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If $\alpha > M$ you have by direct computation that $$\int_{-M + \alpha}^{M + \alpha} \frac{1}{\sqrt{|x|}} \, dx = 2 \sqrt{M + \alpha} - 2 \sqrt{-M + \alpha}.$$ The function $\phi(\alpha) = 2 \sqrt{M + \alpha} - 2 \sqrt{-M + \alpha}$ is decreasing for $\alpha \ge M$ (its first derivative is negative) so that $\alpha \ge M$ implies $\phi(\alpha) \le ... 0 here is a way to show that$\int_0^1 \frac{dx}{\sqrt{1-x^2}} = 1$without the use of trigonometric functions. i will use the fact the area of unit circle is$\pi.$that is$\int_0^1 \sqrt{1-x^2}dx = {\pi \over 4}$twice and integration by parts. \begin{eqnarray} {\pi \over 4} & = & \int_0^1 \sqrt{1-x^2}\ dx \\ & = & ... 0 The integral$\displaystyle\int_1^{x^2} e^{-t^2}\,dt$is positive when$x^2>1$and negative when$0\le x^2<1$. That means it is positive when$x>1$or$x<-1$, and negative when$-1<x<1$. You're multiplying it by$2x$which is positive when$x>0$and negative when$x<0$. So the derivative is negative when$x<-1$, positive when ... 1 Note that$f'(x)=0 \iff x \in \{ -1, 0, 1\} $. It suffices to show that$ \displaystyle\int_{1}^{x^2} e^{-t^2} \, \mathrm{d}t $is positive for$|x|>1$and negative for$|x|<1$. Do you see why this is? 2 You needn't compute it explicitly, but only determine the sign.$f'(x)=0$for$x\in\{-1,0,1\}$. Just check it. Now notice that $$\int_1^{x^2}e^{-t^2}dt$$ is positive for$|x|>1$and negative for$|x|<1$. 0 My replacing$t$with$1-t$we get: $$\int_{0}^{1}\frac{dt}{\sqrt{1-t^2}}=\int_{0}^{1}\frac{dt}{\sqrt{t(2-t)}}\leq\int_{0}^{1}\frac{dt}{\sqrt{t}}=2.$$ 0 For$\forall x\in R, h>0$, by the Mean Value Theorem for integrals, there is$c\in(x-h,x+h)$such that $$\int_{x-h}^{x+h}f(t)dt=f(c)\cdot2h$$ and hence$f(c)=0$. Letting$h\to0$implies$c\to x$. by the continuity of$f$at$x$, we have $$f(x)=\lim_{c\to x}f(c)=0.$$ Since$x$is arbitrary,$f(x)\equiv0$for all$x\in R$. 4 Here's an alternative approach: Using the Euler substitution$t=\sqrt{\frac{1-s}{1+s}}\implies s=\frac{1-t^2}{1+t^2}, the integral is transformed to \begin{align} \int_{0}^{1}\frac{\mathrm{d}s}{\sqrt{1-s^2}} &=\int_{1}^{0}\frac{t+\frac{1}{t}}{2}\,\frac{(-4t)\,\mathrm{d}t}{(1+t^2)^2}\\ &=2\int_{0}^{1}\frac{\mathrm{d}t}{1+t^2},\\ \end{align} ... 0 Iff$is not zero, there is an$x_0 \in \mathbf{R}$($f$is necessarily defined on$\mathbf{R}$) such that$f(x_0) \not = 0$. Switching to$-f$if needed we can assume that$f(x_0)>0$. As$f$is continuous, with$\varepsilon = f(x_0) / 2$we can find$\eta>0$such that$|f(x) - f(x_0)| < \varepsilon$whenever$|x-x0| \leq \eta$. Then for$x\in [x_0 ...

2

Let $a$ a fixed real and $$F(b)=\int_a^bf(x)dx$$ then by the fundamental theorem of analysis we have $$0=F'(b)=f(b),\quad \forall b$$

1

This was my submission. Say, for sake of contradiction, that $f(x)$ is non-zero for some real-valued $x=t$. That is, $f(t) \ne 0$. If $f(x)=c$ for all $x$, then $$\displaystyle\int_{a}^{b} f(x) \, \mathrm{d}x = \displaystyle\int_{a}^{b} f(x) \, \mathrm{d}x = \displaystyle\int_{a}^{b} c \, \mathrm{d}x = c \cdot \left( b - a \right),$$which is non-zero ...

