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11

Let the desired integral be denoted by $I$. Noting that $1-(2t-1)^2=4t(1-t)$ we see that $$ I=4\int_{\square}\frac{-\log(1-(2x-1)^2)+\log(1-(2y-1)^2)}{(2x-1)^2-(2y-1)^2}dxdy\quad\hbox{with $\square=[0,1]^2$}. $$ But, for $t\in(-1,1)$, we have $$ -\log(1-(2t-1)^2)=\sum_{n=1}^\infty\frac{(2t-1)^{2n}}{n} $$ So the integrand becomes, for $(x,y)\in(0,1)^2$, ...


3

You can find the limits of integration by solving the inequalities. But in this case, the inequalities are already in solved form, if you integrate in the order $x,y,z$ (listed from innermost to outermost), so there's nothing to do. If you wanted to integrate in the order $z,y,x$, we could solve in this other direction. In this case, it's pretty easy: we ...


3

One way of viewing the hyperbolic functions is that they are "abbreviations" for combinations of exponential functions, so whatever can be done using hyperbolic functions can equally be done using exponentials - albeit in a less concise manner. Similarly, inverse hyperbolics are really just log functions in various combinations. What this means is that if ...


2

You might think about a diagram like this: region between the white planes for x, region between the pink planes for y, region between the cyan planes for z. The solid is a bit easier to see now, demarcated with blue points


2

Try taking the indefinite integral of both sides instead: \begin{align*} \frac{ds}{s} &= \mu \, dt \\ \int \frac{1}{s} \, ds &= \int \mu \, dt \\ \ln|s| &= \mu t + C &\text{for some constant $C \in \mathbb R$}\\ e^{\ln|s|} &= e^{\mu t + C}\\ |s| &= e^{\mu t} \cdot e^C \\ s &= \underbrace{(\pm e^C)}_{A}e^{\mu t} \\ s(t) &= ...


2

Let $ \displaystyle I(z) = \int_{0}^{\infty}\frac{\arctan \frac{x}{z}}{e^{\pi x}-1} \ dx$. Then $$ I(z) = \int_{0}^{\infty} \int_{0}^{\infty}\frac{1}{e^{\pi x}-1} \frac{\sin xt}{t}e^{-zt} \ dt \ dx$$ $$ = \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin tx}{e^{\pi x}-1} \ dx \ dt$$ $$ = \frac{1}{2}\int_{0}^{\infty} \frac{e^{-zt}}{t} ...


2

Any indefinite integral, which you can integrate to get a known function can be done just based on the fundamental theorem of calculus. All the other tricks like $u$-substitution, partial fractions, integration by parts only make this procedure easier, i.e., it helps us guessing the antiderivative.


2

I think the best justification is that it leads into the general context of substitutions for multiple integration. For example, it can be shown that the Jacobian of a polar transformation $$T(r,\theta)=(r\cos \theta,r\sin \theta)$$ is given by $$\begin{vmatrix}\cos\theta&\sin\theta\\ ...


2

This one is perhaps easier than the arbitrary function $f(x)$ in the example in the comments. First note that: $$ \begin{aligned} &\int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n\\&=\Im{\left[\int_0^1 \int_0^1 \ldots \int_0^1 \text{exp} \bigg(i\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 ...


2

Your intuition is correct: for the first part, it is equal to $\lfloor \frac{x}{\pi} \rfloor \int_{0}^{\pi}\sin (t) dt$, which you can compute. The second one is equal to $\int_{0}^{x-\pi \lfloor \frac{x}{\pi} \rfloor}\sin (t) dt$. As pointed out in comments, we can argue like this since we are looking at $|\sin|$. Morally, $\int_0^{\lfloor ...


1

Well, your function does not depend on $\phi$ for the very beginning therefore it is correct that you have no dependancy in the end :) Some more details that I noticed: "$f(\theta, \phi) = u_0$ for the range $[0 \le \theta < \pi/2]$" <- but you have to integrate over range $[0 \le \theta < \pi]$, what's the $f(\theta, \phi)$ for $[\pi/2 \le \theta ...


1

It looks to me that your problem is: $$\iiint_{D}x^{2}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z,$$ Where $D$ is the region described by: $x^{2}+2y^{2}+z^{2}\leq 2$. Which is a triple integral, rather than the double integral that you describe, however, the process is much the same. We first describe our transformation from Cartesian co-ordinates into the new ...


1

A different manner to answer to the question is to show a graphical representation to the function $f(x)$ for various values of $n$ and a graphical representation of the respective integrals $F(X)$. ( The old saw that “One little picture says more than a long speech”) This makes more understandable the behavior of the function and integral for $n$ tending ...


1

You need to show that there is linear subspace $U\subset C^1(I,\mathbb{C})$ and a vector $v_0\in C^1(I,\mathbb{C})$ such that every solution of $(2)$ lies in $v+U$. We claim that $U$ is a set of solutions of $(1)$, $v_0=f_0$. Indeed, by paragraph $(ii)$ we already know that every solution of $(2)$ is a sum of $f_0$ and some solution of $(1)$. Thus we ...


1

Parts (ii), (iii), and (v) are pure linear algebra. We are given a linear map $$L:\quad C^1(I)\to C^0(I),\qquad f\mapsto f'+a\> f\ .$$ Solving the ODE $(2)$ means finding the $f$'s in $C^1(I)$ with $Lf=b$, where $b\in C^0(I)$ is given in advance. Denote the set of these $f$'s by ${\cal S}$. (ii) When $f$, $f_0\in{\cal S}$ then $L(f-f_0)=Lf ...


1

From the picture it is clear that if you want to change the order of integration you should integrate over the region D. Region D can be divided in two parts: For D1 we have: integrate from $x=1$ to $x=y^2$, and $y\in(1,3)$, and for D2 we integrate from $x=1$ to $x=9$ and $y\in(3,4)$. Calculus for D1: $$ D_1=\int^3_1 dy \int^1_{y^2} dx $$ and for ...


1

Yes, the $Max$ function of two Riemann -integrable functions is Riemann -integrable. Thisis because the $Max$ of two a.e -continuous bounded functions is also a.e -continuous and bounded. For $f,g$ continuous, the function $Max${f,g} is continuous. This implies that (since a Riemann-integral must be a.e -continuous) , that $Max${f,g} is a.e -continuous; the ...


1

The argument looks fine, although the notation is somewhat sloppy: for instance I would write the first step as $$\int_0^{\infty} f(x)\cos(nx)\,dx = \cos(nx)\int_0^x f(y)\,dy\ \biggr\vert_{x=0}^{\infty} + n\int_0^{\infty} f(x)\sin(nx)\,dx.$$ The claim becomes false if $f(x)$ is locally integrable but not $L^1$, i.e. if $\lim_{x\to \infty} \int_0^x ...



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