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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| \ dxWe 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: $$\int_{0}^{\infty} \frac{1}{x(x+1)}\ln(x+1)dx= $$\int_0^{\infty} \frac{u}{e^u-1} du$$$$ Which is similar to this post which shows that $$$\int_0^{\infty} \frac{u}{e^u-1} du$$ = \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.

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