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1

From Hölder: $$I=\int_{a}^{b}(x-x_1)^2\ldots(x-x_n)^2 dx \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^4 \ldots (x-x_n)^4dx \right)^{1/2} \leq \\ \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^8 dx \right)^{1/4}\left(\int_{a}^{b}(x-x_3)^{8}\ldots(x-x_n)^8 dx\right)^{1/4} \leq \ldots \\ \leq ... 1 Here is my evaluation of the quadratic case. The starting point for the derivation is the algebraic identity$$6ab^2=\left(a+b\right)^3+\left(a-b\right)^3-2a^3.$$Then,$$\begin{align} \mathcal{I}_{2} &=\int_{0}^{1}\frac{\ln{\left(\frac{1}{t}\right)}\ln^{2}{\left(t+2\right)}}{t+1}\,\mathrm{d}t\\ ...

2

Write $\dfrac{1}{(x^2-x+1)^2}$ as a Taylor series $\sum_{j=0}^\infty a_j x^j$. The coefficients can be written as $$a_j = \dfrac{j+1}{3} b_j + \dfrac{2}{3} c_j$$ where $b_j$ repeats $1,2,1,-1,-2,-1,\ldots$ for $j = 0,1,2,\ldots$ while $c_j$ repeats $1,1,0,-1,-1,0,\ldots$, both with period $6$. Now $$\int_0^1 \log^2(x)\; x^k\; dx = \dfrac{2}{(k+1)^3}$$ so ...

2

We start with: $\displaystyle \int_0^1 \frac{t^2}{x^2+t^2}\,dt = 1-x\tan^{-1}\frac{1}{x}$ Then, \displaystyle \begin{align} \int_0^{\infty} \log (2\sin x)\left(1-x\tan^{-1}\frac{1}{x}\right)\,dx &= \int_0^{\infty}\int_0^1 \frac{t^2\log (2\sin x)}{t^2+x^2}\,dt\,dx\\&= -\sum\limits_{n=1}^{\infty} \int_0^1 t^2\int_0^{\infty} \frac{1}{n}\frac{\cos ... 2 Let the integral in question be \begin{align}\tag{1} I_{a} = \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+a t)}{1 + t} \, dt. \end{align} For the case ofa=1the following is obtained. By utilizing \begin{align} \int_{0}^{\infty} e^{-s t} \, \ln^{2} t \, dt = \frac{1}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) \end{align} for the changet \to t+1$leads to ... 0 This gives us an intuition as to how to proceed, the blue curve is$5 \cos 5x$and red is the other. The two curves intersect when $$5 \cos 5x = 5 - 5\cos 5x \iff \cos 5x = \frac{1}{2}$$ But $$\cos 5x = \frac{1}{2} \implies x = \frac{\pi}{15}$$ in the range we are concerned in. So the area between the curves is $$\int_0^{\frac{\pi}{15}} 5 \cos 5x - (5 - ... 0 Note that$$5\cos(5x)=5-5\cos(5x)\iff 10\cos(5x)=5\iff \cos(5x)=\frac{1}{2}.$$3 One may show that this integral is equivalent to the integral in this problem as follows: First, rewrite$$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$Then note that$$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x} $$The integrand is then equal to$$\left [\frac{2}{\pi} \tan^{-1}{\left (\frac{2}{\pi} \left ... 5 If one assumes that$f$is sufficiently well-behaved (and in particular, differentiable), one can solve this by differentiating under the integral sign (this is sometimes called Feynman's trick). Define $$I(a) := \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx.$$ Differentiating with respect to$a(and justifying passing the derivative sign through the ... 6 First notice that \begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4 \sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\log(4)}{2} \int_{0}^{\infty} \Big( 1- x \, \text{arccot}(x) \Big) \, dx + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\pi \log(2)}{4} + ... 4 There are possible singularities at the origin and at infinity; by definition the integral is the limit of \int_\delta^A as \delta\to0 and A\to\infty. You have\begin{eqnarray*} \int_\delta^A\frac{f(t)}{t}-\int_\delta^A\frac{f(2t)}{t}&=\int_\delta^A\frac{f(t)}{t}-\int_{2\delta}^{2A}\frac{f(t)}t ... 0 Assume thatf_Y$, the pdf of$Y\ge 0$, exists. In this case the expected value is $$E[Y]=\int_{0}^{+\infty}x\ f_Y(x)\ dx$$ Also, assume that the integral above is finite. It is obvious that$xcan be written as $$x=\int_0^x\ 1\ dy, \ x\ge 0.$$ With this, the expected value is $$E[Y]=\int_{0}^{+\infty}\int_0^x\ 1\ dy\ f_Y(x)\ dx.$$ Since the ... 7 The last series Special Addendum in the OP looks quite interesting! (just an observation, not an answer) We have by symmetry of the series: $$S = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)} = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^n}{k2^k n 2^n (k+n)}$$ Hence, \begin{align}\sum _{k=1}^{\infty } \sum ... 0 This is rather belated, but the formula \arctan(\frac{y}{x}) + \frac{1}{2}z^{2} for the potential of u is correct only for x > 0 (assuming the potential is continuous and takes the value 0 along the positive x-axis). A "formal" potential for u (away from the z-axis) is the multi-valued function \theta + \frac{1}{2}z^{2}, with \theta the ... 2 f(t,t)+\int\limits_{0}^{t} f_{t}(x,t)dx=f(t,t)+\int\limits_{0}^{t} 1 dx =2t+t=3t, which agrees! 2 For the integral from 0 to 1 to exist, we need, as was pointed out in the OP, \alpha\gt -1. We now deal with the integral from 1 to \infty. Rewrite the integrand as \frac{x^\alpha}{e^x}e^x\cos(e^x), and integrate by parts, letting u=\frac{x^\alpha}{e^x} and dv=e^x\cos(e^x)\,dx. Then du=e^{-x}\left(\alpha x^{\alpha-1}-x^\alpha\right)\,dx, ... 2 We must have \alpha>-1, otherwise we have a non-integrable singularity in a right neighbourhood of the origin. Given that, we have: \int_{0}^{M}x^\alpha \cos(e^x)\,dx = \int_{1}^{e^M}\frac{\log^\alpha(t)\cos t}{t}\,dt $$that is not absolutely converging, but is conditionally converging by Dirichlet's test, integral version, since \cos t has a ... 2 Integrating by parts and using the known series results, we get that$$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz=\int_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int_0^1\frac{\log (2-z) \log (z)}{1-z} dz}_{\large \sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^2}=-5/8 \zeta ... 1 The antiderivative is perfectly correct $$\int{\csc^2(4t)\,dt}=-\frac{1}{4} \cot (4 t)$$ but, as already said in comments and answers,\cot (4 t)$shows a discontinuity at$t=\frac \pi 2$and this value is just between the bounds. So, let us consider $$I=\int_1^{\frac \pi 2-\epsilon}{\csc^2(4t)\,dt}=\frac{1}{4} (\cot (4 \epsilon )+\cot (4))$$ ... 1 We going to prove that$|u(x)|\leq c\int_{0}^b |u(t)|dt\Rightarrow u\equiv 0$. For$c=1$. Define for$x\in[0,b]$,$V(x)=\max_{0\leq t\leq x}u(t)$. Suppose by contradiction$u(x)>0$for$0<x\leq b$. We have that$u(t)\leq V(t)$, for all$t\in [0,b]$. Fix$t\in[0,b]$, since$u$is continuous, there is$z\in [0,t]$such that$u(z)=\max_{0\leq s\leq ...

