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2 votes
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When is Laplace transform of a function power-law and relation to the behavior of the function near zero?

Intuitively it should have something to do with the behavior of $f$ near $0$. This is true. Indeed, let $U(x):=\int_0^x du\,f(u)$. Then, according to (say) Tauberian Theorems 3 and 2 in Section XIII....
Iosif Pinelis's user avatar
5 votes
Accepted

Interpretation of an asymptotic result in probability

$\newcommand\ka\kappa$Intuition for this result is as follows. The condition on $h'$ implies that $h(y)=(1+o(1))y$ (as $y\to\infty$). So, for the tail function $T$ given by $T(y):=P(Y\ge y)$ we have $$...
Iosif Pinelis's user avatar
2 votes
Accepted

Nonstationary phase method for oscillatory integral

For stationary phase, you usually consider the integral $$I(\lambda)=\int_a^b f(t) e^{i\lambda g(t)}\,dt$$ with $\lambda>0$ a large parameter. If there are no stationary points inside $[a,b]$, then ...
Dispersion's user avatar
1 vote
Accepted

Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines

Your conjecture is not true. Moreover, $$J_m(y):=\int_y^\infty dx\, e^{-x^2}H_m(x)^2\sim c_m:=\frac12\,\pi^{1/2}\,2^m m! \tag{1}\label{1}$$ whenever $$0\le y=o(m^{1/2}).$$ Indeed, recalling that $H_m^...
Iosif Pinelis's user avatar
1 vote

Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines

Let me first consider the case $x_0=0$, when the integral has a closed form expression: $$\int^{\infty}_{0}|H_m(x)|^2 e^{-x^2}\,dx=\sqrt{\pi } \,2^{m-1} m!\;\;(m\in\mathbb{N}).$$ See, for example, ...
Carlo Beenakker's user avatar
1 vote

Asymptotic behavior of weighted sums involving the fractional part function

Not an answer but a conjectural answer for the value of $C(m)$ supported by extensive numerical evidence: $$C(m)=\dfrac{1}{m-1}\sum_{2\le k\le m-1}\binom{m-1}{k}\zeta(k)+\sum_{2\le k\le m-2}\dfrac{1}{...
Henri Cohen's user avatar
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5 votes
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Mellin transform at $0$

By definition, $\tilde{f}(0)$ exists if and only if $\int_0^\infty f(x)/x\,dx$ exists. As $f(x)$ is smooth and supported on $[0,2]$, we have that that $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\...
GH from MO's user avatar
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