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In general; $$\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f(t) \mathrm{d}t = f(x)$$ $$\frac{\mathrm{d}}{\mathrm{d}t}\int_0^x f(t) \mathrm{d}t = \color{red}{0}$$ Now let's spend some time on this as this is where you did a mistake. Note that $$y(?)=\int_0^x f(t) \mathrm{d}t$$ is a function of $x$!($?=x$)Why? Remember how you deal with definite integrals. You ...

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setting $t=\sqrt{2x}$ we get $x=\frac{t^2}{2}$ and $dx=tdt$ we get the integral $\int te^{t}dt$ in a few minutes i will post the solution setting $u=t$ and $v'=e^{t}$ we obtain $u'=1$ and $v=e^{t}$ thus we have $\int te^{t}dt=te^{t}-e^{t}+C$

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There is a nice Wikipedia page on this: Differentiation under the integral sign. Direct from that page we have $$\frac{\text{d}}{\text{d}x}\left( \int_{a(x)}^{b(x)}f(x,t)\text{d}t \right ) = f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}f_x(x,t)\text{d}t.$$ Then for $f(x,t) = t^2$ we have \frac{\text{d}}{\text{d}x}\left( \int_{0}^{x}t^2\text{d}t \right ... 2 \begin{align} \int_0^\infty \frac{\cos bx -\cos ax}{x}\, dx&=\int_0^\infty \bigg[\cos bx -\cos ax\bigg]\int_0^\infty e^{-xy}\,dy\,\, dx\\ &=\int_0^\infty\int_0^\infty \bigg[e^{-xy}\cos bx -e^{-xy}\cos ax\bigg] \,dx\, dy\\ &=\lim_{s\to\infty}\int_0^s \bigg[\frac{y}{y^2+b^2} -\frac{y}{y^2+a^2}\bigg] \,dy\\ &=\ln\left(\frac{a}{b}\right) ... 3 Using the result from this OP: Integral evaluation \int_{-\infty}^{\infty}\frac{\cos (ax)}{\pi (1+x^2)}dx. We have $$\int_0^\infty\frac{\cos ax}{1+x^2}\,dx=\frac{\pi}{2}e^{-|a|}$$ Thus, our integration is simply \begin{align} \int_0^\infty\frac{x\sin x}{1+x^2}\,dx&=-\lim_{a\to1}\partial_a\int_0^\infty\frac{\cos ... 6 According to the residue theory, \int_0^{+\infty}\frac{1}{s^2+x^2}\mathrm{d}x=\frac{\pi}{2s} ~ , ~ I(\alpha)=\int_0^{+\infty}\frac{x\sin \alpha x}{1+x^2}\mathrm{d}x Laplace transform: \begin{align} \mathcal{L}\left[ I(\alpha)\right] &= \int_0^{+\infty}\frac{x}{1+x^2}\cdot \frac{x}{s^2+x^2}\mathrm{d}x \\ &= ... 1 How to calculate the value of the integrals? Using Wolfram|Alpha Pro, one may obtain\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx=G-\frac{\pi^2}{16}+\frac{\pi}{4}\ln2$$and$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx=\frac{3G}{2}-\frac{\pi^3}{64}-\frac{3\pi^2}{32}+\frac{3\pi}{8}\ln2$$Sorry for the Cleo's style answers, but the answer style is ... 5 Since your integrand is even, this integral is equal to one half of$$ \int _{-\infty}^{\infty}\frac{x\sin x}{x^2+1}dx $$This integral is the imaginary part of$$ \int _{-\infty}^{\infty}\frac{x \cdot e^{ix}}{x^2+1}dx $$This can be solved as the contour integral with contour a half disc of radius R with base on the real axis, letting R go to ... 2 Denote$$I(y)=\int_0^{+\infty}\frac{\cos bx-\cos ax}{x}e^{-yx}dx, y\in[0,+\infty)$$Then$$I'(y)=-\int_0^{+\infty}(\cos bx-\cos ax)e^{-yx}dx=\frac{y}{y^2+a^2}-\frac{y}{y^2+b^2}$$and$$I(+\infty)=0$$So$$I(0)=-\int_0^{+\infty}(\frac{y}{y^2+a^2}-\frac{y}{y^2+b^2})dy=\ln\frac{a}{b}$$2 In general, if f(x) is a continuous function over (0, \infty) such that following two limits exist$$\begin{cases} f_0 &= \lim\limits_{x\to 0} f(x),\\ f_\infty &= \lim\limits_{x\to\infty} f(x) \end{cases}$$then$$\int_0^\infty \frac{f(bx) - f(ax)}{x} dx = (f_0 - f_\infty)\log\frac{a}{b}\tag{*1}$$This is known as Frullani integral. In your ... 1 Use the relation$$ \tan'x = 1 + \tan^2x $$The integral is$$ \int \frac{dx}{9 + x^2} = \frac13\int \frac{\frac 13dx}{1 + \frac {x^2}9} $$now in order to use the relation, substitute$$\frac {x^2}9 = \tan^2 t \implies \frac13x = \tan t $$Then$$ \frac13dx = \tan' t\times dt \\ \frac13\int \frac{\frac 13dx}{1 + \frac {x^2}9} = \frac13\int \frac{\tan' ...

