Articles of integration

Simple Integral Involving Radicals: Why Does Mathematica Fail?

I have $$\int_{d-1}^{3}\textrm{d}x\left(3-x\right)^3 \sqrt{\left(\frac{2(x-1)}{x}\right) \left(x-\left(d-1\right)\right)}$$ but despite this looking like a simple integral involving fractional powers of $x$ with shifts, Mathematica fails, despite restricting $d$ $\in[3,4]$ and $$\left(\frac{2(x-1)}{x}\right) \left(x-\left(d-1\right)\right)>0$$ Can anyone help with the integral? Why is Mathematica failing?

Find new region after applying polar change of coordinates

Let $R$ a region defined by the interior of the circle $x^2+y^2=1$ and the exterior of the circle $x^2+y^2=2y$ and $x\geq 0$, $y\geq 0$ Using polar coordinates $x=r\cos t$, $y=r\sin t$ to determine the region $D$ in $rt$ plane that corresponds to $R$ under this change of coordinate system (polar coordinates) i.e. since $T(r,t)=(r\cos t, […]

Why is the following a solution to the system?

I have the following in my notes, but I can’t remember how it works. Please help! $\nabla^2\psi=0, \quad\psi\to 0\quad\text{as}\quad x^2+y^2\to\infty, \quad\psi (x,y,0)$ is continuous Then by using Green’s function, we get the solution to be $$\psi(x',y',z')={z'\over 2\pi}\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty [(x-x')^2+(y-y')^2+z'^2]^{-3\over 2}\psi(x,y,0)\,\,\,dxdy\;.$$ (This part I am sure about.) The primed $x',y',z'$ are the variables introduced when using the […]

Differential equation with separable, probably wrong answer in book

I have a differential equation: $$ \frac{dy}{dx} = y \log(y)\cot(x)$$ I’m trying solve that equation by separating variables and dividing by $y\log(y)$: $$ dy = y \log(y) \cot(x) dx$$ $$ \frac{dy}{y \log(y)} = \cot(x) dx$$ $$ \cot(x) – \frac{dy}{y \log(y)} = 0 $$ Where of course $ y > 0 $ regarding to division Beacuse: […]

Difference between Riemann and KH integrals

To go from the definition of the Riemann integral ($f$ is Riemann integrable on $[a,b]$ if there exists a real $A$ such that $\forall \epsilon >0, \exists \delta>0$ such that $\forall D=\{([a_i,a_{i+1}],x_i)\}$ tagged partition of $[a,b]$ $h_i=a_{i+1}-a_i < \delta \implies |S_D(f) – A| < \epsilon$, where $S_D(f)$ is the Riemann sum on $D$) to the […]

Washers and Integrals

So I’m working on washers and I was given the equation of $$1/\sqrt{1+x^2}$$ and I am supposed to rotate the solid around the $x$-axis on the interval of $[-1,1]$. I know that I am supposed to use washers, but I can’t figure out how to find the equation for the outer radius and the inner […]

$\int_{\mathbb R^{2}} |\int_{\mathbb R} (f_{r}(t-y)- f_{r}(t)) g(t-x) e^{-2\pi i w\cdot t} dt|dx dw \to 0 $ as $ r\to \infty $?

Let $f\in \mathcal{S}(\mathbb R)$ with $\hat{f}$ has a compact support. For $r>0,$ put $f_{r}(x)= r^{-1}f(x/r), (x\in \mathbb R).$ We note that, $\int_{\mathbb R} |f_{r}(x)| dx = r^{-1} \int_{\mathbb R} |f(x/r)| dx = r^{-1}\int_{\mathbb R} |f(y)| r dy= \int_{\mathbb R} |f(x)| dx;$ and, $\hat{f_{r}} (\xi)= \int_{\mathbb R} f_{r}(x) e^{-2\pi i \xi \cdot x} dx= \hat{f}(r\xi); (\xi […]

Antiderivative of a function involving logarithms and a fraction.

Let $l\ge 0$ be an integer. Using integration by parts and the definition of the polylogarithm we have found the following identity: \begin{equation} \int \frac{[\log(\xi)]^l}{(1+\xi)^2} \log(1+\xi) d \xi = \sum\limits_{p=0}^l l_{(p)} \cdot [\log(\xi)]^{l-p} \cdot \phi_p(\xi) \cdot (-1)^{p+1} \end{equation} where the functions $\phi_p$ are given as follows: \begin{eqnarray} \phi_0(\xi) &:=& \frac{\log (\xi +1)+1}{\xi +1} \\ \phi_1(\xi) […]

Uniqueness of the Solution to Fredholm's Integral Equations of the First Kind

Any one who knows the conditions for Uniqueness of the Solution to Fredholm’s Integral Equations of the First Kind? Thanks.

Generalization of the Beltrami identity to functionals with higher derivatives

Suppose that I have a functional $S[q]=\int_a^b L(t,q(t),q'(t))\,dt$. Such a functional is well-known to extremized by a choice of $q(t)$ satisfying the Euler-Lagrange equation $\dfrac{d}{dt}\dfrac{\partial L}{\partial q’} = \dfrac{\partial L}{\partial q}$. If the integrand has moreover no explicit $t$-dependence, then one can show that the Beltrami identity holds: $L-q’\dfrac{\partial L}{\partial q’}=C$ for some constant $C$. […]