real analysis – Comparing divergent and convergent sums

Let $(x_n)$ be a monotonically decreasing sequence of positive real numbers that is also summable.

Let $(y_n)$ be a sequence of positive real numbers such that
$sum_n x_n y_n$ converges.

Let $(z_n)$ be a monotonically increasing sequence of positive real numbers such that $sum_n x_n z_n =infty.$

Assume that the sequences $y_n$ and $z_n$ are such that $2^{-varepsilon y_n}$ and $2^{-varepsilon z_n}$ are summable for every $varepsilon>0.$ Does it follow that there is some $delta>0$ such that

$$ sum_n Big(2^{-varepsilon y_n}-2^{-varepsilon z_n}Big) ge 0 text{ for all } varepsilon in (0,delta)?$$

The motivation for this statement to be true is that $z_n$ should be larger most of the time than $y_n$ and we capture this most of the time by taking $varepsilon$ small.

NIntegrate of a convergent integral working with large integration limits, but not with infinite integration limits

Code for reproducing is below:

Integrand(x_) := ((x + 1) Abs(x) Log((x + 1)^2/Abs(x)^3))/((x + 1)^2 - Abs(x)^3);
    
NIntegrate(Integrand(x), {x, -10000000, 10000000})
NIntegrate(Integrand(x), {x, -Infinity, Infinity})

The first instance of NIntegrate gives $2.48398$, and one can check by plotting the result of NIntegrate as a function of the integration limits that the integral appears to converge to this answer for very large integration limits. However, the second instance of NIntegrate gives a completely different answer of $1.75434 times 10^{8}$. What’s going on here, and how can I make the two integrals agree? I think something that NIntegrate might be having trouble with is that the convergence of the integral requires the fact that the integrand is essentially odd for very large $|x|$ and these contributions tend to cancel out. Ideally I’d like to find some way to get Mathematica to deal with this properly with the infinite integration range, so that I don’t have to keep remembering to exchange the infinite integration limits with large finite values every time I need to do an integral like this.

real analysis – Order of convergence of a product of two convergent sequences

Let $a_n$ be a sequence that converges to $A$ with order of $n^alpha$, that is $a_n = A + mathcal{O}(n^alpha)$ and $b_n$ is another sequence that converges to B with ordder of $n^beta$; i.e. $b_n = B + mathcal{O}(n^beta)$. What is the order of convergence of $a_n cdot b_n$?

My intuition tells me that:

$$ a_nb_n = AB + mathcal{O}(n^{min(alpha,beta)})$$

but could not formulate the proof. I used the trick:

$$|a_nb_n – a_nB + a_nB – AB| leq |a_n||b_n-B| + |B| |a_n-A| leq K n^{alpha}n^{beta}+ Bn^{alpha}$$

Laplace transform of bounded variation function extends to entire function. Does it imply the integral of the LT is absolutely convergent

I have the following problem. Assume that $u$ is a bounded variation function on $(0,infty)$. We know that
$$h(z)=int_{0}^{infty}e^{-zy}u(y)dy$$
is absolutely convergent for $Re(z)>-1$ and $h$ can be extended to an entire function. Does this imply that $e^{-zy}u(y)$ is absolutely integrable on $(0,infty)$?

I would appreciate any hint and perhaps some reference to a body of literature on Laplace transforms of signed measure provided you are aware of such.

Thanks in advance!

integration – Analytic continuation of Convergent integral

I was trying to solve the following integral.
$$I = oint _{|z|=1}frac{dz}{2 pi i z}int_{0}^{infty} dr dfrac{e^{-tfrac{r^2}{z^2}}r^{2n+1}}{z^2(z-1)} $$

  • The singular structure in the $z$ integral is coming from $z= 0$ and $z = 1$ points.
  • From the term $ e^{-tfrac{r^2}{z^2}}$ the integral over $r$ is giving divergent contribution for $tfrac{pi}{4} leq Arg(z) leq tfrac{3pi}{4} ; ; & ; ; tfrac{5 pi}{4} leq Arg(z) leq tfrac{7pi}{4} $

My question is, can we somehow give some argument like analytic continuation or modify the contour so that we can assign $I$ to some convergent value?