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8

Let $x_0$ be such that : $$\forall x \in [0,1], \quad f(x) - x \leq f(x_0) - x_0.$$ Then : $$\int_0^1 [f(g(x)) - g(x) ] \,dx \leq \int_0^1 [f(x_0) - x_0] \,dx = f(x_0) - x_0$$ And : $$\int_0^1 f(x) \,dx \geq \int_{x_0}^{1} f(x) \,dx \geq (1 - x_0) f(x_0) \geq f(x_0) - x_0$$

5

$$6 x^3+11 x^2 y+6 x y^2+y^3=x$$ convert in polar coordinates $$x=r\cos t;\;y=r\sin t$$ we get $$r^3 \sin ^3 t+6 r^3 \cos ^3 t+11 r^3 \sin t \cos ^2 t+6 r^3 \sin ^2 t \cos t=r \cos t$$ $r$ can be simplified and the equation can be solved wrt $r^2$ $$r^2 \sin ^3 t+6 r^2 \cos ^3 t+11 r^2 \sin t \cos ^2 t+6 r^2 \sin ^2 t \cos t- \cos t=0$$ $$r^2=\frac{\... 5 The close-form result can be expressed as$$\color{blue}{ \int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx = -nG+\frac\pi2 n \ln 2 + \pi \sum_{k=1}^{[\frac n2]}\ln \cos\frac{(n+1-2k)\pi}{4n} } as shown below. Note that \begin{align} I_n = \int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx \overset{x\to\frac1x} == \frac12 J_n - nG \end{align} where  \int_1^\infty\frac{\ln ... 4 As done here by user :Fred By IBP\int_0^1 x^3 f'(x)=0-\int_0^1 3x^2 f(x)$$So$$\int_0^1 x^3 f'(x)=-1$$Now by cauchy schwarz$$1={\left(\int_0^1 x^3 f'(x)\right)}^2\le \int_0^1 x^6\int_0^1 f'(x)^2=\frac{1}{7} \cdot 7$$That means for equality of C-S must occur ie f'(x)=\eta x^3\implies f(x)=\frac{\eta}{4} x^4 + C Now to find \eta substitute it ... 4 No. If G(h)=\int_3^{3+h}e^{t^2}dt, then G'(h)=e^{(3+h)^2}. 3 Both the integral and the series converge if and only if u > 2 \sqrt{g R}, but your formula for T is correct only if u \geq \sqrt{5 g R}. This discrepancy can be understood as follows: The first condition only ensures that the particle has enough energy to reach the height 2R (it can be found by equating 2 R m g and \frac{1}{2} m u^2). But ... 3 Call problem A the one you have drawn: the tracks may only provide a normal force radially inwards. Call problem B the similar problem except that cart and track are replaced by bead on a wire. In A, the cart may fall off the track. In B, the 'cart' may not fall off the 'track'. For problem A: Minimum initial speed to reach height of 2R: at least ... 3 The instantaneous speed is s(t)=\sqrt{x'(t)^2+y'(t)^2}, and the average speed is \frac{1}{T}\int_0^Ts(t)\,\mathrm d t, where T=20 \,\text{s}. Does it make sense? 3 Assuming \epsilon>0, the given integral, after the substitution -ix=z, is$$i\int_{\epsilon-i\infty}^{\epsilon+i\infty}z^{-2}e^z\log z\,dz=-if'(2),\qquad f(s)=\int_{\epsilon-i\infty}^{\epsilon+i\infty}z^{-s}e^z\,dz=\frac{2\pi i}{\Gamma(s)}$$(the last equality is basically Hankel's integral). The final result is then \color{blue}{-2\pi(1-\gamma)}. 3 Not a complete answer, but an elaboration of what I said in the comments. To evaluate$$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$$Write it as$$\int_1^\infty\frac{\ln\left(x^{-2n}+1\right)+\ln\left(x^{2n}\right)}{1+x^2} dx = \int_1^\infty\frac{\ln\left(x^{-2n}+1\right)}{1+x^2} dx+2n\int_1^\infty\frac{\ln\left(x\right)}{1+x^2} dx$$The second integral is ... 3 The lmit is \lim_{h \to 0}\frac {e^{(3+h)^{2}}} 1=e^{9}. 3 You made a mistake in the limit it should be e^{(3+h)^2} 3 No this is false, let f(h)=\int_0^h e^{t^2}dt, then \lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=f'(3)=e^9. You can also use L'Hopital's rule :$$ \lim\limits_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h\rightarrow 0}\frac{f'(3+h)}{1}=f'(3)=e^9 $$3 This is a nice problem.$$f(x)=\underbrace{\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}}_{a}+\underbrace{\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}}_b$$Now note that from the identity (a+b)^3=a^3+b^3+3ab(a+b), and using ab=-\dfrac{1}3, we have:$$(f(x))^3=2x-f(x) \implies (f(x))^3+f(x)=2x $$And from g(x)=x^3+x+1 , we have$$g(f(x))=(f(x))^3+f(x)+1=2x+1$$Therefore ... 3 Note that f^{-1}(x) = \dfrac{x^3+x}{2}\\ \therefore g(x) = 2f^{-1}(x)+1\\ \therefore g(f(g(x))) = 2f^{-1}(f(g(x)))+1\\ and I'm sure you know f^{-1}(f(x)) = x \\ \therefore g(f(g(x))) = 2g(x)+1\\$$\therefore \int_0^4g(f(g(x)))dx = \int_0^4(2g(x)+1)dx = \int_0^4(2x^3+2x+3)dx\\$$3 I think (at least for the time being) tha we need to combine two things$$\frac {x^2}{x^4+1}=\frac{1}{2 \left(x^2+i\right)}+\frac{1}{2 \left(x^2-i\right)}\log(1-x^4)=\log(1+x^2)+\log(1-x^2)$$Each of the four integrals$$\int_0^1\frac{\log \left(1\pm x^2\right)}{2 \left(x^2\pm i\right)}\,dx$$shows a complex expression involving a bunch of ... 2 I think that we can make approximations. First of all, we have$$\int_0^{2\pi} \sinh^{-1}(\text{argument})\,d\phi=2\int_0^{\pi} \sinh^{-1}(\text{argument})\,d\phi$$What I think is that we could expand as (at least) two series : one around \phi=0 to compute the integral between 0 and \frac \pi 2 and another one around \phi=\frac \pi 2 to compute the ... 2 I think you're right! Your domain \Gamma is defined by the inequalities:$$\left\lbrace\begin{matrix} 0 & < & x & < & 1 \\ 0 & < & t & < & T \\ 0 & < & x+t & < & 1 \end{matrix}\right. $$You can translate these into inequalities with (s,t), they then read:$$\left\lbrace \begin{matrix} 0 &...

