@PhilipHoskins Since $f(x) \sim 2\sqrt{x}$ as $x \to 0^+$ we get from Watson's Lemma that $$\int_0^\infty f(x) e^{-kx}\,dx \sim \int_0^\infty 2\sqrt{x} e^{-kx}\,dx = \frac{\sqrt{\pi}}{2k^{3/2}}$$ as $k \to \infty$.
That integral is a gamma function in disguise. Setting $kx = u$ gives $$\int_0^\infty 2\sqrt{x} e^{-kx}\,dx = \frac{2}{k^{3/2}} \int_0^\infty \sqrt{u} e^{-u}\,du = \frac{2}{k^{3/2}} \Gamma\left(\frac{3}{2}\right),$$ and there are formulas for half-integer gamma functions.
Oh I was off by a factor of 1/2
$\Gamma(3/2) = \Gamma(1/2)/2 = \sqrt{\pi}/2$, so actually $$\int_0^\infty f(x) e^{-kx}\,dx \sim \int_0^\infty 2\sqrt{x} e^{-kx}\,dx = \frac{\sqrt{\pi}}{k^{3/2}},$$ and the correct limit is $\sqrt{\pi}$.
@PhilipHoskins That series is an asymptotic series for the integral. It is its own estimate. Each term is smaller than the previous term as $k \to \infty$, so first you have $\sqrt{\pi}$ as the limit, then the next term is an estimate on the error between the integral and $\sqrt{\pi}$ for large $k$, then the next term is an estimate on the error of the error, and so on.
@PhilipHoskins That's good. If I could make a final comment, it's that you do not need to estimate the series, and it does not even need to converge. It is an asymptotic series (this is a conclusion of Watson's Lemma), so you can truncate it after any number of terms to get an estimate.
Yeah I can see that. And d'oh!, My estimate was good because the series starts at 1, so it really does vanish as k-> \infty. I realize that I don't really need to do this estimate, but it was a psychological thing for me
@robjohn ya I saw that .. infact I pointed out the same idea in comment .. but the OP said he's looking for a proof using differentiation under integral sign though.
which is natural enough, since in that form one can write it (up to algebra errors made from doing this off the top of my head) as $$p_n(z)=z^n \frac{z+t}{z-1/z}\left(1+\frac{1}{z^{2n+1}}\frac{t+z}{tz+1}\right)$$
(which actually is in my mental background for the problem i've been working on, namely b/c a periodic jacobi matrix always has a finite number of bands)
so one doesn't have to wrestle with some of those infinities
@Semiclassical That's something I'd like to learn a little about. Certain solutions of it come up in RH stuff for the finite gap potentials, and I was wondering if it's possible to look at things in a more general context to see if other solutions of it can show up in a similar way.
It has these polynomial solutions which some pretty regular zero behavior, and there's a related problem with a generalized Lame' equation and some slightly more interesting zero behavior of its polynomial solutions.
gotcha. the problems i was initially working on, actually, was working with generalizations of mathieu's equation (which isn't finite gap) using WKB i.e. quantization of certain integrals
it's also worth noting that things simplify if $t=1$, since then one has explicitly $$p_n=z^n\left(\frac{z}{z+1}\right)\left(1+\frac{1}{z^{2n+1}}\right)$$
and more generally there's a distinction between $|t|\gtrless 1$, since that determines whether the zero at $z=-t$ is inside or outside of $|z|=1$
it's really the last factor (which goes to unity as $n\to \infty$) which i'm most interested in, since it behaves non-trivially on the value of $n\times (t-1)$
@MaryStar O_o that notation is awfully obscure. Where does it come from ? The usual $\int_\mathcal{S}\vec{g}\cdot\vec{n} dS=\int_\mathcal{V}div(\vec{g})d\tau$ seems much more explicit to me
@MaryStar No. I mean why they hold. A formal proof isn't necessary, but at least grasping how one can understand the formula from physic's point of view would be interesting
He told us for example that the augmentation rate of the mass in $W$ is $-\int_{\partial{W}} \rho \overrightarrow{u} \cdot \overrightarrow{n}dA$. @Hippalectryon
@Masacroso Show me the whole thing. I assure you, the boundary should NOT change since the equality just relies on the fact that $\binom{n}k\frac{1}{k+1}=\binom{n+1}{k+1}\frac{1}{n+1}$
and the integral $\int_{\partial{W}}\rho \overrightarrow{u} ( \overrightarrow{u} \cdot \overrightarrow{n})dA$ is the rate of the outflow of the momentum from the boundary of $W$ @Hippalectryon
@MaryStar And $\vec{u}\cdot\vec{n}$ gives you the particles that really go out (for instance all the particles will go out if $\vec{u}$ and $\vec{n}$ are aligned, none will go out if the speed is perpendicular to the normal vector (=tangent to the surface)
the book says "Ah yes; we fell into the old trap mentioned earlier: We tried to apply symmetry when the upper index could be negative!" but it never can happen! It is not possible that anything would be negative because $k,n\ge 0$
I thought again about the momentum... Isn't it $m \overrightarrow{u}$ and since $m=\int_{W}\rho dV$ shouldn't the momentum be $\int_{W} \rho \overrightarrow{u}dV$ ?
The integral $\int_{\partial{W}}\rho \overrightarrow{u} ( \overrightarrow{u} \cdot \overrightarrow{n})dA$ is the rate of the outflow of the momentum from the boundary of $W$... The rate of the ingoing flows is the same integral with a minus sign? @Hippalectryon
@MaryStar No, inwards/outwards will be a consequence of the speed's direction; my only question is (I don't remember the drawing anymore) does the integral over $W$ include all the open parts (=where the air can come in/out) ?
well, I discovered what is the real reason why the path failed but it doesnt have relation about the book says
well if the boundary changes it makes sense because we go from a binomial that can take value zero on the bottom, ie $\binom{n}{k}$, to someone that never take the value of zero, ie $\binom{n+1}{k+1}$, so maybe the boundary changes to compensate it
The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.
Show that the equation has $0$ as its singular regular point .
Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$
Show that if $a=n$, where $n \in ...
@Hippalectryon, I take the error on the book... all this section that I posted is completely wrong. In the book this deduction is not true: $\frac{1}{n+1}(-1)^n\binom{n-1}{-1}=0$. This is completely false, the nature of binomials is like limits. You cant assume anything if you dont know all variables involved, because when $n=0$ then $\binom{-1}{-1}=1$. So there is no error, LOL.
The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.
Show that the equation has $0$ as its singular regular point .
Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$
Show that if $a=n$, where $n \in ...
