What does this sequence of integral converge to?

Question

I am not a mathematician, I'm a chemist who happened to also be a math enthusiast.

I was playing with error function the other day, and just out of curiosity compute these integrals using Desmos.

$$ \int _{-\infty}^{\infty} e^{-(x^2+x+1)} dx \approx 0.837247915445$$

Then, I just add more terms to it

$$ \int _{-\infty}^{\infty} e^{-(x^4+x^3+x^2+x+1)} dx \approx 0.710530779033$$ $$ \int _{-\infty}^{\infty} e^{-(x^6+x^5+x^4+x^3+x^2+x+1)} dx \approx 0.682543666557$$ $$ \int _{-\infty}^{\infty} e^{-(x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)} dx \approx 0.671744903773$$

And so on. At 20th term (order 40 polynomial), the result is $0.655405978856$

One of the thing that I realize is the difference between terms became smaller, and I think the limit is $\frac{\pi^{0.5}}{e} \approx 0.652049332173292$. But I have no idea why, or if this is even true.

I hope someone can provide me with what the limit is.

Answer 1

The functions $$ f_n(x) = \exp(-(1+x+\cdots + x^{2n}) = \begin{cases}\exp\left( -\frac{x^{2n+1}-1}{x-1}\right) & \text{ if } x \ne 1 \\ \exp\bigl(-(2n+1)\bigr) & \text{ if } x = 1 \end{cases} $$

• decrease to zero if $x \ge 1$,
• decrease to $\exp(-1/(1-x))$ if $0 \le x < 1$,
• increase to $\exp(-1/(1-x))$ if $-1 \le x \le 0$,
• decrease to zero if $x < 1$.

Therefore Beppo Levi's lemma aka “monotone convergence” can be applied on the intervals $(-\infty, -1)$, $[-1, 0]$, $[0, 1]$, and $[1, \infty)$ separately, and it follows that $$ \lim_{n \to \infty} \int_{-\infty}^\infty f_n(x) \,dx = \int_{-1}^1 e^{-1/(1-x)} \, dx \, . $$

According to Wolfram Alpha the value of that integral is $$ Ei\left(-\frac 12\right) + \frac{2}{\sqrt e} \approx 0.653288 $$ where $Ei(x)$ is the exponential integral.

Answer 2

Here is another way using the dominated convergence theorem. The integral can be expressed as $$ \int_{-\infty}^\infty \exp\left(\dfrac{x^{2n + 1} - 1}{1 - x}\right)dx $$ (Small note: Technically, we can ignore $x = 1$ since $\{1\}$ is a null set w.r.t the Lebesgue measure)

As suggested in the comments, we bound the integrand by $$ f(x) = \begin{cases} e^x, \text{ if } x \le -1\\ 1, \text{ if } x \in (-1, 1)\\ e^{-x}, \text{ if } x \ge 1 \end{cases} $$ Notice that $f > 0$ and, $$ \int_{-\infty}^\infty f(x)dx < \infty $$ Thus, by the dominate convergence theorem, \begin{align*} \lim_{n \rightarrow \infty} \int_{-\infty}^\infty \exp\left(\dfrac{x^{2n + 1} - 1}{1 - x}\right)dx &= \int_{-\infty}^\infty \lim_{n \rightarrow \infty}\exp\left(\dfrac{x^{2n + 1} - 1}{1 - x}\right)dx\\ &= \int_{-1}^1 \exp\left(\dfrac{1}{x-1}\right)dx \end{align*}