# Tag Info

3

The main idea is that, near the largest point of the integrand (which occurs at $x=0$), we have $$\frac{1}{\log x} + \frac{1}{1-x} \approx \frac{1}{\log x} + 1.$$ So we split the integral up as $$\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx = \left[\int_0^{1/e} + \int_{1/e}^1 \right] \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx. ... 1 Ah, I guess the problem is that f blows up on the imaginary axis, so z^{v-1/2}f(z) does not go to 0 uniformly as z goes to infinity. 4 Proposition.$$ \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:{\mathrm{d}}x = \frac{45}{512} \zeta(7)-\frac{3\pi^2}{16} \zeta(5)-\frac{5\pi^4}{64} \zeta(3)-\frac{9}{4}\zeta(\bar{5},1,1) $$where \zeta(\bar{p},1,1) is the colored MZV (Multi Zeta Values) function of depth 3 and weight p+2 given by$$ \zeta(\bar{p},1,1) : = \sum_{n=1}^{\infty} ...

0

We can use the following results $$\sum_{n=-\infty}^\infty(-1)^n\frac{1}{bn+a}=\frac{\pi}{b\sin\frac{a\pi}{b}}, \frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ to evaluate the generalization. In fact \begin{eqnarray} \int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=&\int_0^1\frac{x^{a-1}}{1+x^b}dx+\int_0^1\frac{x^{-a-1}}{1+x^b}dx\\ ...

7

We will prove that $$I=-\frac{\pi^4}{2880}.$$ Indeed, let $$J=\int_0^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx$$ It is easy to see that \eqalign{J&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{\pi/4}^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx\cr &=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{0}^{\pi/4}\log^2(\cos ... 6 I am afraid that what you did is wrong : you are not integrating a polynomial expression. Just for your curiosity,\frac{d}{dt}\Big(\frac{e^{t+t^2}}{t^2/2+t^3/3}\Big)=\frac{6 e^{t+t^2} (t+2) \left(4 t^2-3\right)}{t^3 (2 t+3)^2} \neq e^{t+t^2}$$To compute$$I=\int e^{t}.e^{t^2} dt=\int e^{t^2+t} dt$$first complete the square for the exponent and perform a ... 5 We have \lim_{y\to-\infty}\phi(y)=0 and \lim_{y\to\infty}\phi(y)=1. Moreover, the function \phi is continuous. Thus, by the Intermediate Value Theorem, the function \phi is surjective. 1 |S(r)| is the number of y not less than r, so |S(r)|=\sum_{i=1}^n I_{\{y_i \geq r\}}, here I_{\{y_i \geq r\}}=1  if y_i \geq r,$$\int_0^\infty |S(r)|dr=\int_0^\infty\sum_{i=1}^n I_{\{y_i \geq r\}} dr=\sum_{i=1}^n\int_0^\infty I_{\{y_i \geq r\}} dr=\sum_{i=1}^n y_i =1.$$1 Substitute x=\frac{1}{u}. Then \frac{1}{x}dx = \frac{1}{x}\frac{-1}{u^2}du = -\frac{1}{u}du. So$$\int_{1}^\infty \frac{\{u\}}{u\lfloor u\rfloor}du = \int_{1}^\infty\left(\frac{1}{\lfloor u\rfloor}-\frac{1}{u}\right) du = \gamma$$4 We can try the harmonic analysis path. Since:$$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$we have, as an example:$$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$since$$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = ...

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0

Try using $x=\tan\theta$ for the substitution. Then use the second result you have mentioned in your question with new limits. You may want to use the formula for $\tan (A+B)$ at some point.

2

Since you have $x^2 + y^2$, I suggest using polar coordinates. Since $$r^2 = x^2 + y^2$$ $$1 < r^2 < 4$$ So, $$1 < r < 2$$ To find $\theta$ $$\pi/6 < \theta < \pi/3$$ Don't forget to multiply $$rdrd\theta\ \text{in your double integral}$$

2

As it's a little complicated with finding the indefinite integral of $$\int\int(x^2+y^2)^{3/2}dydx$$ I would suggest using polar, as reversing the order of integration isn't really going to make a huge difference. Also the graph is a huge indicator of that.

1

One may say of course that "WLOG" we may assume $r=1$. But leaving $r$ as just $r$ enables us to check that everything is dimensionally correct, i.e. an expression purporting to be a volume is homogeneous of degree $3$ in $r$ and one purporting to be a distance is homogeneous of degree $1$, and in fact we will need to find an area, so that should be ...

1

The direct integration route isn't actually that bad, and obtains in fewer words the same triple integral computed by Kirill. We want to compute $\int_S dV \, |\hat{z}-\mathbf{r}|$ where $S$ is the sphere of radius $r$. By the law of cosines, we can write the separation as $$|\hat{z}-\mathbf{r}| = \sqrt{r^2-2r \rho \cos{\theta}+\rho^2}$$ where $\theta$ is ...

1

Wlog $r=1$. The average is independent of the direction of the special point, so take the average over all directions: the answer is the mean distance between a random point on a sphere and a random point inside it. That distance depends only on how far away the point is from the centre. To calculate the mean distance between a specific point inside and the ...