1

Let $$F(x) = \int_0^x f(t) \, dt.$$ By assumption, $F \equiv 0$. By the fundamental theorem of calculus, $F$ is an antiderivative of $f$, i.e. $$f = F' \equiv 0.$$

1

Here is a proof in single-variable calculus language. On the interval $0\leq s \leq 1$, we have $$\sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \leq \sqrt{2} \sqrt{1-s}$$ So $$\frac{1}{\sqrt{1-s^2}} \geq \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-s}}$$ We would like to use the Comparison Test, comparing $\int_0^1 \frac{ds}{\sqrt{1-s}}$ to $\int_0^1 ... 3 Hint: Suppose by contradiction that$f(c) >0$, continuity provides you a$\delta >0$such that$f(]c-\delta,c+\delta[)\subset \{t \in \Bbb R\mid t> \epsilon\}$for some$\epsilon$small enough. Then get a contradiction. 1 Assume for a contradiciton that$f$is not identically zero, so we have an$x$such that$f(x)\neq 0$, without loss of generality say$f(x)>0$. Since$f$is continuous you can take a small ball of radius$\epsilon$around$x$where$f$takes strictly positive values. Integrate over$[x-\epsilon, x+\epsilon]$and the integral must be strictly positive. ... 2 Hint: for any fixed$x$, take$a=x$,$b=x+h$, and take the limit as$h\to 0$(applying the fundamental theorem of calculus).. 3 By series expansion $$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx=\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$ By some manipulations $$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq ... -5 hint we have$$\int\frac{1}{\sqrt{1-s^2}}ds=\arcsin(s)+C$$3 This is an improper integral and the only problem to treat is at 1. We have$$\frac{1}{\sqrt{1-s^2}}\sim_1\frac1{\sqrt2\sqrt{1-s}}$$and since the integral$$\int_0^1\frac{ds}{\sqrt{1-s}}$$is convergent then the given integral is also convergent. 3$$\int_{-1}^1f(x)dx=\int_{-1}^0f(x)dx+\int_0^1f(x)dx$$By the ranges of definition, this$$=\int_{-1}^0x\ dx+\int_0^1 x^2\ dx$$0 Here's an idea for even n; I haven't looked at any of the links above, so hope this is not the same as that in Strehlke. [Ok, after finishing this, took a look at Strehlke. Approach below is not the same as his.] The sinc function is the Fourier transform of a boxcar. Here I'll work with a boxcar that is 1 for x in -1/2,1/2 and zero otherwise. The ... 0 In the general case, the result cannot be expressed as a combination of a limited number of elementary functions. The closed form involves a special function : http://www.wolframalpha.com/input/?i=integrate+dx%2F%28%28%28ax%2Bb%29%5Ec%29%28%28Ax%2BB%29%5EC%29%29 In particular cases, for example if the powers are integers, or in other cases with particular ... 1$$(x^3)dt+(1-t)[(x^2+1)t-1]dx=0$$X=\frac{1}{x^2}$$X\:dt-(1-t)[(1+X)t-X]\:dX=0$$Y=t-1$$X\:dY+Y[(1+X)(Y+1)-X]\:dX=0X\frac{dY}{dX}+(1+X)Y^2+Y=0$$This is a Bernoulli EDO. Let : F=\frac{1}{Y}$$X\frac{dF}{dX}-F=X+1$$This linear EDO is easy to solve for F(X) 6 since$$2x^3yy'+(1-y^2)(x^2y^2+y^2-1)=0$$then we have$$2x^3yy'=(y^2-1)^2+(y^2-1)x^2y^2=(y^2-1)^2+(y^2-1)x^2(y^2-1)+x^2(y^2-1)$$so$$x^3(y^2-1)'=(y^2-1)^2+(y^2-1)^2x^2+x^2(y^2-1)$$let y^2-1=u,then we have$$\dfrac{du}{dx}=\left(\dfrac{1}{x}+\dfrac{1}{x^3}\right)u^2+\dfrac{u}{x}$$this is Bernoulli equation ... 1 Yes, it's a tough one. There is an integrating factor$$ \mu = \dfrac{1}{x^2 (y^2-1)^2}$$which leads to the implicit solution$$ {\frac {{x}^{2} y^{2}- y^{2}+1}{x \left( y ^{2}-1 \right) }} = c $$but I don't know how you would find these by hand (I used Maple). 1 A set of points on the x -axis is said to have measure zero if the sum of the lengths of intervals enclosing each of the points can be made arbitrarily small. If f(x) is bounded in [a,b] , then a necessary and sufficient condition for the existence of ∫ b a f(x)dx is that the set of discontinuities have measure zero. Great clarification results by ... 