@Chris'ssis @robjohn is there a closed form for $\displaystyle \int_{0}^\infty \frac{cos(ux)}{x}dx$ and $\displaystyle \int_{0}^\infty \frac{sin(ux)}{x}dx$
In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet.
One of those is
: \int_0^\infty \frac{\sin \omega}{\omega}\,d\omega = \frac{\pi}{2}
This integral is not absolutely convergent, and so the integral is not even defined in the sense of Lebesgue integration, but it is defined in the sense of the Riemann integral or the Henstock–Kurzweil integral. The value of the integral (in the Riemann or Henstock sense) can be derived in various ways. For example, the value can be determined from attempts to eva...
@Chris'ssis for motivation, math.stackexchange.com/questions/825869/… and let $\displaystyle I=\int ^\infty_0 \frac{b\sin{ax} - a\sin{bx}}{x}dx$, and see what $I/(ab)$ look like. Rewriting the inside as an integral, I got the computation I asked you
@BalarkaSen have you ever seen the concept of "expected number of different factorizations" in a non-UFD? Like, take $\{z \in\mathbf Z[\sqrt 5]:N(z)<M\}$. Calculate the expected number of factorizations for this set. Let $M\to\infty$ ($N$ is the norm)
@BalarkaSen I think so, I don't know this nomenclature :-) For example, $6$ has two different factorizations: $2\cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5})$ (sorry, I meant $\mathbf Z[\sqrt{-5}]$)
@N3buchadnezzar for which version of the lemma ? for $C^1$ functions it's easy, for just continuous, you need to use density of step functions on $[a,b]$
@N3buchadnezzar Let $\mu = \frac{1}{T}\int_0^T g(t)\,dt$ and $h(t) = g(t) - \mu$. Then $\int_0^T h(t)\,dt = 0$, and you want to show $$\lim_{n\to\infty} \int_a^b f(x) h(nx)\,dx = 0.$$ You do that like Riemann-Lebesgue, assuming $h$ is bounded [e.g. if it is continuous]; take $f\in C^1$ and integrate by parts, approximate general $f$ in $L^1$ by $C^1$ functions. If it's unbounded, a little more care to show it would be necessary.
If we have some closed disc centered at $x$ with points $y \leq r$ from $x$, how can $||z - y|| > r$ not belong to that disk when $y$ is inside that disc and also inside the disc centered at $z$
@nablablah If a point is outside a closed disk (disk isn't important), then it never is "immediately adjacent" to the disk, it always has a positive distance from the disk. For an open disk, you can say that the boundary points (the points of the bounding circle) are immediately adjacent to the disk, but for a closed disk, it's either in the disk, or some distance away from it.
@nablablah Let's for simplicity look at the closed unit disk $D = \{ x : \lVert x\rVert \leqslant 1\}$. Then if $z \notin D$, that means $\lVert z\rVert > 1$, and $z$ is at a distance $\lVert z\rVert-1$ from $D$.
But if a point can be "immediately adjacent" to a disk, isn't that a positive distance of 1? Since the point is 1 "distance" over from the point on the disk?
Ok
@DanielFischer What if we consider something like the natural numbers in $\mathbb{R}$ and we have a like a closed set $1 \leq x \leq 3$. If we consider the point $4$, isn't that a positive distance from $3$?
@nablablah Yes, that is at a positive distance from $3$. But, in the integers, you can only move in discrete steps (of integer size), so one could then say it is immediately adjacent, since there is no integer between $4$ and the set. However, in $\mathbb{R}^n$, we can move arbitrary distances, so whenever we have a positive distance, we have lots of points in between.
Just as I was commenting on someone who followed the first part of my hint and pisted a separate answer ... Although he may have been typing as I posted.
@TedShifrin I would like to ask your opinion. Do you think it is important to teach Krull Schmidt theorem and Maschke theorem in a grad algebra course?
@Jasper: My taste is not typical. I have never followed any single text, and much of what I do is more differential-forms based than any of the standard texts. These days, doCarmo and Petersen are the two most-used texts, but neither fits my personal taste. This last time I proved Chern-Gauss-Bonnet, which is not standard; the previous time, I taught some complex geometry because the students (all undergrads) wanted to see that.
If I get totally bored when I've retired, I may get inspired to TeX up my various grad course notes, but I doubt it.
@GTR, if $f(x)\sim \sum c_ke^{ikx}$, then $f'(x)\sim \sum ikc_ke^{ikx}$, so they're not getting smaller when you differentiate. But if $f$ is $d$ times differentiable, that tells you that you need the $c_k$ to decay sufficiently rapidly for $\sum k^dc_k$ to converge appropriatley @GTR
Of course, the OP had garbage in his post, which I kindly corrected in my comment :) $(\operatorname{Col}(A))^\perp = (\operatorname{Nul} A)^\top$, eh?
@TedShifrin I noted that $\Bbb C(z)$ is the function field of $\Bbb P^1$ and $G = \text{Gal}(L/\Bbb C(z))$ is the deck transformation group of the (branched) galois covering $X \to \Bbb P^1$ for some Riemannsurface $X$. Another covering is $X_0 \to \Bbb P^1$ corresponding to $L'$. One has $\Bbb P^1 \cong X/G$ and $\Bbb P^1 \cong X_0/G$ so $X \cong X_0$.
That was the garbage. But then the kid told the OP to do that in his "comment." Sigh. After I said it was nonsense, he said he hadn't finished editing. @DanielF
Whoa, in general, that can be subtle, @Balarka.
I don't think $X/G = Y/G \implies X=Y$ without something.
But I don't want to think about it now. I have to go back to the kitchen and finish decalcifying my espresso machine :D
You're assuming $G$ acts freely and properly or something? Otherwise, I can take $G$ acting trivially in one case and not in the other, for a stupid counterexample @Balarka.
I know I need to be downvoted sometimes. I told someone on Reddit the other day that if $U = e^{i x}$ then $U U^* = I$ for any $x$ in $\mathbb{C}$ facepalm