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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 = ...


1

Can we decide that whether $\int_{(0, +\infty)} y^{-\frac{1}{2}} (\int_{[0, +\infty)} x^{2m} e^{-x^{2}y}dx)dy$ is finite for $m \in {\Bbb Z}^+$? The change of variable $x\to z=x\sqrt{y}$ shows that the integral is $$\int_0^\infty y^{-1/2}\int_0^\infty z^{2m}y^{-m}\mathrm e^{-z^2}y^{-1/2}\mathrm dz\mathrm dy=\int_0^\infty y^{-1-m}\mathrm ...


0

We can use the geometric series $\frac{1}{1-x}=\sum_{k=0}^\infty x^n$ for $|x|<1$ to evaluate: \begin{eqnarray} \int_0^\infty\frac{1}{1+x^n}dx&=&\int_0^1\frac{1+x^{n-2}}{1+x^n}dx\\ &=&\sum_{k=0}^\infty(-1)^k\int_0^1(1+x^{n-2})x^{nk}dx\\ &=&\sum_{k=0}^\infty(-1)^k\left(\frac{1}{nk+1}+\frac{1}{nk+n-1}\right)\\ ...


0

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5

Use the substitution: $x = \tan\theta$. The integral is then equal to: $$I= \int_{0}^{\pi/4} \ln(1+\tan\theta) \ d\theta (*)$$ Also,we know the property: $$\int_{0}^{b} f(x) \ dx = \int_{0}^{b} f(b-x) \ dx$$ so we have $$I = \int_{0}^{\pi/4} \ln\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} ...


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 ...


1

Two thirds of the way: Forget about $K$ for the moment. Assume we are given a continuous $f:\ [0,1]\to{\mathbb R}$, and we want to construct a $g$ fulfilling the given conditions. From $g'''=f$ we conclude that $$g''(x)=\int_0^x f(t)\>dt +c\ ,$$ where $c$ is as yet undetermined. This leads to $$g'(x)=g'(0)+\int_0^x g''(t)\>dt=\int_0^x\left(\int_0^t ...


1

I have decided to post a partial answer because a paper describing the solution to the integral has already been found. Begin with a simple substitution of $x = e^{-u}$ to arrive at the equivalent integral $$\int_0^\infty \!\frac{\log(u)}{e^u-1+e^{-u}} \mathrm{d}u$$ I will attempt to find a closed form for this integral. First, I will consider integrals ...


5

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0

In fact, $$ \biggl|\int_{1}^{x}\dfrac{\sin t}{t}dt\biggr| \leq \int_{1}^{x}\biggl|\dfrac{\sin t}{t}\biggr|dt = \int_{1}^{x}\dfrac{|\sin t|}{|t|}dt = \int_{1}^{x}\dfrac{|\sin t|}{t}dt \leq \int_{1}^{x}\dfrac{dt}{t} = \ln x $$ Note that $|\sin x| \leq 1$ for all $x \in \mathbb{R}$.


0

You're on the right track. What do you know about the bounds of $\sin(t)$?


1

Note that $\int_{1}^{x}\lvert\frac{\sin(t)}{t}\rvert dt\le\int_{1}^{x}\frac{1}{t}dt$. Now integrate.


1

Recall Binet's formula $$ \log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) + \int_0^{\infty} \! \left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x,\quad \Re z >0 $$ which, upon making $x = - \log v$, can be written as $$ \log \Gamma(z)= \! \left( z-\frac{1}{2}\right)\log z - z + ...


2

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, $$ \log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a. $$


5

As stated in the comments, by following the Feynman's approach to integrals we have: $$\begin{eqnarray*} I'(s) &=& \int_{0}^{1}\left(1+\frac{\log x}{1-x}-\frac{\log x}{2}\right)\frac{x^s}{1-x}dx\\ &=& \frac{1}{2}\zeta(2,s+1)+\int_{0}^{1}\left(1+\frac{\log ...