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No. The function $\ln z$ is not meromorphic on any open set containing the circle $\{|z|\le1\}$. $z=0$ is a branching point.

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This integral is zero, since if $f(t) = e^{2t} u(\tau - t) t^2$ then $$\int_{- \infty}^{+ \infty} f(t) \delta(t) \, dt = f(0) = e^{0} u(\tau) 0^2 = 0.$$

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There are 2 proofs: a short one and a longer one. First Proof: f is bounded on [0,1] since |sin 1/x| is less than or equal to 1. f is continuous except at x=0 (since limit as x-> 0 doesn’t exist), so the set of discontinuities is the single point x=0 and this set has measure zero. There is a well known Thm. that if a function is bounded and continuous ...

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I suppose that $0<\delta<b-a$. Your last equality $$\int_{0}^{\delta}\left(\int_{a}^{b}( g(x+t)-g(x))dx\right)dt=\int_{a}^{b}\left(\int_{0}^{\delta} (g(x+t)-g(x))dt\right)dx$$ is true by Fubini's theorem. Now we have: $$\int_{a}^{b}( g(x+t)-g(x))dx=\int_{a+t}^{b+t}g(u)du-\int_a^b g(x)dx=\int_b^{b+t}g(u)du-\int_a^{a+t}g(u)du$$ Or: $$\int_{a}^{b}( ... 3 Just evaluate the definite integral of:$$\int_{-\pi}^{\pi}\cos[(m+n)x]dx$$You'll get:$$\frac{1}{m+n}*sin[(m+n)x]\Big|_{-\pi}^{\pi} = \frac{1}{m+n}*(sin[(m+n)*\pi]-sin[(m+n)*-\pi])$$What can you say about this? 8 Let :$$\eqalign{ & A = \int_{ - \pi }^\pi {\sin (mx)\sin (nx)} dx \cr & B = \int_{ - \pi }^\pi {\cos (mx)\cos (nx)} dx \cr} $$Then :$$\left\{ \matrix{ B + A = \int_{ - \pi }^\pi {\cos [(m - n)x]} dx = 0 \cr B - A = \int_{ - \pi }^\pi {\cos [(m + n)x]} dx = 0 \cr} \right. \to A = B = 0$$5$$ \cos A \cos B = \frac12(\cos (A+B) + \cos (A-B) ) and remember you need m \ne n 1 Hint: Try using product to sum formulae. \cos(a-b)-\cos(a+b)=2 \sin a \sin b and \cos(a+b)+ \cos(a-b)=2\cos a \cos b 1 Here is a result, though I think it's not complete. As in my previous solution, I'll consider only the z-dependence of each function. Let the differential operator in this problem and the linked question be written as \mathcal{L}=1-2\lambda D_z. Then the results of the linked question may be summarized as: If \mathcal{L}W=G, then \displaystyle ... 7 You may apply Coordinate Rotation to simplify the process: \begin{align} \iint\limits_D{e^{x+y}}d\sigma \mathrel{\mathop{\xrightarrow[u=\frac{1}{\sqrt2}(x+y)]{v=\frac{1}{\sqrt2}(x-y)}}}\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}dv\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}e^{\sqrt2u}du=\left[ e^{\sqrt2u} ... 1 Remember that E(A)^{2}=E(A)=E(A)^{\star}. So \mu_{x,x}=(E(A)x,x)=(E(A)^{2}x,x)=(E(A)x,E(A)^{\star}x)=(E(A),E(A)x)=\nu_{x}. $$It is interesting that \mu_{x,y} << \mu_{x,x} and \mu_{x,y} << \mu_{y,y} because$$ |\mu_{x,y}(A)| \le \mu_{x,x}(A)^{1/2}\mu_{y,y}(A)^{1/2}. $$Radon-Nikodym gives you something interesting, ... 2 You can also use the Bioche rules: Take t = \tanh x/2:$$ dt = \frac {dx}{2\cosh^2x/2} \\ \int \frac{e^x dx}{1 + e^x} = \int \frac{e^{x/2} dx}{e^{-x/2} + e^{x/2}} = \int \frac{e^{x/2} dx}{2\cosh x/2} = \frac12 \int \frac{1 + e^{x} dx}{2\cosh^2 x/2} $$Then use the expression of \tanh^{-1}:$$ \tanh^{-1}(t) = \frac 12\log\frac{1+t}{1-t} \\ \implies ...

