# Tag Info

0

$\cos x$ goes from $1$ (at $x=0$) to $0$ (at $x=\frac{\pi}{2}$) and $-1$ (at $x=\pi$). And x goes from $0$ to $\pi$ being positive, so the signe of $x\cos x$ depends of the signe of $\cos x$ in the interval $[0, \pi]$. We have $$\int _{\frac{\pi}{2}}^{\pi}x\cos x\approx \frac{\pi}{2}(-\pi)=-\frac{\pi^2}{2}$$ and $$\int_0^{\frac{\pi}{2}} x\cos ... 0 Notice that$$\int_0^{\pi} x \cos xdx =\int_0^{\frac{\pi}{2}}x\cos xdx+\int_{\frac{\pi}{2}}^{\pi} \cos x dx$$Now, the first integral is positive and the second one is negative. But for every x\in [0, \pi/2) we have$$x<\pi -x$$Because of the monotonicity of the integral$$\int_0^{\frac{\pi}{2}} x\cos x dx<\int_0^{\frac{\pi}{2}} (\pi-x)\cos ...

0

$\int_{-1}^1 P_n P_{n+1}^{'}dx = \int_{-1}^1 P_n ((2n+1)P_n + P_{n-1}^{'})dx = (2n+1)\int_{-1}^1 P_n^2dx + \int_{-1}^1 P_n P_{n-1}^{'}dx$ = 2 Here I used the following: $$(2n+1)P_n = P_{n+1}^{'} - P_{n-1}^{'}$$ $$\int_{-1}^1 P_n^2dx = \frac{2}{2n+1}$$ (see Legendre Polynomials: proofs) And the fact that $P_{n-1}^{'}$ is a polynomial of degree $n-2$ and ...

2

If $f(x)=|x|$ we have \begin{eqnarray} \sup_{x \in [-1,1]}\left| f_n(x)-f(x) \right|&=&\sup_{x \in [-1,1]}\left| \sqrt{x^2+\frac1n\left( \cos \left( x^n \right) \right)^2} -|x|\right| \\ &\leq& \sup_{x \in [-1,1]}\left| \sqrt{x^2+\frac1n}-\sqrt{x^2} \right|\\ &=& \sup_{x \in [-1,1]}\frac{\left| \sqrt{x^2+\frac1n}-\sqrt{x^2} ...

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You have edited the question: Ok now you may try the following way : $\int (tue^{tu}) - \int u^2 e^t$ then use integration by parts for the first part and the second part is easy integration.

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Since the question is tagged Laplace transform, I suspect that the convolution should be $$t^2e^{-2t}*te^t=\int_0^t u^2e^{-2u}(t-u)e^{t-u}\,du.$$ Using linearity (and the fact that the $t$-dependence can go outside the integral), this integral can be written as $$te^t\int_0^tu^2e^{-3u}\,du-e^t\int_0^tu^3e^{-3u}\,du.$$ I suggest that you start to ...

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I would do the following way : Let, $t-u = p ==> dt = dp$ $\int_{-u}^0 p e^p dp$ Now try integration by parts. $\int_{-u}^0 p e^p dp = pe^p|_{-u}^0 - \int_{-u}^0 e^p dp = -e^{-u} - 1+e^{-u}$ The answer is: $-e^{p-t} - 1+e^{p-t}$ Hope this might help!

1

$\begin{array}\\ \int_0^{\pi} x\cos(x)dx &=\int_0^{\pi/2} x\cos(x)dx+\int_{\pi/2}^{\pi} x\cos(x)dx\\ &=\int_0^{\pi/2} x\cos(x)dx+\int_{0}^{\pi/2} (x+\pi/2)\cos(x+\pi/2)dx\\ &=\int_0^{\pi/2} (\pi/2-x)\cos(\pi/2-x)dx+\int_{0}^{\pi/2} (x+\pi/2)\cos(x+\pi/2)dx\\ &=\int_0^{\pi/2} (\pi/2-x)\cos(\pi/2-x)dx-\int_{0}^{\pi/2} (x+\pi/2)\cos(x-\pi/2)dx\\ ... 2 $$\int_0^{\pi} x\cos(x)dx=\int_0^{\frac \pi 2} x\cos(x)dx+\int_{\frac \pi 2} ^{\pi} x\cos(x)dx$$ let$u=\pi-x$in the second integral $$\int_0^{\pi} x\cos(x)dx=\int_0^{\frac \pi 2} x\cos(x)dx - \int_0^{\frac \pi 2} (\pi-u)\cos(u)du$$ $$=\int_0^{\frac \pi 2} (2u-\pi)\cos(u)du$$ which is negative because$\cos(u)\ge 0$and$(2u-\pi )\le 0 $whenever$0 ...