3

I couldn't resist as if your limits were same for each variable say varying $a$ to $b$, for any $a$ and $b$ , the following trick I saw somewhere, is applicable, but in your case, there is no short cut, I guess. $$I=\int_{a}^b\int_{a}^b \int_{a}^b \int_{a}^b \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_{a}^b\int_{a}^b \int_{a}^b ... 2 The book is wrong. Plain and simple. They transposed the last two digits, which is a simple typo. Originally, I was going to write the following, and I include it here for numerical reference. You need to do all the rounding right at the end. Since we have calculators today, which usually carry 12-16 digits of precision, round-off error is the least of our ... 4 The sum$$\frac{1}{m+1} \sum_{j=0}^{n-1} \gamma^{m+1}(t_{j+1}) - \gamma^{m+1}(t_{j})$$telescopes since$$(\gamma^{m+1}(t_{1}) - \gamma^{m+1}(t_{0})) + (\gamma^{m+1}(t_{2}) - \gamma^{m+1}(1)) + \cdots + (\gamma^{m+1}(t_{n}) - \gamma^{m+1}(t_{n-1}))=-\gamma^{m+1}(t_0) + (\gamma^{m+1}(t_1) - \gamma^{m+1}(t_1)) + \cdots + (\gamma^{m+1}(t_{n-1}) - ...

1

Because $\gamma(a)=\gamma(b)$ since $\gamma$ is a closed curve.

