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1d
comment Computation of an iterated integral
(+1 from me) I modified your answer to include a limit. There's no way to interpret the integral except as some improper integral, and this interpretation is the only sensible one; and it makes your solution the best one.
1d
revised Computation of an iterated integral
added 23 characters in body
2d
comment Fourier transform of $\frac{\sin(6\pi t)}{t}$
@lucad93 : I like this approach because you don't have to assume much of anything. There is a scaling property of $\frac{dt}{t}$ being used, where $t=\rho s$ gives $\frac{dt}{t} = \frac{ds}{s}$, but that's fairly straightforward.
Feb
3
comment Use the Fourier transform to find value of definite integral from negative infinity to infinity
$\Im (e^{ix}-1)=\Im e^{ix}=\sin(x)$. Any real constant $a$ gives $\Im(e^{ix}-a)$, but you want the integral to converge, which means $a=1$ is the correct choice.
Feb
3
answered Question about proof of Fourier Transform of derivative
Feb
3
answered Fourier transform of $\frac{\sin(6\pi t)}{t}$
Feb
3
answered Use the Fourier transform to find value of definite integral from negative infinity to infinity
Feb
3
answered Solve problem using Fourier's transform
Feb
1
comment Neumann Series for integral equation with inhomogeneous term zero
If the Neumann Series converges, then it should converge to a solution. And it does converge in this case to the $0$ solution.
Feb
1
comment Reference request: about inverse Laplacian operator
That's what an orthonormal basis of eigenfunctions is. It's a Fourier basis. Take a look at the theorems in the section of Evan's text that I mentioned.
Feb
1
comment Neumann Series for integral equation with inhomogeneous term zero
If $f=0$, then $\phi=0$ is a solution.
Feb
1
answered Extension of Fourier transform to complex analytic function
Jan
31
answered Self-adjoint operators, projections, and resolutions of the identity.
Jan
31
comment Positive self-adjoint operators and norm resolvent convergence
Your name is an oxymoron.
Jan
31
answered Positive self-adjoint operators and norm resolvent convergence
Jan
31
comment Example of operator with spectrum equal to $\mathbb{C}$?
@JanWesterdiep : You're welcome. Differentiation is a strange example of several things. If you compute the resolvent of $\frac{1}{i}\frac{d}{dt}$ on the domain with periodic conditions $f(0)=f(2\pi)$, you get first order poles at the integers, and the sum of the residues of $-(T-\lambda I)^{-1}f$ is the Fourier series for $f$.
Jan
31
comment Example of operator with spectrum equal to $\mathbb{C}$?
@user1952009 : The domain $\mathcal{D}(T)$ is dense in the underlying space $L^2[0,1]$. That is, $\mathcal{D}(T)^c=L^2[0,1]$, where $\mathcal{D}(T)^c$ is the topological closure of $\mathcal{D}(T)$ in $L^2[0,1]$.
Jan
31
comment Mathematical meaning of certain integrals in physics
@Self-teachingDavide : Now look at the Helmholtz Theorem, which allows you to work with gradient and curl. By the way, the Laplace result is needed to boostrap up to the Helmholtz Theorem. $\nabla\cdot B=0$.
Jan
31
answered Example of operator with spectrum equal to $\mathbb{C}$?
Jan
31
comment Image of a dense set through unbounded operator
@Confusio : A restriction of $I+T^{\star}T$ to a subspace $\mathcal{D}$ has a dense range iff $S=I+T^{\star}T$ on $\mathcal{D}(S)=\mathcal{D}$ is essentially selfadjoint, meaning that the closure of $S$ is selfadjoint. This is because $(Sx,x) \ge \|x\|^2$ for all $x\in\mathcal{D}$. In fact, the closure of such an $S$ with dense range will always be $I+T^{\star}T$.