2 If c and f are positive integers, evaluate I(b,e)=\displaystyle\int\frac1{(ax+b)(dx+e)}dx in terms of b and ~\quad~ e, then differentiate the obtained result c times with regard to b, and f times with regard to e. If c and f are negative integers, let t=ax+b, for instance, then expand the other term using the binomial theorem, and switch the ... 1 In order to split a sum of integrals, you need them to each converge to a finite number, or for there to be just one of +\infty and -\infty which all of the infinite ones converge to. Here this may not necessarily happen, because you can have, for instance, the case f(x)=a, in which case the original integral is zero but the sum is the indeterminate ... 5 If you're still interested in approaches that use contour integration, consider the function$$f(z) = \frac{\log(1+z) \log(-z)}{1+z^{2}}.$$Using the principal branch of the logarithm, there is a branch cut along [0,\infty) and a branch cut along (-\infty, -1]. Then integrating counterclockwise around a keyhole contour deformed around the branch cuts ... 2 Hint: U-Substitution first$$ u = x^2 $$Therefore$$du = 2xdx $$then$$ \frac{du}{2x} = dx \int 3x \cos x^2 \, dx =\int 3x \cos x^2 \frac{du}{2x} = \int \frac{3}{2} \cos u \, du = \frac{3}{2}\int \cos u\, du $$Now you can figure integral of cos out by yourself... 5 Let$$ I(a)=\int_0^\infty\frac{\ln(1+at)}{1+t^2}dt. $$Then I(0)=0 and \begin{eqnarray} I'(a)&=&\int_0^\infty\frac{t}{(1+at)(1+t^2)}dt\\ &=&\frac{1}{1+a^2}\int_0^\infty\left(\frac{a+t}{1+t^2}-\frac{a}{1+at}\right)dt\\ &=&\frac{1}{2(1+a^2)}(2a\arctan t-2\ln(1+at)+\ln(1+t^2))\bigg|_{0}^\infty\\ &=&\frac{a\pi-2\ln a}{2(1+a^2)}. ... 0$$ \int 3x \cos(x^2) \, dx = \frac{3}{2} \sin(x^2) + C $$as can be checked by$$ \left[\frac{3}{2} \sin(x^2) + C \right]' = \frac{3}{2} \, \cos(x^2)\, 2x = 3x \cos(x^2) $$If you do not see it, you could use the substitution u = x^2:$$ \int 3x \cos(x^2) \, dx = \int 3x \cos(u) \, \frac{du}{2x} = \int \frac{3}{2} \cos(u) \, du= \frac{3}{2} \sin(u) + ... 4 HINT: Let$t=x^2$,$dt=2x\,dx$and the rest should be simple. So: yes, there is a mistake in your solution. Your answer cannot be identical with the correct one, because the former function is periodic and the latter not. 3 Another way is to note that the numerator is a constant times the derivative of expression within the square root in the denominator, i.e. $$\frac{d}{d\theta}(\cos^2\theta+8)=-2\cos\theta\sin\theta$$ so that the indefinite integral is simply $$-(\cos^2\theta+8)^{\frac{1}{2}}$$ You can verify this by differentiating the above, using the chain rule. 0 Let $$I_n(a)=\int_0^{\frac{\pi}{2}}\sum_{k=1}^n\left(\frac{\sin(akx)}{k}\right)^2dx.$$ Then$I_n(0)=0$and \begin{eqnarray} I_n'(a)&=&\int_0^{\frac{\pi}{2}}\sum_{k=1}^n\frac{x\sin(2akx)}{k}dx\\ &=&\sum_{k=1}^n\frac{-ak\pi\cos(ak\pi)+\sin(akx)}{4a^2k^3}\\ &=&-\frac{d}{da}\sum_{k=1}^n\frac{\sin(ak\pi)}{4ak^3} \end{eqnarray} and hence ... 5 Your answer is correct; however, because you have only described your method of solution rather than shown it, it is not possible at this time to confirm whether or not your actual computation was without methodological or conceptual flaws. We can only say that the value of the quantity you wrote is in fact$\pi/4$as claimed. 0 Let's suppose that we are aware of the integral $$\int_{C_k} \prod_{i=1}^k x_i^{\alpha_i-1} dx_i = \frac{\prod_{i=1}^k \Gamma(\alpha_i)}{\Gamma\left(\sum_{i=1}^k \alpha_i\right)}$$$(\textbf{1})$(See here for a proof). It turns out that only thing we need is to rewrite properly the inner one of the two products: $$P(k)=\prod_{j=i}^k (x_i ... 3 We have:$$-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$$for any x such that |x|<1, hence:$$-\frac{\log(1-x)}{1-x}=\sum_{n\geq 1}H_n x^n.$$and since \frac{d}{dx}\log^2(1-x) = -2\frac{\log(1-x)}{1-x} we have:$$\log^2(1-x) = 2\sum_{n\geq 1}\frac{H_n}{n+1}x^{n+1}\tag{1}.