!$ or $e^x \geq x^5/5!$. This actually holds for any terms. We can therefore say $\forall m\geq 0, \forall x \geq 0, e^x \geq x^m/m!$.

With this little theorem in hand we can then say : $$ \int_1^\infty \frac{x^i}{e^x} \overset{(*)}{\leq} \int_1^\infty \frac{x^i}{x^{i+2}/(i+2)!} = (i+2)! \cdot \int_1^\infty \frac{1}{x^2} = (i+2)! \cdot 1 < \infty \tag{2} $$ $(*)$ Here I chose to only retain the $(i+2)$'th term of the "exponential sum".

Putting $(2)$ and $(1)$ together should give you the conclusion that $ \int_1^\infty \frac{p(x)}{e^x}$ is simply a sum of perfectly defined integral that all converges to sum finite value. Since the sum is also finite (up to $n$), it converges.

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$$p(x)=\sum_{k=0}^n a_kx^k$$ $$(*)\ \int_1^{\infty}\frac{p(x)}{e^x}dx=\sum_{k=0}^n a_k\int_1^{\infty}\frac{x^k}{e^x}$$ This equation is always true, so left integral converges if and only if every integral on the right converges and it can be easily estimated using series representation. For all $1\le x$, $m\in \mathbb{Z}_+$: $$ e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}\ge \frac{x^{m+2}}{(m+2)!}$$ $$ 0\le \frac{x^m}{e^x}\le \frac{(m+2)!x^m}{x^{m+2}}=\frac{(m+2)!}{x^{2}}$$ Now the integral $$ \int_1^{\infty}\frac{(m+2)!}{x^2}dx = (m+2)!\int_1^{\infty}\frac{1}{x^2}dx$$ is convergent (exercise). $$ \int_1^{\infty}\frac{x^m}{e^x} \le \int_1^{\infty}\frac{(m+2)!}{x^2}dx$$ Thus every integral on the right side of $(*)$ is convergent.

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Let $p$ be a polynomial. We have that $$\frac{|p(x)|}{e^{kx}} \to 0$$ for any $k>0$. In particular, for $k=1/2$. Therefore, there exists $M>0$ such that if $x>M$, then $$\frac{|p(x)|}{e^{x/2}} < 1,$$ from which follows that if $x>M$, $$\frac{|p(x)|}{e^x}<\frac{1}{e^{x/2}}.$$ Since the latter is integrable, the former also is.

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