4

Continuing from O.L.'s answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin 2x}{x} \text{Ci}(x) \ dx .$$ First notice that by making the substitution $\displaystyle u = \frac{t}{x}$, $$\text{Ci}(x) = - \int_{x}^{\infty} \frac{\cos t}{t} \ dt = - \int_{1}^{\infty} \frac{\cos xu}{u} \ du.$$ Therefore, $$\int_{0}^{\infty} \frac{\sin ... 2 I bet that the answers to the first and third question (about convergence and zeroes) are affirmative, since the function$$ f(z) = \left(1-\frac{x^2}{4}+\frac{x^4}{64}\right)\mathbb{1}_{[0,1]}(x)+\sqrt{\frac{2}{\pi x}}\cos(x-\pi/4)\mathbb{1}_{[1,+\infty)}(x),$$by following Abramowitz and Stegun, is a very good approximation for J_0(x), but I do not think ... 0 My answer is different from that you gave. Let$$ I(a)=\int_0^{\frac{\pi}{2}}\arctan(a\tan^2x)dx. $$Than I(0)=0 and \begin{eqnarray} I'(a)&=&\int_0^{\frac{\pi}{2}}\frac{\tan^2x}{1+a^2\tan^4x}dx\\ &=&\int_0^\infty\frac{u^2}{(1+u^2)(1+a^2u^4)}du\\ ... 3 Let x=e^{-u}. Then \begin{eqnarray} \int_0^1\frac{\ln\ln\left(\frac{1}{x}\right)}{x^2-x+1}dx&=&\int_0^\infty\frac{e^{-u}\ln u}{e^{-2u}-e^{-u}+1}du\\ &=&\int_0^\infty\frac{e^{-u}(1+e^{-u})\ln u}{1+e^{-3u}}du\\ &=&\int_0^\infty\sum_{n=0}^\infty(-1)^ne^{-(3n+1)u}(1+e^{-u})\ln u\ du\\ ... 1 Here is a result avoiding differentiation with respect to a parameter. Set$$ I(\alpha):= \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(\frac{2\alpha \:\sin^2 x}{\alpha^2-1+\cos^2 x}\right)\: \mathrm{d}x, \quad \alpha>0. $$Then$$ I(\alpha)= \pi \arctan \left(\frac{1}{2\alpha}\right) \quad ({\star}) $$With  \alpha:=1, we get$$ ...

0

I will denote $$I = \int_a^b \frac{1}{x} \mathrm{d}x$$ Break the interval into partitions $(t_{i-1},t_i)$ such that $$a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b$$ Let $\lambda$ denote the mesh of this partition or the length of the longest partition ($\max(t_i-t_{i-1})$). We have broken our interval up into $n$ partitions and now we must ...

1

Using harmonic numbers, we can write $$(\frac{b-a}{n})\sum_{i=1}^n \frac{1}{a+\frac{(b-a)}{n}i}=H_{\frac{b n}{b-a}}-H_{\frac{a n}{b-a}}$$ Now, since we look at the limit for an infinite value of $n$, consider the expansion $$H_k=\left(\gamma -\log \left(\frac{1}{k}\right)\right)+\frac{1}{2 k}+O\left(\left(\frac{1}{k}\right)^2\right)$$ Now replace ...

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Alternative proof Let $$F(a)=\int_1^{a^b}\frac{dx}{x}$$ then by the fundemental theorem of analysis $$F'(a)=\frac{1}{a^b}\times ba^{b-1}=\frac ba$$ so we integrate we find $$F(a)=b\log a+C$$ and using that $F(1)=0$ we conclude that $$F(a)=\log(a^b)=b\log a$$

1

$$\log(a^b) = \int_1^{a^b} \frac{dx}x.$$ Suppose $w^{\,b}=x$. Then $bw^{b-1}\,dw= dx$, so $b\dfrac{dw}{w} = \dfrac{dx}x$. As $x$ goes from $1$ to $a^b$, then $w$ goes from $1$ to $a$. Hence, $$\log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a.$$

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It is always much easier to use the Mean Value Theorem on the function $g(x) = 4x - x^2$ ni every interval on the partition. That is for each $i$ there exists $c_i \in [t_{i- 1}, t_i]$ such that $$\dfrac{g(t_i) - g(t_{i-1})}{ t_i - t_{i- 1} } = g'(c_i) = (4 - 2x)|_{x = c_i} = 4 - 2c_i = f(c_i)$$ Now try to use the fact that $m_i \ge f(c_i) \ge M_i$ ...

0

Hint: if you are integrating over an interval $[a,b]$ then $$\Delta t_i = t_i-t_{i-1} = \frac{b-a}{n}.$$ Also, your function is a decreasing function then you should know what $m_i$ and $M_i$ are.

0

Hint: Since you know that $s(f,P)<S(f,P')$ for any two divisions $P,P'$, it is enough to find, for each $\epsilon$, such a division $P$ that $$-9-\epsilon < s(f,P)< S(f,P) < -9+\epsilon$$

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