$$Since, by partial summation:$$\sum_{n=1}^{N}\frac{H_n}{n}= ... 4 Provided$\nabla f$is non-vanishing, the formula is $$\int_{\Bbb R^n}\delta(f(x))g(x) \,dx = \int_{f^{-1}(0)} \dfrac{g(x)}{|\nabla f (x) |}\,d\sigma$$ where$d\sigma$is the surface measure$^*$on$f^{-1}(0)$. This is basically just the change of variables ($u$-substitution) for integration. On a volume, you're going to have trouble. The gradient will ... 2 As $$e^t=\sum_{n=0}^\infty\frac{t^n}{n!},$$ we have $$e^{-t^2}=\sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{n!}.$$ Integrating term to term (why is possible?), $$\frac{\sqrt\pi}{2}\text{erf}(z)=\sum_{n=0}^\infty\frac{(-1)^nt^{2n+1}}{(2n+1)n!}.$$ 4$\sqrt[3]{2x^2-x^3} = x\sqrt[3]{\dfrac{2}{x}-1} \to u = \sqrt[3]{\dfrac{2}{x} - 1} \to u^3 = \dfrac{2}{x} - 1 \to x = \dfrac{2}{u^3+1}$. Can you take it from here? 0 For functions obeying just some very fairly modest restrictions, you differentiate under the integral sign by evaluating the function being integrated at the endpoint of the integral; this is the fundamental theorem of integral calculus. So $$\frac{d}{dz} \frac{2}{\pi} \int_0^z e^{-t^2}dt =\frac{2}{\pi} e^{-z^2}$$ At$z = 0$, this is$\frac{2}{\pi}$And ... 3 For$\alpha_1, \alpha_2 \gt0$We have $$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \text{B}(\alpha_1,\alpha_2) \tag{1}$$ Where$\text{B}(a,b)is Beta Function \begin{align} \Gamma(\alpha_1) & =\int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \tag{2}\\ \Gamma(\alpha_2) & =\int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y ... 9 Here is an elementary way to evaluate the integral without involving any special functions or advance formulas. Notice that\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}Hence we have \begin{align} \int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\int_0^\infty\int_0^\infty \frac{e^{-y}\sin(xy)}{e^{2\pi ... 0 As f is bounded on [a,b], we have |f(x)| \leqslant M for all x \in [a,b]. Choose x_1 such thatx_1 - a < \frac{\epsilon}{4M}.As f is integrable over [x_1,b], for every \epsilon > 0 there is a partition P': x_1 < x_2 < \ldots < x_n = b such that the difference between the upper and lower Darboux sums satisfies ... 8 A Generalisation: \begin{align} \int^\infty_0\frac{x}{(x^2+w^2)(1+e^{2\pi x})}{\rm d}x\tag1 =&\int^\infty_0\frac{xe^{-x}}{(x^2+4\pi^2w^2)(1+e^{-x})}{\rm d}x\\ \tag2 =&\int^\infty_0\frac{x}{x^2+4\pi^2w^2}\left(\sum^\infty_{n=1}(-1)^{n-1}e^{-nx}\right){\rm d}x\\ \tag3 ... 0 I don't know the definition of jacobi elliptic functions defined by parametrising the ellipse. But i know that the the case k^2=2 is impossible, because |k|<1. Look up here, here or here. 2 Let's write \begin{align} x &= a(\rho\sin(\phi)\sin(\theta))^3 \\ y &= b(\rho\sin(\phi)\cos(\theta))^3 \\ z &= c(\rho\cos(\phi))^3, \end{align} for then, (x/a)^{2/3} + (y/b)^{2/3} + (z/c)^{2/3} = \rho^2. Furthermore, the Jacobian of change of variables is648 \rho ^8 \sin ^2(\theta ) \cos ^2(\theta ) \sin^5(\varphi ) \cos ^2(\varphi ... 2 If you were integrating over the volume $$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} + \left(\frac{z}{c}\right)^{2} \leq 1$$ you would use spherical polars withx = ar\sin\theta\cos\phi$,$y = br...$,$z = cr...$. Now try and modify those so they fit your shape by taking the appropriate powers. Then calculate the Jacobian.... 1 For a function$f(x)=\frac{1}{\pi}\frac{a}{a^2+x^2}\$ (also kwnown as Lorentzian function) the Fourier transform is: $$\int_{-\infty}^{\infty}\frac{1}{\pi}\frac{a}{a^2+x^2}e^{-itx}dx=\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{-itx}dx$$ Now we will use Cauchy's integral formula. Although we only need to solve our integral for the real axis, ...

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