5 votes
Accepted

$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$

Here's an approach using differentiation under the integral sign: $$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\ &=\left.\frac{(x+1)\ln\...
user avatar
  • 3,092
3 votes

Does a vertical asymptote mean it is not a function?

When defining a function, it is crucial to consider the set that we are inputting (the domain) and the set that we are outputting (the codomain). The notion of "undefined" is not a unique ...
user avatar
3 votes
Accepted

Asymptotic behavior of the integral $\int_{0}^{\infty} e^{-\rho \cosh(R)}\cosh(Rq)dR$

Assume that $\Re \rho>0$ and $q \in \mathbb C$ is fixed. Your integral is $$ K_q (\rho ) = e^{ - \rho } \int_0^{ + \infty } {e^{ - \rho (\cosh (R) - 1)} \cosh (Rq)dR} . $$ Now perform the change of ...
user avatar
  • 19.7k
2 votes

Speed of Convergence for some series (double sum)

We will try to find the asymptotics of the sum at $n\to\infty$. Let's denote $$S(a,n)=\sum_{i=1}^n\sum_{j=n-i+1}^ni^{-a}j^{-a}$$ Using the relation $j^{-a}=\frac{1}{\Gamma(a)}\int_0^\infty t^{a-1}e^{-...
user avatar
  • 5,807
2 votes
Accepted

Can someone help find methods to study the asymptotic behavior of these integrals as $\rho \to \infty$?

Watson's Lemma will still apply, since the dominant contribution is from the left hand end of the interval. You can shift the integral to start at zero: \begin{align*} I &= \int_{-1}^1 e^{-\rho t}...
user avatar
  • 111
2 votes

Solution verification of an approximation of an integral

If you want a straight answer, you can observe that indefinite integral equals to $$ -(e^{-t x} (\cos(t) + x \sin(t)))/(1 + x^2) $$ and the definite integral behaves in the infinity like $1/x^2$, ...
user avatar
2 votes
Accepted

Speed of convergence of continued radicals with constant term

Hint: Notice that from your inequality (1), $\sum_n(\ell_a-x_n)$ converges since $0<x_n<\ell_a$, $n\geq1$, $\ell_a>1$, which means that $\sum_n\ell^{-n}_a<\infty$. This in turn, implies ...
user avatar
  • 26.2k
1 vote

The largest $f(x)$ for which $f(x)/g(x)\to 0$ as $x \to \infty$

Realistically, there is no largest such $f(x)$. Simply take $$f(x)=\frac{g(x)}{r(x)}$$ for any $r(x)$ such that $$\lim_{x\to\infty}r(x)=\infty$$ This then gives $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\...
user avatar
  • 9,365
1 vote

Can someone help find methods to study the asymptotic behavior of these integrals as $\rho \to \infty$?

If you let $t=\cos(x)$ and if $q$ is a positive integer you find something which is quite close to the integral formulae of the modified Bessel functions of the first kind (have a look here). Using $...
user avatar
1 vote

Can someone help find methods to study the asymptotic behavior of these integrals as $\rho \to \infty$?

Here are some coarse ideas. Note that $t^2-1>0$ on $t > 1$ and $1-t^2>0$ on $|t|<1.$ Escribe $e^{-\rho t} = e^{-\rho t/2} e^{-\rho t/2} \leq e^{-\rho/2} e^{-\rho t/2}$ for $t > 1.$ Then,...
user avatar
  • 6,147
1 vote

Doubts on asymptotic criterion for $\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}n^{a}\tan^{-1}\bigg(\frac{1}{n^a}\bigg)-e^{1/n}$ with $a>0$.

First note that $$ \arctan x = x - \frac{{x^3 }}{3} + \mathcal{O}(x^5 )\quad \text{ and }\quad e^x = 1 + x + \mathcal{O}(x^2 ) $$ as $x\to 0$. Thus, \begin{align*} a_n &= n^a \left( {\frac{1}{{n^...
user avatar
  • 19.7k
1 vote
Accepted

Low convergence of a perturbation solution to a non-linear ODE

Writing the solution as $$ y(x) = \gamma(\epsilon) (1-x) \ \ \ \quad \mathrm(0) $$ we see that $\gamma$ must satisfy $$ \gamma^3\frac{\epsilon}{40}+\frac{\gamma}{12} -1=0. \ \ \ \quad \mathrm(1) $$ ...
user avatar
  • 1,834
1 vote
Accepted

Speed of Convergence for some series (double sum)

We will show that $S_n=o(n^{-(\alpha-1)})$. We claim $$n^{(\alpha-1)}S_n \approx n^{(\alpha-1)} \sum_{i=1}^n i^{-\alpha}\left[(n-i+1)^{-(\alpha-1)}-n^{-(\alpha-1)}\right] =\sum_{i=1}^n i^{-\alpha}\...
user avatar
  • 315
1 vote

Asymptotic for a binomial sum

I followed the same suggestion as @Maxim of summing over $r=n-k$. The main reason is that for large $n$, ie $n\gg a, 1/|x|$, the largest terms are with $k$ close to $n$: having them for small $r$ is ...
user avatar

Only top scored, non community-wiki answers of a minimum length are eligible