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Using the Fourier series of $\ln(\tan{x})$, \begin{align} &\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\ =&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\ =&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\ ...

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Consider the case of an non closed interval. $$\int_a^b f(x) dx = \lim_{a'\downarrow a, b'\uparrow b} \int_{a'}^{b'}f(x)dx$$ ans the integrals $$\int_{a'}^{b'}f(x)dx$$are definite Riemann integrals (so $f$ is bounded on $[a',b']$). Now make the change in the definite integrals: $$\int_{a'}^{b'}f(x)dx = \int_{g^{-1}(a')}^{g^{-1}(b')} f(g(y)) g'(y) ... 0 You've messed something up...I can't quite see it, but this is how I would solve the problem: $$I=\int_0^\frac{\pi}{2}\cos^2\left(x\right)dx=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}+\frac{\cos\left(2x\right)}{2}\right)dx$$ via using the trigonometric identity. Your u-substitution is wrong btw...you cannot substitute ... 0 Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by y=\sqrt{9-x^2} as you have, and the bottom of the rectangle is given by y=-11. If we take the density \delta=1, we have that \displaystyle ... 0 if \sin(x)=u we get x=\arcsin(u) and dx=\frac{1}{\sqrt{1-u^2}}du thus we get \int \cos(x)^2dx=\int 1-\sin(x)^2dx=\int \sqrt{1-u^2}du 2 The problem is in using \cos x = \cos \sin^{-1} x. Arcsine is not a single valued function on x=[0,\pi]. In the conventional sense, it's not defined outside of [-\pi/2,\pi/2]. You can only do a u substitution using a one-to-one function. 5$$ \begin{align} \lim_{d\to1^-}\int_0^d\left[\frac{t^{a-1}}{1-t}-\frac{c\,t^{b-1}}{1-t^c}\right]\mathrm{d}t &=\lim_{d\to1^-}\int_0^d\left[\frac{t^{a-1}}{1-t}-\frac{t^{b/c-1}}{1-t}\right]\mathrm{d}t\tag{1}\\ &-\lim_{d\to1^-}\int_d^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t\tag{2}\\ ...