2

Actually more is true: the Fourier transform of $J_0(ax)J_0(x)^2$ for any $a\in \mathbb{R}$ can be expressed in elliptic integral $K(m) = \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$. Here is the explicit form of the already nontrivial case $a=1$: let $$\tag{1}G(w) = \frac{2 \sqrt{4-3 t}}{\pi ^2 w(1-t)}K(m_+)K(m_-)$$ with $$2v = 5-w^2+\sqrt{(w^2-9)(w^2-1)} \... 2 Not an answer, but if you want your cross-correlation integral to look like a convolution, define:$$g_2(x) = f_2(-x)H(-x) $$where H(x) is the Heaviside unit step function. Then$$\begin{align*} I(\alpha) &= \int_\alpha^\infty f_1(x)f_2(x-\alpha) dx\\ \\ &= \int_{-\infty}^\infty f_1(x)f_2(x-\alpha)H(x-\alpha) dx\\ \\ &= \int_{-\infty}^\infty ...

2

That is not a gaussian integral/integrand. That is: $$\int\limits_{c}^{\infty}e^{-a(x+b)^2}\,\mathrm{d}x=-\dfrac{\sqrt{{\pi}}\left(\operatorname{erf}\left(\sqrt{a}\left(c+b\right)\right)-1\right)}{2\sqrt{a}}$$ But if that's really what you meant, I don't think there is a closed form for this. If you choose $b=0$, one can write $$\int\limits_{c}^{\infty}\frac{... 2 Make the appropriate change of variable and$$\int K_0(t)\,dt=\frac{\pi}{2} \, t \,(\pmb{L}_{-1}(t) K_0(t)+\pmb{L}_0(t) K_1(t))$$where appear the modified Struve functions. Learn about them since they appear in most integrals of Bessel functions. 2 Note that each curve represents a set of three equally spaced pedals with each pedal subtending a polar angle of \frac\pi3. Thus, the area is$$\frac{3}{2}\int_0^{\frac\pi3} (f(\theta)^2-g(\theta)^2)d\theta = \frac{3}{2}\int_0^{\frac\pi3} (64\sin^2(3\theta)-25\sin^2(3\theta))d\theta=\frac{39\pi}4 $$2 Just to put in more closed form what @Varu Vejalla has evaluated I=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx \int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=I+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx \int_1^\infty\... 2 Here is the method to evaluate this integral. Source:- Some very challenging calculus problems by Joseph Breen 2$$n\pm(-1)^n\ge n-1$$shows divergence. 2 Your question is how we can evaluate the integral$$\int\limits_{t}^{\infty}x\left(\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{x^2}{2} \right\}\right)\,\mathrm{d}x = \frac{1}{\sqrt{2\pi}}\int\limits_{t}^{\infty}x\exp\left\{-\frac{x^2}{2} \right\}\,\mathrm{d}x.$$Substitute u=-x^2/2. The differential is going to be \mathrm{d}x=-x^{-1}\mathrm{d}u. This will ... 1 You said numerical methods is an option so I'm only posting this because of that. This just uses the generic quad method which I think is Gauss-Legendre integration. from scipy.integrate import quad from numpy import sqrt,arcsinh, pi,cos def integrand(x, h, s,R): return arcsinh(h/sqrt(s**2 +R**2 -2*s*R*cos(x))) ### parameters h = 1 s = 1 R = 10 I = ... 1 By Fubinis theorem:$$I_p = \int\limits_0^{+\infty}\,\int\limits_{-\infty}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}x\,\mathrm{d}c =\int\limits_{-\infty}^{+\infty}\,\int\limits_{0}^{+\infty}\,e^{-p\,[c+(c^2+1)\cosh x]}\,\mathrm{d}c\,\mathrm{d}x.$$We can solve the integral with respect to c analytically:$$\int\limits_{0}^{+\infty}\,e^{-p\,[c+(c^2+...

1

Let $$H(t):=e^{-T}\int_{t-T}^t x(\tau)x^\top (\tau) d\tau-\int_{t-T}^t e^{\tau-t}x(\tau)x^\top (\tau) d\tau.$$ The goal is to show that $H(t)$ is a negative definite matrix for all $t\geq T$, i.e., we have $v^\top H(t) v\leq 0$ for all $v\ne 0$. Using the definition of $H(t)$, we have \begin{align} v^\top H(t) v&=\int_{t-T}^t (e^{-T}-e^{\tau-t}) (v^\top ...

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