Search Results

Absent details about the function space, I'll assume that everything is sufficiently smooth such that we can infer from the integral equation that $g(0)=0$. Then, there is a whole set of solutions of …

Transforming to polar coordinates, $$ \int_{D}\dfrac{1}{\sqrt{(x-(1-\epsilon))^2 +y^2}} \frac{1}{\sqrt{1-\sqrt{x^2+y^2}}} \, dx \, dy = $$ $$ \int_{0}^{1} dr \frac{1}{\sqrt{1-r} } \int_{0}^{2\pi } d\p …

The integral formula as written does not continue to hold. Let's construct a counterexample. First, note that, in general, the $i$-th component of the integral in question is $$ \begin{equation} \left …

One can use, for example, $n \in \mathbb{Z}^3$ with the following restrictions (edited upon comment by Alexei Kulikov to be careful about cases with $n_i =0 $): $n_1 >0 \ \ \lor \ (n_1 =0 \land n_2 > …

The answer is no. A counterexample is $$ f(t,x) = x - \frac{1}{2} -\frac{1}{24} t^2 $$ $$ B(t,s,x) = \left( x-\frac{1}{2} \right) (t-s) $$ (Method: I obtained this by expanding $f$ and $B$ into power …

I asked Mathematica about the boundary behavior. First, the $P^{1/4}_{\nu } $ solution: We have, for $\epsilon \searrow 0$, $$ (\sin \epsilon )^{1/4} P^{1/4}_{\nu } (\cos \epsilon ) = \frac{2^{1/4} }{ …

A differential equation for ${\cal A} (x) $ can be obtained as follows, $$ \frac{d^3}{dx^3 } {\cal A} (x) = \int_{-\infty }^{\infty } dk\, (-ik^3 ) e^{ikx} e^{-k^4 } = \frac{x}{4} \int_{-\infty }^{\in …

The extended operator can be treated along similar lines as the $c=0$ case. One merely has to modify the algebra a little. Again, denote the standard harmonic oscillator eigenfunctions (i.e., the eige …

First some heuristics, before constructing the complete answer - this looks a bit more transparent if one considers $$ A^2 = \begin{pmatrix} -\partial_{x}^{2} +x^2 & 1 \\ 1 & -\partial_{x}^{2} +x^2 \e …

You seem to have a typo in your value of $a$ - evaluating the sum as is yields something more like 1.76. Converting to an integral, $$ \lim_{n\rightarrow \infty } \frac{1}{n^2 } \sum_{i=1}^{n} \sum_{j …

We can obtain power series in $a$ by expanding in geometric series. Since the specific treatment depends on the integration region, let's introduce lower and upper limits, $y_0 <y$, $$ I = \int_{y_0 } …

By noting that $-i\partial_{x_1} $ is diagonalized by $e^{ik_1 x_1} $ and $-i\partial_{x_2} $ by $e^{ik_2 x_2} $, the problem reduces to a $2\times 2$ diagonalization for each $(k_1,k_2)$-block. The r …

$$ \left( \begin{array}{c} 1 \\ \frac{k_x - ik_y }{\sqrt{k_x^2 + k_y^2 } } \end{array} \right) e^{i(k_x x + k_y y)} \ \ \ \mbox{with eigenvalue} \ \ \ i\sqrt{k_x^2 + k_y^2 } $$ and $$ \left( \begin{ar …

If you allow for special functions, you of course allow for quite some beasts. Generally, your expression is $$ I = \int_{0}^{1} dx \left( \sum_{k=0}^{n} (f(x))^k \right)^{\alpha } = \int_{0}^{1} dx \ …

I can give an answer to your first question - below, I'll give a translation of the "Descartessche Regel" from a suitably old German book (to make the experience authentic). Before getting to that, t …