2

You may just integrate by parts: $$\int_0^{\pi} x\cos x\:dx=[x \sin x]_0^\pi-\int_0^{\pi} \sin x\:dx=0-\int_0^{\pi} \sin x\:dx<0$$ since $\sin x \geq0$ over $[0,\pi]$.

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It's more convenient to use cylindical coordinates; set $x=r\cos\phi$ and $z=r\sin\phi$. Then $\phi\in[0,2\pi)$ and $y\in[0,4]$. To find the limits of $r$, note that $r=\sqrt{x^2+z^2}$, so $0\leq r\leq y$. Also, the Jacobian is equal to $r$, therefore the integral is equal to $$\int_0^{2\pi}\int_0^4\int_0^yr\cdot ... 0 Let$$F(x) = \int_a^x f(t) dtG(x) = F(x^2)-F(x)G'(x) = 2xf(x^2)-f(x)$$If f(x)=1, G(x) = \int_x^{x^2} f(x)dx = x^2-x So G'(x) = 2x-1 = 2xf(x^2)-f(x) 1 \int_a^b \! f(x) \, \mathrm{d}x=g(b)-g(a) is a result obtained directly from the fundamental theorem of calculus. Say there's a function f(x) Now the area under the curve from 0 up til x can be found using Reimann sums. Let A(x) = \int_0^x f(t) dt where A(x) is the area function that gives you the area under the curve of f(x) from 0 up to an ... 4 The first question is the fundamental theorem of calculus: it says that if there is a differentiable function F with \frac{dF}{dx} = f, and f is Riemann integrable on [a,b], then \int_{a}^b f(x) dx = F(b) - F(a). This can be proved from the definitions relatively easily, but not obviously. It will be covered in any book on elementary real analysis, ... 2 You cannot associate any physical interpretation to \int_b^a f(x) \ dx You're right, the sum would be the same irrespective of whether you sum it forward or backward. But \int_b^a f(x) dx doesn't connote taking the sum in the opposite manner. If it were then \int_b^a f(x) dx would have been equal to \int_a^bf(a+b-x)dx which doesn't happen. However, ... 2 Since x=e^{\ln x}, we are integrating \exp(-(p\ln x)/x). Since \frac{\ln x}{x}\to 0 as x\to\infty, the integrand has limit 1, and therefore in particular is after a while greater than \frac{1}{2}. It follows that the improper integral diverges to \infty. 1 Hint1: Taylor series expansion. Hint2: Treat as complex function (asymptotic stability). :) 0 We would have$$\int_0^x \lfloor t \rfloor dt = \int_0^{\lfloor x \rfloor} \lfloor t \rfloor dt + \int_{\lfloor x \rfloor}^x \lfloor t \rfloor dt = \frac{\lfloor x \rfloor \cdot (\lfloor x \rfloor - 1)}{2} + \lfloor x \rfloor (x - \lfloor x \rfloor) = \frac{\lfloor x \rfloor (2x - \lfloor x \rfloor - 1)}{2}$$0 The integration range is small: [0, 1] so one of the method could be to use Taylor series for the \arctan(x) function:$$\arctan^2(x) = \sum_{n = 0}^{+\infty}\sum_{m = 0}^{+\infty}\frac{(-1)^n (-1)^m}{(2n+1)(2m+1)}x^{2n+1}x^{2m+1}$$And write \sqrt{x} = x^{\frac{1}{2}} to obtain:$$I = \int_0^1\sum_{n = 0}^{+\infty} \sum_{m = ...

2

The integrand has a closed-form antiderivative in terms of elementary functions and dilogarithms. Mathematica can find it if we help it by first converting the arctangent into a combination of logarithms: $$\arctan(x)=\frac i2\ln(1-i x)-\frac i2\ln(1+ix)$$ After some simplifications it takes this form. Its correctness can be checked manually using direct ...

0

Hint: for $n\neq -1$ $$\int At^ndt=A\frac{t^{n+1}}{n+1}$$ so $$\int 34^3t^0dt+\int5t^{-3}dt$$

