$\int_{0}^{\infty}\frac{e^{-at^2}}{\sqrt{t}}\prod_{i=1}^n\left|\frac{1}{\sqrt{t}}-r_i\right|^2 dt$; $r_i\in\mathbb{R},a>0$.

Question

Does the following integral converge or diverge?

$$\int_{0}^{\infty}\frac{e^{-at^2}}{\sqrt{t}}\prod_{i=1}^n\left|\frac{1}{\sqrt{t}}-r_i\right|^2 dt$$

Here $r_i\in\mathbb{R}$ and $a>0$.

Answer 1

I'm hoping the integral doesn't change again, because I have an answer.

I claim that $$\int_0^\infty \left|\frac{e^{-at^2}}{\sqrt{t}}\prod_{i=1}^n \left(\frac{1}{\sqrt{t}} - r_i\right)\right|^2 \text{d}t$$ diverges for every $n \geq 1$, $\forall r_i\in \mathbb{R}$, $1\leq i\leq n$.

What follows is an (almost complete) argument to support this claim, but is not a complete proof (though could be extended to one with a bit of work).

Let's rewrite this integral a bit: \begin{align*} &\int_0^\infty \left|\frac{e^{-at^2}}{\sqrt{t}}\prod_{i=1}^n \left(\frac{1}{\sqrt{t}} - r_i\right)\right|^2 \text{d}t \\ &=\int_0^\infty e^{-at^2}\left|\frac{1}{\sqrt{t}} \left(\frac{1}{(\sqrt{t})^n} - \left(\sum_{i=1}^n r_i\right)\frac{1}{(\sqrt{t})^{n-1}} + \cdots + (-1)^n\left(\prod_{i=1}^n r_i \right) \right) \right|^2\text{d}t\\ &= \int_0^\infty e^{-at^2} \left| \frac{1}{(\sqrt{t})^{n+1}} - \left(\sum_{i=1}^n r_i\right)\frac{1}{(\sqrt{t})^{n}} + \cdots +(-1)^n \left(\prod_{i=1}^n r_i \right) \frac{1}{\sqrt{t}}\right|^2\text{d}t \end{align*} The $\cdots$ here represent homogeneous polynomials in the $r_i$. For example, the $\frac{1}{(\sqrt{t})^{n-1}}$ coefficient would be $$2\sum_{1 \leq i < j\leq n} r_i r_j$$ and the others would be expressed similarly.

In general, the $\frac{1}{(\sqrt{t})^{n-j}}$ term would have coefficient $$(-1)^{j+1} \left(\begin{array}{c} n\\ j+1\end{array}\right) \sum_{1\leq i_1 < \cdots < i_{j+1} \leq n} r_{i_1}\cdots r_{i_{j+1}}$$ for $1\leq j \leq n-1$.

Since $e^{-at^2}$ is decreasing on $[0,\infty)$ for any $a > 0$, then in a neighborhood of $0$ (say, on $[0,1]$), the integrand is bounded from below by $$e^{-a}\left|\frac{1}{(\sqrt{t})^{n+1}} - \left(\sum_{i=1}^n r_i\right)\frac{1}{(\sqrt{t})^{n}} + \cdots +(-1)^n \left(\prod_{i=1}^n r_i \right) \frac{1}{\sqrt{t}} \right|^2$$ Making use of this leads to the following string of inequalities. \begin{align*}&\int_0^\infty \left|\frac{e^{-at^2}}{\sqrt{t}}\prod_{i=1}^n \left(\frac{1}{\sqrt{t}} - r_i\right)\right|^2 \text{d}t \\ &\geq \int_0^1 \left|\frac{e^{-at^2}}{\sqrt{t}}\prod_{i=1}^n \left(\frac{1}{\sqrt{t}} - r_i\right)\right|^2 \text{d}t\\ &\geq e^{-a} \int_0^1 \left| \frac{1}{(\sqrt{t})^{n+1}} - \left(\sum_{i=1}^n r_i\right)\frac{1}{(\sqrt{t})^{n}} + \cdots +(-1)^n \left(\prod_{i=1}^n r_i \right) \frac{1}{\sqrt{t}} \right|^2\text{d}t\\ &= e^{-a}\int_0^1 \left[\left(\frac{1}{|t|^{n+1}}\right) + O\left(\frac{1}{|t|^{n+\frac{1}{2}}}\right)\right]\,\text{d}t \end{align*} which diverges.

Realistically, this is almost a proof, but there are two steps omitted. Namely, one must first prove that the integrand is (as claimed) $$\frac{1}{|t|^{n+1}} + O\left(\frac{1}{|t|^{n+\frac{1}{2}}}\right)$$ but this is relatively clear, since the behavior of a sum of functions near $0$ is dominated by the worst behavior of any of the summands.

The proof would be completed once one proves that the integral over $[0,1]$ of any function of that form must diverge.

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A quick note about the dependence on $r_i$:

The only thing $r_i = 0$ $\forall 1\leq i\leq n$ would do is eliminate the term $O\left(\frac{1}{|t|^{n+\frac{1}{2}}}\right)$ from the integrand, which wouldn't change the fact that the integral diverges (in fact, the proof that this integral diverges is quite short).

The other case worth considering is if each $r_i$ is nonzero, distinct. But this does not effect the strength of the singularity at the origin, hence cannot change the fact that the integral diverges.