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:
$$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\
&=\left.\frac{(x+1)\ln\...
- 3,092
3
votes
Does a vertical asymptote mean it is not a function?
When defining a function, it is crucial to consider the set that we are inputting (the domain) and the set that we are outputting (the codomain).
The notion of "undefined" is not a unique ...
- 31
3
votes
Accepted
Asymptotic behavior of the integral $\int_{0}^{\infty} e^{-\rho \cosh(R)}\cosh(Rq)dR$
Assume that $\Re \rho>0$ and $q \in \mathbb C$ is fixed. Your integral is
$$
K_q (\rho ) = e^{ - \rho } \int_0^{ + \infty } {e^{ - \rho (\cosh (R) - 1)} \cosh (Rq)dR} .
$$
Now perform the change of ...
- 19.7k
2
votes
Speed of Convergence for some series (double sum)
We will try to find the asymptotics of the sum at $n\to\infty$. Let's denote
$$S(a,n)=\sum_{i=1}^n\sum_{j=n-i+1}^ni^{-a}j^{-a}$$
Using the relation $j^{-a}=\frac{1}{\Gamma(a)}\int_0^\infty t^{a-1}e^{-...
- 5,807
2
votes
Accepted
Can someone help find methods to study the asymptotic behavior of these integrals as $\rho \to \infty$?
Watson's Lemma will still apply, since the dominant contribution is from the left hand end of the interval. You can shift the integral to start at zero:
\begin{align*}
I &= \int_{-1}^1 e^{-\rho t}...
- 111
2
votes
Solution verification of an approximation of an integral
If you want a straight answer, you can observe that indefinite integral equals to
$$
-(e^{-t x} (\cos(t) + x \sin(t)))/(1 + x^2)
$$
and the definite integral behaves in the infinity like $1/x^2$, ...
- 13.4k
2
votes
Accepted
Speed of convergence of continued radicals with constant term
Hint:
Notice that from your inequality (1),
$\sum_n(\ell_a-x_n)$ converges since
$0<x_n<\ell_a$, $n\geq1$,
$\ell_a>1$, which means that $\sum_n\ell^{-n}_a<\infty$.
This in turn, implies ...
- 26.2k
1
vote
The largest $f(x)$ for which $f(x)/g(x)\to 0$ as $x \to \infty$
Realistically, there is no largest such $f(x)$. Simply take
$$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)}=\...
- 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 $...
- 213k
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,...
- 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^...
- 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
$$
\gamma^3\frac{\epsilon}{40}+\frac{\gamma}{12} -1=0. \ \ \ \quad \mathrm(1)
$$
...
- 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}\...
- 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 ...
- 8,202
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