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Because $\frac{x^2e^x}{(e^x+1)^2}=\frac{x^2}{\left(e^{x/2}+e^{-x/2}\right)^2}$ is even, we can integrate by parts and use the Dirichlet eta function: \begin{align} \int_{-\infty}^\infty\frac{x^2e^x}{(e^x+1)^2}\,\mathrm{d}x &=2\int_0^\infty\frac{x^2e^x}{(e^x+1)^2}\,\mathrm{d}x\\ &=-2\int_0^\infty x^2\,\mathrm{d}\frac1{e^x+1}\\ ... 0 HINT:\int\left(34^3-5t^{-3}\right)\space\text{d}t=\int 34^3\space\text{d}t-\int 5t^{-3}\space\text{d}t=34^3\int 1\space\text{d}t-5\int t^{-3}\space\text{d}t$$1 HINT: I assume you are familiar with the power rule, which states that$$\frac{\text{d}}{\text{d}x}(x^n)=nx^{n-1}$$Integrating both sides and dividing by n, we have$$\int{\frac{\text{d}}{\text{d}x}(x^n)}\text{d}x=\int{nx^{n-1}}\text{d}x\implies \frac{x^n}{n}=\int{x^{n-1}}\text{d}x$$Noting that integration distributes over addition, you can ... 0 The answer for this problem would be =\frac{5}{2t^2}+39304t+C For this section of the textbook you are working with the anti-derivative. So basically you do the opposite of what you do for finding the derivative. If the integral was for 9, your answer would be 9x+C ( C just being a general representation of a a constant) 6 Hint. Observe that we have$$ \begin{align} \int_{-\infty}^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx&=\int_{-\infty}^0 \frac{x^2 e^x}{(e^x+1)^2}\:dx+\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx\\\\ &=2\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx\\\\ &=2\int_0^\infty \frac{x^2e^{-x} }{(1+e^{-x})^2}\:dx\\\\ ...

6

Often, with such complicated primitives, I suggest not to calculate the primitive, but to use some trick, like introducing a parameter and differentiate with respect to it. Read below if you are interested in such a way to calculate this integral. Tell me if this is far from what you looked for. First, since the integrand is even, your integral equals $$... 0 HINT:$$\int\frac{s^2+\sqrt{s}}{s^2}\space\text{d}s=$$Substitute u=\sqrt{s} and \text{d}u=\frac{1}{2\sqrt{s}}\space\text{d}s:$$2\int\frac{u^3+1}{u^2}\space\text{d}u=2\int\left(\frac{1}{u^2}+u\right)\space\text{d}u=2\int\frac{1}{u^2}\space\text{d}u+2\int u\space\text{d}u=2\int u^{-2}\space\text{d}u+2\int u\space\text{d}u$$1 \int \dfrac{s^{2} + \sqrt{s}}{s^{2}}ds=\int (1+ s^{-3/2})ds 0 HINT:$$\int \frac{s^2+\sqrt s}{s^2}\ ds=\int \left(\frac{s^2}{s^2}+\frac{\sqrt s}{s^2}\right)\ ds=\int \left(s^{0}+s^{-3/2}\right)\ ds $$Use can now use \large \int x^n\ dx=\frac{x^{n+1}}{n+1} 0 I used your suggestion, tired, and it led to some fun results involving Euler- trig sums and polylogs. I am confident that the use of these will eventually lead to the solution to the integral listed above as well as other integral(s). I list several I derived using the famous identities for H_{n}. I was mainly interested in the real parts due to the cos ... 8$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$0 We want to find \int_0^1 x\log(1+x)\,dx-\int_0^1 x\ln(1-x)\,dx. Since the second integral is under suspicion, let us look at it. Let u=\ln(1-x) and dv=x\,dx. Then we have du=-\frac{1}{1-x}\,dx, and we can take v=\frac{x^2}{2}-\frac{1}{2}. (We used a little trick here!) So our antiderivative is$$\ln(1-x)\left(\frac{x^2}{2}-\frac{1}{2}\right) ...

1

First, just calculate the integral on $[0,1-\varepsilon]$: $$\lim_{\varepsilon\to 0}\int_{0}^{1-\varepsilon}x\ln\frac{1+x}{1-x}\mathrm{d}x=\boxplus$$ Since: $$\int (1+x)\ln(1+x)=\frac{1}{2} (1+x)^2 \log (1+x)-\frac{(1+x)^2}{4}+C$$ $$\int \ln(x+1)=(x+1)\ln (x+1)-(x+1)+C$$ Using the logarithm-identities: $$\boxplus=\lim_{\varepsilon\to ... 0 Is it even possible to evaluate it analytically ? Yes. The integral evaluates to \dfrac{2K(a)+3E(a)}{240},~ where a=\dfrac{\sqrt5}3.~ See elliptic integrals for more information. I only want to know if there's a relatively easy way to solve this. Of course. Just let x=2t, and use 1-\cos2t=2\sin^2t, in conjunction with the definition ... 9 We have$$ \int_{0}^{1}{\dfrac{1-x}{\log x}(x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots)}\:dx=-\log 3. \tag1 $$Proof. One may recall that, using Frullani's integral, we have$$ \int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{\log x}\:dx=\log\frac ab \quad (a,b>0). \tag2 $$Considering a finite sum in the integrand, we get$$ \begin{align} ...

1

$ax^4+c$ is symmetric either side of $x=0$. It is called an even function. So the area under the curve is the same on either side, and the total area equals twice the area on the right-hand side. $bx$ is an odd function, because the value at $-x$ is the negative of the value at $+x$. So the integral for negative $x$ cancels out the integral for positive ...

