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6

Note $z=x^2+y^2>0$, therefor $$V=\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\int_{x^2+y^2}^{\sqrt{2-x^2-y^2}}dzdydx$$ Now apply cylinder coordinate.


4

Through the substitution $x=e^{-t}$ the original integral equals: $$ I=\int_{0}^{+\infty}\left(2\frac{e^{-t}-1}{t^2}+\frac{e^{-t}+1}{t}\right)e^{-t}\,dt \\=2\color{purple}{\int_{0}^{+\infty}\frac{e^{-t}-1+t}{t^2}\,e^{-t}\,dt}+\color{blue}{\int_{0}^{+\infty}\frac{e^{-t}-1}{t}\,e^{-t}\,dt}$$ where the blue integral is yet manageable through Frullani's theorem (...


4

We can write the interal in $(4)$ as $$I=-\int_{0}^{1}\frac{1-x+\log\left(x\right)}{\log^{2}\left(x\right)}dx $$ now define $$I\left(\alpha\right)=-\int_{0}^{1}\frac{x^{\alpha}\left(1-x+\log\left(x\right)\right)}{\log^{2}\left(x\right)}dx,\,\alpha\geq0 . $$ We have $$I''\left(\alpha\right)=-\int_{0}^{1}x^{\alpha}\left(1-x+\log\left(x\right)\right)dx=-\...


4

I don't know if this counts as a "closed form", but if $a\in(0,1)$ we have: $$ \int_{0}^{a}\frac{\log(1+x)}{x}\,dx = \int_{0}^{a}\sum_{n\geq 1}\frac{(-1)^{n-1} x^{n-1}}{n}\,dx = \sum_{n\geq 1}\frac{(-1)^{n-1} a^n}{n^2}. \tag{1}$$ that can be written as $-\text{Li}_2(-a)=\text{Li}_2(a)-\frac{1}{2}\text{Li}_2(a^2).$ By taking the limit as $\alpha\to 1^-$ we ...


3

The question specifically asks to find an equation for the constant $c$ that minimizes this expression. The typical method here is to let this expression be a function of $c$ and find the minimum of this function with respect to $c$. Thus, we have $$ f(c) = \int_0^1 |e^x - c| \ dx$$ We have to be careful here, as we are not sure if the expression inside of ...


3

You can rephrase your question as finding $c$ to minimize $E[|e^X-c|]$ where $X$ is a uniform random variable. In general, The $c$ that minimizes $E[|Y-c|]$ is the median of the distribution of $Y$. Some intuition can be found in this answer. Here, we see the median of $X$ is $1/2$, and since $x \mapsto e^x$ is increasing, the median of $e^X$ is $e^{1/...


3

Use substitution $u=\frac{x}{x+1}$. Then $x=1-\frac{1}{1-u}$ and \begin{align} \int_{0}^{\infty}\frac{1}{x(x+1)}\ln(x+1)dx=\int_{0}^{1}\frac{1}{u}\ln\left(\frac{1}{1-u}\right)du=Li_{2}(1)=\frac{\pi^{2}}{6} \end{align} where $Li_{2}(z)$ is dilogarithm.


3

Let $I$ denote the integral. Substituting $u = \ln(x+1)$, we get: $$I = \int_0^{\infty} \frac{u}{e^u -1 }du$$ So: $$I = -\int_0^{\infty} \frac{ue^{-u}}{1- e^{-u}} du = - \int_0^{\infty} ue^{-u} \sum_{n=0}^{\infty} (e^{-u})^n du = -\sum_{n=0}^{\infty} \int_0^{\infty} ue^{-(n+1)u}du$$ Integrating by parts, $$\int_0^{\infty} ue^{-(n+1)u}du = -\frac1{(n+1)^...


3

It is comfortable to describe your solid in cylindrical coordinates since both the sphere and the paraboloid have the $z$-axis as an axis of symmetry (both surfaces are surfaces of revolution around the $z$-axis). In cylindrical coordinates, the sphere is described by $\rho^2 + z^2 = 2$ and the paraboloid is described by $\rho^2 = z$. The solid bounded ...