1

Here is a completely elementary (i.e., nothing beyond basic integration) proof. Taking advantage of the symmetries of $\cos$, $I_{n} =\int\limits_{0}^{2\pi} \cos^{2n}(x)\,{\rm d}x =4\int\limits_{0}^{\pi/2} \cos^{2n}(x)\,{\rm d}x =4\int\limits_{0}^{\pi/2} (\cos^2(x))^{n}\,{\rm d}x =4\int\limits_{0}^{\pi/2} (1-\sin^2(x))^{n}\,{\rm d}x$. Since $\sin(x) \ge ... 0 setting$g=u^n$and$f'=e^{au}$we get$g'=nu^{n-1}$and$f=\frac{e^{au}}{a}$then you will obtain your formula for$a\ne 0$1 Hint: Shifting by$u=x+\frac{\pi}{12}$, $$\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=\int_{0}^{\pi/6}\ln \tan\left(\frac{\pi}{4}+u\right)\,du.$$ The integrand can be expressed via trigonometric series as: $$\frac12\ln\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}=\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[(2k-1)x\right]}}{2k-1}$$ ... 3 Rewriting the Lobachevsky functions in terms of dilogarithms, we get $$\mathcal{I}=-\frac12\Im\left[\operatorname{Li}_2\left(e^{\pi i/6}\right)+\operatorname{Li}_2\left(e^{5\pi i/6}\right)\right]=\frac12\color{blue}{\Im\left[\operatorname{Li}_2\left(e^{-\pi i/6}\right)-\operatorname{Li}_2\left(e^{5\pi i/6}\right)\right]}=-\frac23\mathbf{G},$$ where the blue ... 5$\qquad\qquad\qquad\qquad$Hello, there! Cleo just asked me to post this: $$\int_0^\tfrac\pi{12}\ln(\tan x)~dx=-\dfrac23\cdot\text{Catalan}$$ 0 This is called: Integration by Substitution. It is a consequence of the Chain Rule for the derivative of the composition of functions. 1 I believe it's a bit beyond just being a "general" integral as$f\left(x\right)$matters. For example, if we have $$\int\left(\log\left|x\right|\right)^2dx$$ the methods we would use to integrate are somewhat different than if you were trying to solve $$\int\sin^2\left(x\right)dx$$ or $$\int\left(ax^3+bx^2+cx+d\right)^2dx,$$ given that$a,b,c,d\in\mathbb{R}... 1 It is with little difficulty to show that \begin{align} I_{n} = \int_{0}^{2 \pi} \cos^{2n}\theta \, d\theta = 2 B\left(\frac{1}{2}, n + \frac{1}{2} \right). \end{align} Using Stirling's approximation it can be shown that \begin{align} I_{n} \rightarrow \sqrt{\frac{4 \pi}{n+1}} \hspace{5mm} n \rightarrow \infty \end{align} which leads to \begin{align} ... 1 I think that 2) holds if you are referring to property P as one of the convergence theorems like monotone convergence or dominated convergence theorem. The point is that Fatou's lemma remains valid if we replace a.e. convergence with convergence in measure. Try to prove this. 2 Since|\cos(x)| \leq 1$, we can use the Dominated convergence theorem on the sequence of functions:$f_n(x)=(\cos(x))^{2n}$. But$f_n \rightarrow 0$a.e. on$[0,2\pi]$, and so by DCT we have that$\lim\limits_{n \rightarrow \infty}I_n=0$3 Using$t^c\mapsto t, we will arrive to the following equation \begin{align} \int_0^1 \left[\frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\right]\ dt&=\int_0^1 \left[\frac{t^{a-1}}{1-t}-\frac{t^{\large\frac{b-c}{c}}}{1-t}\right]\ dt\\ &=\lim_{x\to0}\left[{\rm{B}}\left(a,x\right)-{\rm{B}}\left(\frac{b}{c},x\right)\right]\\ ... 0 $$\frac{3}{\left(\frac 3 2\right)} = 3\cdot\frac 2 3 \ne 3\cdot \frac 3 2.$$ 1 Consider that $$\frac{d}{dt} (\mathbf{v} \times\mathbf{\dot{v}} ) = \mathbf{\dot{v}} \times\mathbf{\dot{v}} +\mathbf{v} \times\mathbf{\ddot{v}}$$ What is the cross product of a vector and itself? 2 In your penultimate step,\frac 23$.$\frac 93=\frac {18}{9}=2$, not 6. 1 You failed in$\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{18}{3}$, actually$\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{2}{3}\cdot3 = 2$. 1$\frac{2}{3} \cdot \frac{9}{3}=2$. not$6$0 You can do the partial fraction expansion in a more efficient and less error prone way than the standard high school method, which also helps to safe effort in cases where the degree of the numerator is larger than that of the denominator. For a given rational function Q(x) you consider the behavior near its singular points$x_1$,$x_2$, etc. $$Q(x) = ... 1 Another slightly different approach, using$$\int\frac{f'}{\sqrt f}dx=2\sqrt f+C\;\;:\int\frac{u}{\sqrt{R^2+r^2-2rRu}}du=-\frac1{2rR}\int\frac{(R^2+r^2-2rRu)'}{\sqrt{R^2+r^2-2rRu}}du==-\frac1{rR}\sqrt{R^2+r^2-2rRu}+C$$3 Let v=R^2+r^2-2Rru. It is simple to write u and du in terms of v and dv. You get$$\int\frac{A+Bv}{\sqrt{v}}dv.$$Expand this out before integrating. 2 First, we may note that the$x,y$dependence is irrelevant, along with the shift in the argument of$g(x,y,z)$. To that end I'll introduce$W(z)=w(x,y,z)$and$G(z)=g(x,y,z+h)\to g(z)$. Then we just have a simplified ODE given as$(1-\lambda D_z)W(z)=G(z)\$. This can indeed be handled by an integrating factor by observing ...

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