8

Let $I(a)$ be the integral $$I(a)\equiv \int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx$$ Taking a derivative with respect to $a$ reveals that \begin{align} I'(a)&=\frac{d}{da}\int_0^{\infty}\log\left(1+\frac{a^2}{x^2}\right)dx\\\\ &=2a\int_0^{\infty}\frac{dx}{x^2+a^2}\\\\ &=\pi \end{align} Integrating shows that $I(a)=\pi a +C$. ...

6

Using integration by parts: $$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)& du=\frac{\frac{-2a^2}{x}}{x^2+a^2}\,dx \\ dv=dx& v=x\end{bmatrix}\\=\underbrace{\left.x\cdot \ln\left(1+\frac{a^2}{x^2}\right)\right|_0^{\infty}}_{L}+\underbrace{\int_0^{\infty}\frac{2a^2}{x^2+a^2}\,dx}_{I}$$ ...

5

$$\int_{0}^{+\infty}\log\left(1+\frac{a^2}{x^2}\right)\,dx = a\int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx$$ then by setting $x=\tan\theta$ we have: $$\int_{0}^{+\infty}\log\left(1+\frac{1}{x^2}\right)\,dx = -2\int_{0}^{\pi/2}\frac{\log\sin\theta}{\cos^2\theta}\,d\theta=2\int_{0}^{\pi/2}\cot(\theta)\tan(\theta)\,d\theta=\color{red}{\pi},$$ where ...

0

It's not that the second derivative being unbounded means that the midpoint rule does not converge. But it does mean that this method of estimating the error doesn't tell you anything about the convergence. In such cases, you must either use a different approximation method or be more clever about your current application. In practice, approximating ...

2

For $\int e^{\sin t}~dt$ , $\int e^{\sin t}~dt$ $=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n)!}dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$ $=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$ For $n$ is any natural number, ...

3

&=&4\int_{0}^{\pi/4}\frac{dx}{\sin(2x)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{\pi/4}\frac{dy}{\cos(2y)+\frac{\sqrt{3}}{2}}\\&=&4\int_{0}^{1}\frac{dt}{2+\left(\frac{\sqrt{3}}{2}-1\right)(1+t^2)}\\&=&8\int_{0}^{1}\frac{dt}{\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)t^2}\\&=&8\,\text{arctanh}(2-\sqrt{3})=\color{red}{2\log ...

2

Your solution is OK. My (clumsier) solution, just to reinforce yours, is as follows We can approach the integral $$\int_a^bf'(x)f''(x)\ dx$$ via integration by part. The mnemonic is that $\int u'v=uv-\int uv'.$ Let $u'=f''$ and let $v=f'$. Then $u=f'$ and $v'=f''$. So $$\int_a^b f'(x)f''(x) dx=\left[(f'(x))^2 \right]_a^b-\int_a^bf'(x)f''(x)\ dx.$$ ...

0

No, this one is among the functions that do not admit primitives expressible as a composition of elementary functions. The definite integral, though, can still be computed, through other means than the Leibniz-Newton theorem. Nevertheless, the result is ugly, since it is expressed with the aid of special functions, so just forget about it.

3

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos (x-\frac{\pi}{3})\cdot\cos (x-\frac{\pi}{6})}\mathrm{d}x=\int_0^{\pi/2}\frac{4}{\sqrt{3}+2\sin(2x)}dx$$ and the change of variables... $$=\lim_{M\rightarrow\infty}\int_{\sqrt 3}^{\sqrt{3}+M}\frac{4\sqrt 3}{u(2\sqrt{3}+u)}du=\cdots$$ $$=\lim_{M\rightarrow\infty}\left[2\ln(3)+2\ln\left(\frac{M+\sqrt ... 0 HINT: You could try the variable change u = sin(t). Therefore you would have du = (\frac{\pi}{2}-u)dt, and :$$ \int_0^{\pi \over 2}e^{c \sin^2t}dt = \int_0^1{e^{c u^2} \over \frac{\pi}{2}-u}dt $$Which takes away the sin 2 Hint: Use the fact that \dfrac{1-\cos2x}2=\sin^2x, in conjunction with the integral formulas for Bessel functions. 1 As x \to \infty,$$\log \left( \frac{x}{\sqrt{x^2+1}} \right) \to 0.$$6 A first step. Let:$$ I_k = \int_{0}^{1}\frac{(x-x^2)^k}{1-x+x^2}\,dx.$$We have I_0=\frac{2\pi}{3\sqrt{3}}, I_1=-1+\frac{2\pi}{3\sqrt{3}} and:$$ I_{n+1} = \int_{0}^{1}\frac{(x-x^2)(x-x^2)^n}{1-x+x^2}\,dx = I_n-\int_{0}^{1}x^n(1-x)^n\,dx = I_n - \frac{\Gamma(n+1)^2}{\Gamma(2n+2)}$$hence:$$ ...