2

Let $u = \ln(x+1) \implies x = e^u-1,dx = e^udu, x+1 = e^u $ The integral becomes: $$\int_{0}^{\infty} \frac{1}{x(x+1)}\ln(x+1)dx= \begin{equation} \int_0^{\infty} \frac{u}{e^u-1} du \end{equation}$$ Which is similar to this post which shows that $\begin{equation} \int_0^{\infty} \frac{u}{e^u-1} du \end{equation} = \frac{\pi^2}{6}$


2

By the change of variable $x=u^{1/4}$, $dx=\dfrac14u^{-3/4}du$ you get $$ \int_0^\infty\sin{(x^4)}dx=\frac14\int_0^\infty\frac{\sin{u}}{u^{3/4}}du $$ then it is clearer how to apply the Dirichlet test for integral.


2

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2

The computation of sum of squares is wrong. $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$


2

They're not asking for all the coefficients, just which ones are zero. Your friend seems to be making arguments to linearity. In the first two (top) graphs, they split into the "1" part and the "sin" part, each 0 on the half of the period. Then he writes the first part as a sum again: a constant 1/2, and an alternating-sign 1/2. (The first two graphs on the ...


2

This is similar to the answer from anomaly, but I hope to make it easier to understand by moving the essential point out of a little subscript and into the text. The first equation in the question, $A=\iint dx'\,dy'$ is fine when the integral is over $A$, as you said. But the next bit, $\dots=\iint G\,dx\,dy$ requires that the integral be over a different ...


1

It's only in cartesian (rectangular) coordinates that $A = \iint_\Omega{\rm d}x\,{\rm d}y$ denotes the area of the region $\Omega$. When we change to a different coordinate-system then the Jacobian describes how the area of a small element of size ${\rm d}x'\,{\rm d}y'$ in the new coordinates is related to the corresponding area of size ${\rm d}x\,{\rm d}y$ ...


1

Here is a slightly different variation to OPs example, followed by another one. Suppose $p$ is an even function, i.e. $p(x)=p(-x)$ and $q(x)q(-x)=1$. Then \begin{align*} \int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\int_{0}^{a}p(x)\,dx \end{align*} A proof of the statement together with an application can be found in this answer. Note: This technique ...


1

Let the integral be denoted as $I$. Make the substitution $u={x+1}$ and $du=dx$ we have $$I=\int_{1}^{\infty}\frac{\ln(u)}{(u-1)u}du.$$ Now let $u=\frac{1}{z}$ so that $du=\frac{-1}{z^2}dz.$ Now $I=\int_{0}^{1} \frac{\ln(z)}{z-1}dz.$ Now rewrite the integrand as a geometric series as such $$\frac{\ln(z)}{z-1}=-\sum_{n=0}^{\infty}\ln(z)z^n.$$ Replace this ...


1

$$\begin{align}\int_0^\infty\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x &= \int_0^1\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x + \int_1^\infty\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x \\ &= \int_0^1\frac{\ln(1+x)}{x(1+x)}\mathrm{d}x + \int_0^1\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}(1+\frac{1}{x})x^2}\mathrm{d}x\tag{1} \\ &= \int_0^1\frac{\ln(1+x)}{x}-\frac{\ln(1+x)}{1+x}\...


1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


1

Let $t=x^2+1$ then your integral becomes $$\frac{1}{2}\int \frac{t-1}{t^2-7}\,dt=\frac{1}{2}\int \frac{tdt}{t^2-7}- \frac{1}{2}\int \frac{dt}{t^2-7}.$$ Now $$\int \frac{tdt}{t^2-7}=\frac{1}{2}\ln(t^2-7)$$ and $$\int \frac{dt}{t^2-7}=-\frac{1}{\sqrt{7}}\mbox{arctanh}(t/\sqrt{7})$$


1

The solution is not hard: I think: $$a_{n} = \frac{1}{\pi} (\int\limits_{-\pi}^{0}-1cos(nx)dx + \int\limits_{0}^{\pi}sin(x)cos(nx) dx) = \frac{1}{\pi} \int\limits_{0}^{\pi}sin(x)cos(nx) dx = \frac{1}{\pi} \frac{cos(n\pi)+1}{1-n^2}$$ $$b_{n} = \frac{1}{\pi} (\int\limits_{-\pi}^{0}-1sin(nx)dx + \int\limits_{0}^{\pi}sin(x)sin(nx) dx)=\frac{1}{\pi} (\frac{...



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