2

Related question. As I mentioned there in a comment, using the substitution $t=x-\pi/4$ you can rewrite your integral as $$I=\sqrt{2}\sin\frac{2015\pi}{4}\int_0^{\pi/4}\frac{\cos 2015x}{\cos x}\,dx=-\int_0^{\pi/4}\frac{\cos 2015x}{\cos x}\,dx=-I_{2015},$$ where $$I_n=\int_0^{\pi/4}\frac{\cos nx}{\cos x}\,dx.$$ Now let us recall the identity $\cos nx=2\cos ...


2

As you see in other posts, the answer may vary. I know a test which is depend to some conditions: Let $\lim_{x\to 0^+}x^p\ln(f(x))=A$. If $p<1$ and you found $A<\infty$ so the integral is converges and if $p\ge1$ while $A\neq 0$ (or $A=\infty)$ then the integral is diverges. We have this when we know that the integrand is unbounded only at the lower ...


2

If $f(x)=1$ for every $x$ in $(0,1)$ then both limits of $f$ at $0$ and at $1$ exist and the integral converges. If $f(x)=\mathrm e^{-1/x}$ for every $x$ in $(0,1)$ then both limits of $f$ at $0$ and at $1$ exist and the integral diverges.


1

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1

Let's put it with real numbers. We have $$ \int_{0}^{\pi/2} \ln \left(1+t\sin^4 x\right) \mathrm{d}x = \pi \ln \left( \frac{1}{4} \sqrt{ 1 + \sqrt{1+t} } \left( \sqrt{1 + \sqrt{1+t} } + \sqrt{2} \right) \right), \quad t \geq -1. $$


2

Hint: $\quad1+t^2\sin^4x~=~(1-ti\sin^2x)\cdot(1+ti\sin^2x),\qquad\log(ab)=\log a+\log b,\quad$ and $\Re(\pm~ti)>-1$.


1

If $x^2+2x-2=0$, then $$ P=x^3+x^2+ax+1=(x-1)\cdot(x^2+2x-2)+(4+a)x-1=(4+a)x-1$$ So at least if $a=-4$ we obtain $P=-1$ for all (i.e. both) such $x$. In principle, some other $a$ might work as well, but even without computing $x_1,x_2$ explicitly, we know that $x_1+x_2=-2$, hence $P(x_1)+P(x_2) = -2(a+4)-2$, i.e. $a$ must be rational; and then $(4+a)x$ is ...


1

If $x^2 + 2x - 2 = 0$ then $$P-(x^2+2x-2)=P-0=P$$ which tells us that $$P = x^3 + (a-2)x +3$$ Now we need to plug in $x_1$ and $x_2$ in order to solve for $a$. $$P=(-1+\sqrt{3})^3 + (a-2)(-1+\sqrt{3}) + 3 = (-1+3\sqrt{3}-9+3\sqrt{3})+(a-2)(-1+\sqrt{3}) + 3$$ $$=(-5) + 4\sqrt{3} + a(-1 + \sqrt{3})$$ $$=(-5) + (4+a)\sqrt{3} - a$$ Notice that we need to get ...


1

After several attempts, I nearly managed to compute this integral with a decent approach. (I have removed my previous answers that were not up to par) Firstly, we split the integral up to simplify matters. \begin{align} I &=-\int^{\sqrt{2}}_1\frac{\ln{x}}{x}dx+\int^{\sqrt{2}}_1\frac{\ln{((x^2-1)^2+1)}}{x}dx-\int^{\sqrt{2}}_1\frac{\ln{((x-1)^2+1)}}{x}dx\\ ...


3

The contributions from fellow users Mhenni Benghorbal and oks pave the way. Using equidistant spacings one can perform the summation like this: $$ \begin{align} s(f,P') &= \sum_{i=1}^n(4-2t_i)(t_i-t_{i-1}) \\ &= \left(\sum_{i=1}^n 4 - 2 \left(a + \frac{b-a}{n} i \right) \right)\frac{b-a}{n} \\ &= \left(4n - 2 \left(a n + (b-a) \frac{n+1}{2} ...


0

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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