2

You want to use $\displaystyle V=\int_3^72\pi r(y)h(y)dy=\int_3^7 2\pi(y+1) \cdot2\sqrt{4-(y-5)^2}dy=4\pi\int_3^7(y+1)\sqrt{4-(y-5)^2}dy$ $\hspace{.4 in}$since $x=\pm\sqrt{4-(y-5)^2}\implies h(y)=2\sqrt{4-(y-5)^2}$. (Notice that we can't double the volume generated by the top half of the circle, since the volume generated by the top half is greater than ...

1

Yes, that is accurate. Another way to do this problem, which acts as a check on your method, is to translate the entire graph to the right so that the axis of rotation is the $y$-axis, as is required by your original formula $S=2\pi xf(x)\,dx$. We do that by replacing $x$ in that formula by $x-2$ and moving the bounds from $0..4$ to $2..6$, so the integral ...

1

You're basically on the right track, but you've set up the integral a little bit wrong (and omitted which variable you're integrating with). You've sort of plugged all your variables in, but not correctly. Firstly, your integral needs to be of the form: $$\int_{R}^{\infty}w(x) \,dx$$ since you start at position $R$ and end at position $\infty$ while ...

0

You want $$Work \ Integral = \int_{R_E}^\infty Force(x). dx = \int_{R_E}^\infty \frac{GM_Em}{x^2} dx = \frac{GM_Em}{R_E}$$ where $R_E$ is the radius of the Earth, $M_E$ is the mass of of the Earth, $G$ is the universal gravitational constant and $m$ is the mass of the object you are moving from the surface of the Earth to 'infinity'.

2

as @achille hint,I have post it since $$I=\int_{0}^{+\infty}\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=(\int_{0}^{1}+\int_{1}^{\infty})\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=I_{1}+I_{2}$$ since $$I_{2}=-\int_{0}^{1}x\sin^3{(\dfrac{1}{x}-x)^5}dx$$ so ...

3

The integral is Riemann improper integrable. To save the argument that whether it converges or not. Let us look at following integral as a function of cutoff $\Lambda$. I(\Lambda) \stackrel{def}{=}\int_{1/\Lambda}^\Lambda \sin^3\left[\left(x - \frac{1}{x}\right)^5\right] \frac{dx}{x^3} = \left(\int_{1/\Lambda}^1 + \int_1^\Lambda\right) ...

0

me again. I was very tired when I asked the question, the answer is trivial $\int_a^b f(x)dx = \Phi(b) - \Phi(a)$, where $\Phi(x) = \int f(x) dx$ is the corresponding definite integral. So the problem boils down to being able to compute a definite integral, and then being able so solve the resulting ordinary equation.

2

No, in order for an integral to converge both $\int_0^{\infty} f(x) dx$ and $\int_{-\infty}^{0} f(x) dx$ must converge. $\int_{-\infty}^{\infty}\sin(x)dx$ converges in principal value however. $\int_{-\infty}^{\infty}\sin(x)dx$ is DEFINED as the sum $\int_{-\infty}^{0} f(x) dx+\int_0^{\infty} f(x) dx$ where both converge, so no $\int_{-\infty}^{\infty} ... 3 Performing integration by parts by taking$u=\ln^2(1+x)$and$\mathrm dv=\dfrac{\ln x}{x}\ \mathrm dx, then \begin{align} I&=\int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \mathrm dx\\ &=\ln x\ln^2(1+x)\Bigg|_0^1-\int_0^1\frac{\ln^2{x}\ln(1+x)}{1+x}\,\mathrm dx\\ &=-\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^2x\,\mathrm dx\tag{1}\\ ... 1 For sure, as pbs answered, adding the surface of the trapezoids is a way to go. However, you can do slightly better if, by chance, you have an odd number of data points. The method consists in fitting three data points by a parabola and integrate the resulting equation. Consider points(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$and you want the area between ... 1 Use the trapezoid rule. You can find information on this here: https://en.wikipedia.org/wiki/Trapezoidal_rule The integral can be interpreted as the area "under" the curve. You can approximate this area using the trapezoid rule, which will give you a value. 1$\displaystyle\int_0^1\frac{x^{1/2}(1-x)^{1/2}}{x+2}dx=\int_0^1\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{2^{n+1}}x^{1/2}(1-x)^{1/2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\int_0^1x^{n+\frac{1}{2}}(1-x)^{\frac{1}{2}}dx$... 1 Hint: Blue$\rightarrow y=\sqrt{4x-x^2}$Purple$\rightarrow y=x\sqrt{4x-x^2}\$

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