0

Not an answer, but an attempt to push you in a good way for it First of all, let's simplify a bit the integral by collecting $2$ at the numerator and $4$ at the denominator: $$\int_0^{\pi} \frac{2(5 - \cos\theta)}{\sqrt{[4^3\cdot (26 - 10 \cos\theta)^3]}} \text{d}\theta$$ namely $$\frac{1}{4} \int_0^{\pi} \frac{(5 - \cos\theta)}{\sqrt{ (26 - 10 ... 1 Couldn't you just differentiate both sides to get (rpm = r)$$r'(t) = \frac{T}{m} + D r(t)r(0)=0$$Then we may easily solve this using any of several methods. The result is$$r(t) = \frac{T}{m D} \left (e^{D t}-1 \right )$$Then$$\int_0^t dt' \, r(t') = \frac{T}{m D} \left (\frac{e^{D t}-1}{D} - t \right ) $$1 This is more a comment/hint than a real answer, but it's too long for the comment section. First do the substitution e^u = x, so that dx = e^udu and your integral becomes$$\color{red}{-}\int_{-\infty}^\infty\color{red}{(-}e^u\color{red}{)}e^{-e^u}\frac{1}{a^2+u^2}du.$$Now you could try to substitute using the Lambert W function to get rid of the ... 1 Sorry this answer is not a complete on but I have not enough reputation to put this as comment. If some admin cares to change this to a comment, please do so. The standard approach to such a thing would be to look at the integral as part of a closed path integral. Let R_0, R_1>0, R_0\rightarrow 0,\;\; R_1 \rightarrow \infty and integrate from ... 2 While x\in[-1,2] so |x^2+x-6|=-x^2-x+6 and so$$||x^2+x-6|-6|\to|-x^2-x|=|x^2+x|$$and when x\in[2,4] so |x^2+x-6|=+x^2+x-6 and so$$||x^2+x-6|-6|\to|x^2+x-12|=|(x+4)(x-3)|$$As the integrand is a integrable function on [-1,4] so we get$$\int_{-1}^4||x^2+x-6|-6|dx=\int_{-1}^2|x^2+x|dx+\int_{2}^4|(x+4)(x-3)|dx$$Now consider the same way for each ... 2 The integrand has a closed-form antiderivative in terms of elementary functions and polylogarithms. It can be found using Mathematica after expressing inverse trig functions through logarithms of complex arguments, and can be manually checked for correctness using differentiation. After subtracting its limits at \infty and 0 and simplification, we can ... 1 The above answer is very good to understand why the integrand is odd. I will instead focus on how to prove the value of the integral. Note that for all integers m,n\in\mathbb{Z} one have$$ \int_0^{2\pi} \sin mx \cos nx \,\mathrm{d}x = 0 $$This can be shown by rewriting$$ \sin mx \cos nx = \frac{1}{2}\sin(m+n) x - \frac{1}{2}\sin(m-n) x $$... 1 The expression (16 \sin \phi \cos^2\phi-4\sin^{2}\phi+8\sin\phi) is not an odd function of \phi. Consider each of the terms one-by-one: $$\int^{2\pi}_{0} 16 \sin \phi \cos \phi\ d \phi = 0$$ because \sin \phi \cos \phi is antisymmetric (odd) about the \phi=\pi line. \int^{2\pi}_{0} -4\sin^{2}\phi\ ... 0 This is a problem of functional optimization. T depends on F and p is given (or at least could be known using polls). Let us first follow @Ian's comment by assuming that F is absolutely continuous. This means that \mathrm d F(y)=f(y)\mathrm dy, where f is the probability density of the price. As T depends not on a single variable, but on all ... 0 Hint: I'm not sure to interpret correctly your question, but, for r=e^{-\theta^2} we have:$$ \int_0^\infty r d\theta=\frac{\sqrt{\pi}}{2} $$and also$$ \int_0^\infty r^2 d\theta=\frac{1}{2}\sqrt{\frac{\pi}{2}} $$2 I understand what you're asking now. If you want to integrate a function f\colon [0,\infty) \to [0,\infty), then you can write \int_{0}^{\pi/2} \frac12 r^2(\theta) d\theta. You're asking if \int_{0}^{\infty} \frac12 r^2(\theta) d\theta=\infty. Well unless r=0 then yes of course. You're adding up a positive number infinitely many times. You are ... 1 If you want your integral to have the meaning of an area, then you should not integrate the polar function itself but its square:$$\frac12\int_{\phi=0}^{2\pi}r(\phi)^2d\phi$$Of course, if you decide to evaluate that integral over a larger interval then you are running around the origin multiple times. For a closed curve and an integration interval that ... 2 No.$$\int 0 = 0 no matter of the domain.

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